id stringlengths 11 15 | title stringlengths 7 171 | problem stringlengths 9 4.33k | solution stringlengths 6 19k | problem_wikitext stringlengths 9 4.42k | solution_wikitext stringlengths 7 19.1k | proof_title stringlengths 9 171 | theorem_url stringlengths 34 198 | proof_url stringlengths 36 198 | categories listlengths 0 9 | theorem_references listlengths 0 36 | proof_references listlengths 0 253 |
|---|---|---|---|---|---|---|---|---|---|---|---|
proofwiki-19800 | Dilation of Open Set in Topological Vector Space is Open | Let $K$ be a topological field.
Let $X$ be a topological vector space over $K$.
Let $\lambda \in K \setminus \set {0_K}$.
Let $U$ be a open set in $X$.
Then $\lambda U$, the dilation of $U$ by $\lambda$ is open. | Let $c_\lambda$ be the dilation by $\lambda$ mapping.
From Dilation Mapping on Topological Vector Space is Homeomorphism, $c_\lambda$ is a homeomorphism.
Hence the image $c_\lambda \sqbrk U$ is open.
That is, $\lambda U$ is open.
{{qed}}
Category:Topological Vector Spaces
bszuwbeeyz3zg2eyhh6t4yivewu94pb | Let $K$ be a [[Definition:Topological Field|topological field]].
Let $X$ be a [[Definition:Topological Vector Space|topological vector space]] over $K$.
Let $\lambda \in K \setminus \set {0_K}$.
Let $U$ be a [[Definition:Open Set (Topology)|open set]] in $X$.
Then $\lambda U$, the [[Definition:Dilation of Subset... | Let $c_\lambda$ be the [[Definition:Dilation Mapping|dilation by $\lambda$ mapping]].
From [[Dilation Mapping on Topological Vector Space is Homeomorphism]], $c_\lambda$ is a [[Definition:Homeomorphism|homeomorphism]].
Hence the [[Definition:Image of Set under Mapping|image]] $c_\lambda \sqbrk U$ is [[Definition:Open... | Dilation of Open Set in Topological Vector Space is Open | https://proofwiki.org/wiki/Dilation_of_Open_Set_in_Topological_Vector_Space_is_Open | https://proofwiki.org/wiki/Dilation_of_Open_Set_in_Topological_Vector_Space_is_Open | [
"Topological Vector Spaces"
] | [
"Definition:Topological Field",
"Definition:Topological Vector Space",
"Definition:Open Set/Topology",
"Definition:Linear Combination of Subsets of Vector Space/Dilation",
"Definition:Open Set/Topology"
] | [
"Definition:Central Dilatation Mapping",
"Dilation Mapping on Topological Vector Space is Homeomorphism",
"Definition:Homeomorphism",
"Definition:Image (Set Theory)/Mapping/Subset",
"Definition:Open Set/Topology",
"Definition:Open Set/Topology",
"Category:Topological Vector Spaces"
] |
proofwiki-19801 | Translation Mapping on Topological Vector Space is Continuous | Let $K$ be a topological field.
Let $X$ be a topological vector space over $K$.
Let $x \in X$.
Let $T_x$ be the translation by $x$ mapping.
Then $T_x$ is continuous. | From the definition of a topological vector space, the mapping $X \times X \to X$ defined by $\tuple {y, x} \mapsto y + x$ is continuous.
From Horizontal Section of Continuous Function is Continuous, it follows that the $\paren {-x}$-horizontal section $T_x : X \to X$ with $y \mapsto y - x$ is continuous.
{{qed}}
Categ... | Let $K$ be a [[Definition:Topological Field|topological field]].
Let $X$ be a [[Definition:Topological Vector Space|topological vector space]] over $K$.
Let $x \in X$.
Let $T_x$ be the [[Definition:Translation in Vector Space|translation by $x$ mapping]].
Then $T_x$ is [[Definition:Continuous Mapping (Topology)|c... | From the definition of a [[Definition:Topological Vector Space|topological vector space]], the [[Definition:Mapping|mapping]] $X \times X \to X$ defined by $\tuple {y, x} \mapsto y + x$ is [[Definition:Continuous Mapping (Topology)|continuous]].
From [[Horizontal Section of Continuous Function is Continuous]], it foll... | Translation Mapping on Topological Vector Space is Continuous | https://proofwiki.org/wiki/Translation_Mapping_on_Topological_Vector_Space_is_Continuous | https://proofwiki.org/wiki/Translation_Mapping_on_Topological_Vector_Space_is_Continuous | [
"Translation Mappings (Vector Spaces)",
"Topological Vector Spaces"
] | [
"Definition:Topological Field",
"Definition:Topological Vector Space",
"Definition:Translation Mapping/Vector Space",
"Definition:Continuous Mapping (Topology)"
] | [
"Definition:Topological Vector Space",
"Definition:Mapping",
"Definition:Continuous Mapping (Topology)",
"Horizontal Section of Continuous Function is Continuous",
"Definition:Horizontal Section of Function",
"Definition:Continuous Mapping (Topology)",
"Category:Translation Mappings (Vector Spaces)",
... |
proofwiki-19802 | Translation Mapping on Topological Vector Space is Homeomorphism | Let $K$ be a topological field.
Let $X$ be a topological vector space over $K$.
Let $x \in X$.
Let $T_x$ be the translation by $x$ mapping.
Then $T_x$ is a homeomorphism. | From Translation Mapping on Topological Vector Space is Continuous, both $T_x$ and $T_{-x}$ are continuous.
It is therefore sufficient to establish that $T_{-x}$ is the inverse mapping of $T_x$.
For all $y \in X$, we have:
{{begin-eqn}}
{{eqn | l = \map {\paren {T_x \circ T_{-x} } } y
| r = \map {T_x} {y - \paren... | Let $K$ be a [[Definition:Topological Field|topological field]].
Let $X$ be a [[Definition:Topological Vector Space|topological vector space]] over $K$.
Let $x \in X$.
Let $T_x$ be the [[Definition:Translation in Vector Space|translation by $x$ mapping]].
Then $T_x$ is a [[Definition:Homeomorphism|homeomorphism]... | From [[Translation Mapping on Topological Vector Space is Continuous]], both $T_x$ and $T_{-x}$ are [[Definition:Continuous Mapping (Topology)|continuous]].
It is therefore sufficient to establish that $T_{-x}$ is the [[Definition:Inverse Mapping|inverse mapping]] of $T_x$.
For all $y \in X$, we have:
{{begin-eqn}}
... | Translation Mapping on Topological Vector Space is Homeomorphism | https://proofwiki.org/wiki/Translation_Mapping_on_Topological_Vector_Space_is_Homeomorphism | https://proofwiki.org/wiki/Translation_Mapping_on_Topological_Vector_Space_is_Homeomorphism | [
"Translation Mappings (Vector Spaces)",
"Topological Vector Spaces"
] | [
"Definition:Topological Field",
"Definition:Topological Vector Space",
"Definition:Translation Mapping/Vector Space",
"Definition:Homeomorphism"
] | [
"Translation Mapping on Topological Vector Space is Continuous",
"Definition:Continuous Mapping (Topology)",
"Definition:Inverse Mapping",
"Definition:Identity Mapping",
"Definition:Inverse Mapping"
] |
proofwiki-19803 | Sum of Set and Open Set in Topological Vector Space is Open | Let $K$ be a topological field.
Let $X$ be a topological vector space over $K$.
Let $A$ and $B$ be subsets of $X$, with $A$ an open set.
Then the linear combination $A + B$ is open. | It can be shown that:
:$\ds A + B = \bigcup_{b \mathop \in B} \paren {A + b}$
If:
:$\ds v \in \bigcup_{b \mathop \in B} \paren {A + b}$
then $v = a + b$ for some $a \in A$ and $b \in B$.
So $v \in A + B$, giving:
:$\ds \bigcup_{b \mathop \in B} \paren {A + b} \subseteq A + B$
Conversely, suppose that $v \in A + B$.
... | Let $K$ be a [[Definition:Topological Field|topological field]].
Let $X$ be a [[Definition:Topological Vector Space|topological vector space]] over $K$.
Let $A$ and $B$ be [[Definition:Subset|subsets]] of $X$, with $A$ an [[Definition:Open Set (Topology)|open set]].
Then the [[Definition:Linear Combination of Subs... | It can be shown that:
:$\ds A + B = \bigcup_{b \mathop \in B} \paren {A + b}$
If:
:$\ds v \in \bigcup_{b \mathop \in B} \paren {A + b}$
then $v = a + b$ for some $a \in A$ and $b \in B$.
So $v \in A + B$, giving:
:$\ds \bigcup_{b \mathop \in B} \paren {A + b} \subseteq A + B$
Conversely, suppose that $v \in A... | Sum of Set and Open Set in Topological Vector Space is Open | https://proofwiki.org/wiki/Sum_of_Set_and_Open_Set_in_Topological_Vector_Space_is_Open | https://proofwiki.org/wiki/Sum_of_Set_and_Open_Set_in_Topological_Vector_Space_is_Open | [
"Topological Vector Spaces"
] | [
"Definition:Topological Field",
"Definition:Topological Vector Space",
"Definition:Subset",
"Definition:Open Set/Topology",
"Definition:Linear Combination of Subsets of Vector Space",
"Definition:Open Set/Topology"
] | [
"Definition:Set Equality",
"Translation of Open Set in Topological Vector Space is Open",
"Definition:Open Set/Topology",
"Definition:Set Union",
"Definition:Open Set/Topology",
"Definition:Topology",
"Definition:Set Union",
"Definition:Open Set/Topology",
"Category:Topological Vector Spaces"
] |
proofwiki-19804 | Image of Translation of Set under Linear Transformation is Translation of Image | Let $K$ be a field.
Let $X$ and $Y$ be vector spaces over $K$.
Let $T : X \to Y$ be a linear transformation.
Let $E \subseteq X$ be a non-empty set.
Let $x \in X$.
Then:
:$\map T {E + x} = \map T E + T x$
where $E + x$ denotes the translation of $E$ by $x$. | We have:
:$y \in \map T {E + x}$
{{iff}}:
:$y = \map T {u + x}$ for some $u \in E$.
From the linearity of $T$, this is equivalent to:
:$y = T u + T x$ for some $u \in E$.
This is equivalent to:
:$y \in \map T E + T x$
So by the definition of set equality we have:
:$\map T {E + x} = \map T E + T x$
{{qed}}
Category:Li... | Let $K$ be a [[Definition:Field (Abstract Algebra)|field]].
Let $X$ and $Y$ be [[Definition:Vector Space|vector spaces]] over $K$.
Let $T : X \to Y$ be a [[Definition:Linear Transformation|linear transformation]].
Let $E \subseteq X$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Set|set]].
Let $x \in X$... | We have:
:$y \in \map T {E + x}$
{{iff}}:
:$y = \map T {u + x}$ for some $u \in E$.
From the [[Definition:Linear Transformation|linearity]] of $T$, this is equivalent to:
:$y = T u + T x$ for some $u \in E$.
This is equivalent to:
:$y \in \map T E + T x$
So by the definition of [[Definition:Set Equality|set e... | Image of Translation of Set under Linear Transformation is Translation of Image | https://proofwiki.org/wiki/Image_of_Translation_of_Set_under_Linear_Transformation_is_Translation_of_Image | https://proofwiki.org/wiki/Image_of_Translation_of_Set_under_Linear_Transformation_is_Translation_of_Image | [
"Linear Transformations",
"Translation of Subsets of Vector Spaces"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Vector Space",
"Definition:Linear Transformation",
"Definition:Non-Empty Set",
"Definition:Set",
"Definition:Translation Mapping/Vector Space"
] | [
"Definition:Linear Transformation",
"Definition:Set Equality",
"Category:Linear Transformations",
"Category:Translation of Subsets of Vector Spaces"
] |
proofwiki-19805 | Image of Dilation of Set under Linear Transformation is Dilation of Image | Let $K$ be a field.
Let $X$ and $Y$ be vector spaces over $K$.
Let $T : X \to Y$ be a linear transformation.
Let $E \subseteq X$ be a non-empty set.
Let $\lambda \in K$.
Then:
:$\map T {\lambda E} = \lambda \map T E$
where $\lambda E$ denotes the dilation of $E$ by $\lambda$. | We have:
:$y \in \map T {\lambda E}$
{{iff}}:
:$y = \map T {\lambda x}$ for some $x \in E$.
From the linearity of $T$, this is equivalent to:
:$y = \lambda T x$
This is equivalent to:
:$y \in \lambda \map T E$
So by the definition of set equality we have:
:$\map T {\lambda E} = \lambda \map T E$
{{qed}}
Category:Dilat... | Let $K$ be a [[Definition:Field (Abstract Algebra)|field]].
Let $X$ and $Y$ be [[Definition:Vector Space|vector spaces]] over $K$.
Let $T : X \to Y$ be a [[Definition:Linear Transformation|linear transformation]].
Let $E \subseteq X$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Set|set]].
Let $\lambda ... | We have:
:$y \in \map T {\lambda E}$
{{iff}}:
:$y = \map T {\lambda x}$ for some $x \in E$.
From the [[Definition:Linear Transformation|linearity]] of $T$, this is equivalent to:
:$y = \lambda T x$
This is equivalent to:
:$y \in \lambda \map T E$
So by the definition of [[Definition:Set Equality|set equality]]... | Image of Dilation of Set under Linear Transformation is Dilation of Image | https://proofwiki.org/wiki/Image_of_Dilation_of_Set_under_Linear_Transformation_is_Dilation_of_Image | https://proofwiki.org/wiki/Image_of_Dilation_of_Set_under_Linear_Transformation_is_Dilation_of_Image | [
"Dilations of Subsets of Vector Spaces",
"Linear Transformations"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Vector Space",
"Definition:Linear Transformation",
"Definition:Non-Empty Set",
"Definition:Set",
"Definition:Linear Combination of Subsets of Vector Space/Dilation"
] | [
"Definition:Linear Transformation",
"Definition:Set Equality",
"Category:Dilations of Subsets of Vector Spaces",
"Category:Linear Transformations"
] |
proofwiki-19806 | Characteristic of Extending Operation | Let $E$ be an extending operation.
Then there exists a mapping $F$ on the class of all ordinals $\On$ such that:
:$\forall \alpha \in \On: F \restriction \alpha^+ = \map E {F \restriction \alpha}$
where:
:$F \restriction \alpha$ denotes the restriction of $F$ to $\alpha$
:$\alpha^+$ denotes the successor ordinal of $\a... | Let $M$ be the class which is minimally superinductive under $E$.
Recall the Transfinite Recursion Theorem:
{{begin-axiom}}
{{axiom | n = 1
| lc= Zeroth Ordinal:
| q =
| m = M_0 = \O
}}
{{axiom | n = 2
| lc= Successor Ordinal:
| q = \forall \alpha \in \On
| m = M_{\alpha... | Let $E$ be an [[Definition:Extending Operation|extending operation]].
Then there exists a [[Definition:Class Mapping|mapping]] $F$ on the [[Definition:Class of All Ordinals|class of all ordinals]] $\On$ such that:
:$\forall \alpha \in \On: F \restriction \alpha^+ = \map E {F \restriction \alpha}$
where:
:$F \restrict... | Let $M$ be the [[Definition:Class (Class Theory)|class]] which is [[Definition:Minimally Superinductive Class|minimally superinductive]] under $E$.
Recall the [[Transfinite Recursion Theorem]]:
{{begin-axiom}}
{{axiom | n = 1
| lc= [[Definition:Zeroth|Zeroth]] [[Definition:Ordinal|Ordinal]]:
| q =
... | Characteristic of Extending Operation | https://proofwiki.org/wiki/Characteristic_of_Extending_Operation | https://proofwiki.org/wiki/Characteristic_of_Extending_Operation | [
"Extending Operations"
] | [
"Definition:Extending Operation",
"Definition:Mapping/Class Theory",
"Definition:Class of All Ordinals",
"Definition:Restriction/Mapping",
"Definition:Successor Ordinal"
] | [
"Definition:Class (Class Theory)",
"Definition:Minimally Superinductive Class",
"Transfinite Recursion Theorem",
"Definition:Zeroth",
"Definition:Ordinal",
"Definition:Successor Ordinal",
"Definition:Limit Ordinal",
"Definition:Ordinal",
"Definition:Element/Class",
"Definition:Well-Ordered Class u... |
proofwiki-19807 | Translation of Union of Subsets of Vector Space | Let $K$ be a field.
Let $X$ be a vector space over $K$.
Let $\family {E_\alpha}_{\alpha \mathop \in I}$ be an indexed family of subsets of $X$.
Let $x \in X$.
Then:
:$\ds \paren {\bigcup_{\alpha \mathop \in I} E_\alpha} + x = \bigcup_{\alpha \mathop \in I} \paren {E_\alpha + x}$
where $E_\alpha + x$ denotes the transl... | We have:
:$\ds v \in \paren {\bigcup_{\alpha \mathop \in I} E_\alpha} + x$
{{iff}}:
:$v = u + x$ for some $\ds u \in \bigcup_{\alpha \mathop \in I} E_\alpha$.
This is equivalent to:
:there exists $\alpha \in I$ and $u \in E_\alpha$ such that $v = u + x$.
That is:
:there exists $\alpha \in I$ such that $v \in E_\alpha... | Let $K$ be a [[Definition:Field (Abstract Algebra)|field]].
Let $X$ be a [[Definition:Vector Space|vector space]] over $K$.
Let $\family {E_\alpha}_{\alpha \mathop \in I}$ be an [[Definition:Indexed Family of Subsets|indexed family]] of [[Definition:Subset|subsets]] of $X$.
Let $x \in X$.
Then:
:$\ds \paren {\bi... | We have:
:$\ds v \in \paren {\bigcup_{\alpha \mathop \in I} E_\alpha} + x$
{{iff}}:
:$v = u + x$ for some $\ds u \in \bigcup_{\alpha \mathop \in I} E_\alpha$.
This is equivalent to:
:there exists $\alpha \in I$ and $u \in E_\alpha$ such that $v = u + x$.
That is:
:there exists $\alpha \in I$ such that $v \in E... | Translation of Union of Subsets of Vector Space | https://proofwiki.org/wiki/Translation_of_Union_of_Subsets_of_Vector_Space | https://proofwiki.org/wiki/Translation_of_Union_of_Subsets_of_Vector_Space | [
"Translation of Subsets of Vector Spaces",
"Set Union"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Vector Space",
"Definition:Indexing Set/Family of Subsets",
"Definition:Subset",
"Definition:Translation of Subset of Vector Space"
] | [
"Definition:Set Equality",
"Category:Translation of Subsets of Vector Spaces",
"Category:Set Union"
] |
proofwiki-19808 | Distance between Disjoint Compact Set and Closed Set in Metric Space is Positive | Let $\struct {X, d}$ be a metric space.
Let $A$ and $B$ be disjoint subsets of $X$ such that $A$ is compact and $B$ is closed.
Then $\map d {A, B} > 0$, where $\map d {A, B}$ is the distance between $A$ and $B$. | Define $f : X \to \R$ by:
:$\map f x = \map d {x, B}$
for each $x \in X$.
From the definition of the distance between $A$ and $B$, we have:
:$\ds \map d {A, B} = \inf_{a \mathop \in A} \map f a$
From Compact Subspace of Hausdorff Space is Closed and Metric Space is Hausdorff, $A$ is closed and hence contains all its... | Let $\struct {X, d}$ be a [[Definition:Metric Space|metric space]].
Let $A$ and $B$ be [[Definition:Disjoint Sets|disjoint]] [[Definition:Subset|subsets]] of $X$ such that $A$ is [[Definition:Compact Subspace|compact]] and $B$ is [[Definition:Closed Set (Metric Space)|closed]].
Then $\map d {A, B} > 0$, where $\map ... | Define $f : X \to \R$ by:
:$\map f x = \map d {x, B}$
for each $x \in X$.
From the definition of the [[Definition:Distance between Element and Subset of Metric Space|distance between $A$ and $B$]], we have:
:$\ds \map d {A, B} = \inf_{a \mathop \in A} \map f a$
From [[Compact Subspace of Hausdorff Space is Clos... | Distance between Disjoint Compact Set and Closed Set in Metric Space is Positive | https://proofwiki.org/wiki/Distance_between_Disjoint_Compact_Set_and_Closed_Set_in_Metric_Space_is_Positive | https://proofwiki.org/wiki/Distance_between_Disjoint_Compact_Set_and_Closed_Set_in_Metric_Space_is_Positive | [
"Compact Metric Spaces",
"Distance Function"
] | [
"Definition:Metric Space",
"Definition:Disjoint Sets",
"Definition:Subset",
"Definition:Compact Topological Space/Subspace",
"Definition:Closed Set/Metric Space",
"Definition:Distance/Sets/Metric Spaces"
] | [
"Definition:Distance/Sets/Metric Spaces",
"Compact Subspace of Hausdorff Space is Closed",
"Metric Space is T2",
"Definition:Closed Set/Metric Space",
"Definition:Limit Point/Metric Space",
"Point at Zero Distance from Subset of Metric Space is Limit Point or Element",
"Distance from Point to Subset is ... |
proofwiki-19809 | Dilation of Open Set in Normed Vector Space is Open | Let $\Bbb F$ be a subfield of $\C$.
Let $\struct {X, \norm \cdot}$ be a normed vector space over $\Bbb F$.
Let $U \subseteq X$ be an open set.
Let $\lambda \in \Bbb F \setminus \set 0$.
Then:
:$\lambda U$ is open. | Let:
:$v \in \lambda U$
Then:
:$\dfrac v \lambda \in U$
Since $U$ is open, there exists $\epsilon > 0$ such that whenever $v' \in X$ and:
:$\ds \norm {\frac v \lambda - \frac {v'} \lambda} < \epsilon$
we have $v'/\lambda \in U$.
That is, $v' \in \lambda U$.
So, whenever:
:$\ds \norm {v - v'} < \epsilon \cmod \lambda... | Let $\Bbb F$ be a [[Definition:Subfield|subfield]] of $\C$.
Let $\struct {X, \norm \cdot}$ be a [[Definition:Normed Vector Space|normed vector space]] over $\Bbb F$.
Let $U \subseteq X$ be an [[Definition:Open Set in Normed Vector Space|open set]].
Let $\lambda \in \Bbb F \setminus \set 0$.
Then:
:$\lambda U$ i... | Let:
:$v \in \lambda U$
Then:
:$\dfrac v \lambda \in U$
Since $U$ is [[Definition:Open Set in Normed Vector Space|open]], there exists $\epsilon > 0$ such that whenever $v' \in X$ and:
:$\ds \norm {\frac v \lambda - \frac {v'} \lambda} < \epsilon$
we have $v'/\lambda \in U$.
That is, $v' \in \lambda U$.
So, w... | Dilation of Open Set in Normed Vector Space is Open | https://proofwiki.org/wiki/Dilation_of_Open_Set_in_Normed_Vector_Space_is_Open | https://proofwiki.org/wiki/Dilation_of_Open_Set_in_Normed_Vector_Space_is_Open | [
"Dilations of Subsets of Vector Spaces",
"Open Sets (Normed Vector Spaces)"
] | [
"Definition:Subfield",
"Definition:Normed Vector Space",
"Definition:Open Set/Normed Vector Space",
"Definition:Open Set/Normed Vector Space"
] | [
"Definition:Open Set/Normed Vector Space",
"Definition:Open Set/Normed Vector Space",
"Category:Dilations of Subsets of Vector Spaces",
"Category:Open Sets (Normed Vector Spaces)"
] |
proofwiki-19810 | Translation of Convex Set in Vector Space is Convex | Let $\Bbb F \in \set {\R, \C}$.
Let $X$ be a vector space over $\Bbb F$.
Let $C \subseteq X$ be convex.
Let $x \in X$.
Then $C + x$, the translation of $C$ by $x$, is convex. | Let $t \in \closedint 0 1$ and $u, v \in C + x$.
Then there exists $u', v' \in C$ such that:
:$u = u' + x$
and:
:$v = v' + x$
Then, we have:
{{begin-eqn}}
{{eqn | l = t u + \paren {1 - t} v
| r = t u' + t x + \paren {1 - t} v' + \paren {1 - t} x
}}
{{eqn | r = \paren {t u' + \paren {1 - t} v'} + \paren {t + \paren ... | Let $\Bbb F \in \set {\R, \C}$.
Let $X$ be a [[Definition:Vector Space|vector space]] over $\Bbb F$.
Let $C \subseteq X$ be [[Definition:Convex Set (Vector Space)|convex]].
Let $x \in X$.
Then $C + x$, the [[Definition:Translation of Subset of Vector Space|translation of $C$ by $x$]], is [[Definition:Convex Set (V... | Let $t \in \closedint 0 1$ and $u, v \in C + x$.
Then there exists $u', v' \in C$ such that:
:$u = u' + x$
and:
:$v = v' + x$
Then, we have:
{{begin-eqn}}
{{eqn | l = t u + \paren {1 - t} v
| r = t u' + t x + \paren {1 - t} v' + \paren {1 - t} x
}}
{{eqn | r = \paren {t u' + \paren {1 - t} v'} + \paren {t + \... | Translation of Convex Set in Vector Space is Convex | https://proofwiki.org/wiki/Translation_of_Convex_Set_in_Vector_Space_is_Convex | https://proofwiki.org/wiki/Translation_of_Convex_Set_in_Vector_Space_is_Convex | [
"Translation of Subsets of Vector Spaces",
"Convex Sets (Vector Spaces)"
] | [
"Definition:Vector Space",
"Definition:Convex Set (Vector Space)",
"Definition:Translation of Subset of Vector Space",
"Definition:Convex Set (Vector Space)"
] | [
"Definition:Convex Set (Vector Space)",
"Definition:Convex Set (Vector Space)",
"Category:Translation of Subsets of Vector Spaces",
"Category:Convex Sets (Vector Spaces)"
] |
proofwiki-19811 | Sum of Convex Sets in Vector Space is Convex | Let $\Bbb F \in \set {\R, \C}$.
Let $X$ be a vector space over $\Bbb F$.
Let $A$ and $B$ be convex subsets of $X$.
Then:
:$A + B$ is convex. | Let $x, y \in A + B$ and $t \in \closedint 0 1$.
Then there exists $a, a' \in A$ and $b, b' \in B$ such that:
:$x = a + b$
and:
:$y = a' + b'$
Then:
{{begin-eqn}}
{{eqn | l = t x + \paren {1 - t} y
| r = t a + t b + \paren {1 - t} a' + \paren {1 - t} b'
}}
{{eqn | r = \paren {t a + \paren {1 - t} a'} + \paren {t b ... | Let $\Bbb F \in \set {\R, \C}$.
Let $X$ be a [[Definition:Vector Space|vector space]] over $\Bbb F$.
Let $A$ and $B$ be [[Definition:Convex Set (Vector Space)|convex subsets]] of $X$.
Then:
:$A + B$ is [[Definition:Convex Set (Vector Space)|convex]]. | Let $x, y \in A + B$ and $t \in \closedint 0 1$.
Then there exists $a, a' \in A$ and $b, b' \in B$ such that:
:$x = a + b$
and:
:$y = a' + b'$
Then:
{{begin-eqn}}
{{eqn | l = t x + \paren {1 - t} y
| r = t a + t b + \paren {1 - t} a' + \paren {1 - t} b'
}}
{{eqn | r = \paren {t a + \paren {1 - t} a'} + \paren... | Sum of Convex Sets in Vector Space is Convex | https://proofwiki.org/wiki/Sum_of_Convex_Sets_in_Vector_Space_is_Convex | https://proofwiki.org/wiki/Sum_of_Convex_Sets_in_Vector_Space_is_Convex | [
"Convex Sets (Vector Spaces)",
"Sum of Convex Sets in Vector Space is Convex"
] | [
"Definition:Vector Space",
"Definition:Convex Set (Vector Space)",
"Definition:Convex Set (Vector Space)"
] | [
"Definition:Convex Set (Vector Space)",
"Definition:Convex Set (Vector Space)",
"Category:Convex Sets (Vector Spaces)",
"Category:Sum of Convex Sets in Vector Space is Convex"
] |
proofwiki-19812 | Sum of Convex Sets in Vector Space is Convex/Corollary | Let $\Bbb F \in \set {\R, \C}$.
Let $X$ be a vector space over $\Bbb F$.
Let $A$ and $B$ be convex subsets of $X$.
Let $\lambda, \mu \in \Bbb F$.
Then:
:$\lambda A + \mu B$ is convex. | From Dilation of Convex Set in Vector Space is Convex, we have:
:$\lambda A$ and $\mu B$ are convex.
From Sum of Convex Sets in Vector Space is Convex, we have:
:$\lambda A + \mu B$ is convex.
{{qed}}
Category:Sum of Convex Sets in Vector Space is Convex
nzn7p472agtl9fr99dz8pqzkp5rtt93 | Let $\Bbb F \in \set {\R, \C}$.
Let $X$ be a [[Definition:Vector Space|vector space]] over $\Bbb F$.
Let $A$ and $B$ be [[Definition:Convex Set (Vector Space)|convex subsets]] of $X$.
Let $\lambda, \mu \in \Bbb F$.
Then:
:$\lambda A + \mu B$ is [[Definition:Convex Set (Vector Space)|convex]]. | From [[Dilation of Convex Set in Vector Space is Convex]], we have:
:$\lambda A$ and $\mu B$ are [[Definition:Convex Set (Vector Space)|convex]].
From [[Sum of Convex Sets in Vector Space is Convex]], we have:
:$\lambda A + \mu B$ is [[Definition:Convex Set (Vector Space)|convex]].
{{qed}}
[[Category:Sum of Convex... | Sum of Convex Sets in Vector Space is Convex/Corollary | https://proofwiki.org/wiki/Sum_of_Convex_Sets_in_Vector_Space_is_Convex/Corollary | https://proofwiki.org/wiki/Sum_of_Convex_Sets_in_Vector_Space_is_Convex/Corollary | [
"Sum of Convex Sets in Vector Space is Convex"
] | [
"Definition:Vector Space",
"Definition:Convex Set (Vector Space)",
"Definition:Convex Set (Vector Space)"
] | [
"Dilation of Convex Set in Vector Space is Convex",
"Definition:Convex Set (Vector Space)",
"Sum of Convex Sets in Vector Space is Convex",
"Definition:Convex Set (Vector Space)",
"Category:Sum of Convex Sets in Vector Space is Convex"
] |
proofwiki-19813 | Closed Convex Set in terms of Bounded Linear Functionals | Let $X$ be a normed vector space over $\R$.
Let $X^\ast$ be the normed dual of $X$.
Let $C$ be a closed convex subset of $X$.
Then:
:$\ds C = \bigcap_{f \in X^\ast} \set {x \in X : \map f x \le \sup_{c \in C} \map f c}$ | {{finish|A few of these links only apply to finite sups/infs so will need to be replaced. they remain as placeholders}}
Clearly if $x \in C$ then:
:$\ds \map f x \le \sup_{c \in C} \map f c$
for all $f \in X^\ast$, by the definition of supremum.
That is:
:$\ds C \subseteq \bigcap_{f \in X^\ast} \set {x \in X : \map f... | Let $X$ be a [[Definition:Normed Vector Space|normed vector space]] over $\R$.
Let $X^\ast$ be the [[Definition:Normed Dual Space|normed dual]] of $X$.
Let $C$ be a [[Definition:Closed Set of Normed Vector Space|closed]] [[Definition:Convex Set (Vector Space)|convex]] subset of $X$.
Then:
:$\ds C = \bigcap_{f \i... | {{finish|A few of these links only apply to finite sups/infs so will need to be replaced. they remain as placeholders}}
Clearly if $x \in C$ then:
:$\ds \map f x \le \sup_{c \in C} \map f c$
for all $f \in X^\ast$, by the definition of [[Definition:Supremum of Real-Valued Function|supremum]].
That is:
:$\ds C \s... | Closed Convex Set in terms of Bounded Linear Functionals | https://proofwiki.org/wiki/Closed_Convex_Set_in_terms_of_Bounded_Linear_Functionals | https://proofwiki.org/wiki/Closed_Convex_Set_in_terms_of_Bounded_Linear_Functionals | [
"Normed Dual Spaces",
"Convex Sets (Vector Spaces)"
] | [
"Definition:Normed Vector Space",
"Definition:Normed Dual Space",
"Definition:Closed Set/Normed Vector Space",
"Definition:Convex Set (Vector Space)"
] | [
"Definition:Supremum of Mapping/Real-Valued Function",
"Rule of Transposition",
"Finite Topological Space is Compact",
"Definition:Compact Topological Space/Subspace",
"Definition:Disjoint Sets",
"Hahn-Banach Separation Theorem/Compact Convex Set and Closed Convex Set",
"Definition:Infimum of Mapping/Re... |
proofwiki-19814 | Transfinite Recursion Theorem/Formulation 2 | Let $\On$ denote the class of all ordinals.
Let $S$ denote the class of all ordinal sequences.
Let $g$ be a mapping such that $S \subseteq \Dom g$.
Then there exists a '''unique''' mapping $F$ on $\On$ such that:
:$\forall \alpha \in \On: \map F \alpha = \map g {F \restriction \alpha}$
where $F \restriction \alpha$ den... | First we establish the following | Let $\On$ denote the [[Definition:Class of All Ordinals|class of all ordinals]].
Let $S$ denote the [[Definition:Class (Class Theory)|class]] of all [[Definition:Ordinal Sequence|ordinal sequences]].
Let $g$ be a [[Definition:Class Mapping|mapping]] such that $S \subseteq \Dom g$.
Then there exists a '''[[Definitio... | First we establish the following | Transfinite Recursion Theorem/Formulation 2 | https://proofwiki.org/wiki/Transfinite_Recursion_Theorem/Formulation_2 | https://proofwiki.org/wiki/Transfinite_Recursion_Theorem/Formulation_2 | [
"Transfinite Recursion Theorem"
] | [
"Definition:Class of All Ordinals",
"Definition:Class (Class Theory)",
"Definition:Ordinal Sequence",
"Definition:Mapping/Class Theory",
"Definition:Unique",
"Definition:Mapping/Class Theory",
"Definition:Restriction/Mapping"
] | [] |
proofwiki-19815 | Transfinite Recursion Theorem/Formulation 4 | Let $\On$ denote the class of all ordinals.
Let $g$ be a mapping defined for all sets.
Let $c$ be a set.
Then there exists a '''unique''' $\On$-sequence $S_0, S_1, \dots, S_\alpha, \dots$ such that:
{{begin-axiom}}
{{axiom | n = 1
| q =
| ml= S_0
| mo= =
| mr= c
}}
{{axiom | n = 2
... | This is a special case of the Transfinite Recursion Theorem: Formulation $3$.
Recall:
{{:Transfinite Recursion Theorem/Formulation 3}}{{qed|lemma}}
Set $\map f x = \bigcup x$, so that:
:$\map f {F \sqbrk \lambda}$
is then:
:$\ds \bigcup_{\alpha \mathop < \lambda} \map F \alpha$
Let $S_\alpha$ be the set $\map F \alpha$... | Let $\On$ denote the [[Definition:Class of All Ordinals|class of all ordinals]].
Let $g$ be a [[Definition:Class Mapping|mapping]] defined for all [[Definition:Set|sets]].
Let $c$ be a [[Definition:Set|set]].
Then there exists a '''[[Definition:Unique|unique]]''' [[Definition:Ordinal Sequence|$\On$-sequence]] $S_0,... | This is a special case of the [[Transfinite Recursion Theorem/Formulation 3|Transfinite Recursion Theorem: Formulation $3$]].
Recall:
{{:Transfinite Recursion Theorem/Formulation 3}}{{qed|lemma}}
Set $\map f x = \bigcup x$, so that:
:$\map f {F \sqbrk \lambda}$
is then:
:$\ds \bigcup_{\alpha \mathop < \lambda} \map ... | Transfinite Recursion Theorem/Formulation 4 | https://proofwiki.org/wiki/Transfinite_Recursion_Theorem/Formulation_4 | https://proofwiki.org/wiki/Transfinite_Recursion_Theorem/Formulation_4 | [
"Transfinite Recursion Theorem"
] | [
"Definition:Class of All Ordinals",
"Definition:Mapping/Class Theory",
"Definition:Set",
"Definition:Set",
"Definition:Unique",
"Definition:Ordinal Sequence",
"Definition:Class (Class Theory)",
"Definition:Limit Ordinal"
] | [
"Transfinite Recursion Theorem/Formulation 3",
"Definition:Set",
"Transfinite Recursion Theorem",
"Definition:Unique"
] |
proofwiki-19816 | Intersection of Zero Loci is Zero Locus | Let $k$ be a field.
Let $n \in \N_{>0}$.
Let $A = k \sqbrk {X_1, \ldots, X_n}$ be the ring of polynomials in $n$ variables over $k$.
Let $\mathbb S \subseteq \powerset A$ be a subset of the power set of $A$.
Then:
:$\ds \bigcap _{S \mathop \in \mathbb S} \map V S = \map V {\bigcup \mathbb S}$
where $\map V \cdot$ denot... | {{begin-eqn}}
{{eqn | l = x
| o = \in
| r = \bigcap_{S \mathop \in \mathbb S} \map V S
}}
{{eqn | ll= \leadstoandfrom
| q = \forall S \in \mathbb S
| l = x
| o = \in
| r = \map V S
}}
{{eqn | ll= \leadstoandfrom
| q = \forall S \in \mathbb S: \forall f \in S
| l = \map f ... | Let $k$ be a [[Definition:Field (Abstract Algebra)|field]].
Let $n \in \N_{>0}$.
Let $A = k \sqbrk {X_1, \ldots, X_n}$ be the [[Definition:Ring of Polynomials|ring of polynomials]] in $n$ [[Definition:Variable of Polynomial Ring|variables]] over $k$.
Let $\mathbb S \subseteq \powerset A$ be a [[Definition:Subset|sub... | {{begin-eqn}}
{{eqn | l = x
| o = \in
| r = \bigcap_{S \mathop \in \mathbb S} \map V S
}}
{{eqn | ll= \leadstoandfrom
| q = \forall S \in \mathbb S
| l = x
| o = \in
| r = \map V S
}}
{{eqn | ll= \leadstoandfrom
| q = \forall S \in \mathbb S: \forall f \in S
| l = \map f ... | Intersection of Zero Loci is Zero Locus | https://proofwiki.org/wiki/Intersection_of_Zero_Loci_is_Zero_Locus | https://proofwiki.org/wiki/Intersection_of_Zero_Loci_is_Zero_Locus | [
"Algebraic Geometry"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Polynomial Ring",
"Definition:Polynomial Ring/Indeterminate",
"Definition:Subset",
"Definition:Power Set",
"Definition:Zero Locus of Set of Polynomials"
] | [
"Category:Algebraic Geometry"
] |
proofwiki-19817 | Union of two Zero Loci is Zero Locus | Let $k$ be a field.
Let $n \in \N_{>0}$.
Let $A = k \sqbrk {X_1, \ldots, X_n}$ be the ring of polynomials in $n$ variables over $k$.
Let $S,T \subseteq A$.
Then:
:$\ds {\map V S} \cup \map V T = \map V {S \cdot T}$
where:
:$\map V \cdot$ denotes the zero locus
:$S \cdot T := \set { fg : f \in S, g \in T }$ | Let $x \in {\map V S} \cup \map V T$.
That is, either $x \in \map V S$ or $x \in \map V T$.
Then, if $x \in \map V S$:
:$\forall \paren {f,g} \in S \times T : \map f x \map g x = 0 \map g x = 0$
else:
:$\forall \paren {f,g} \in S \times T : \map f x \map g x = \map f x 0 = 0$
Therefore:
:$x \in \map V {S \cdot T}$
On... | Let $k$ be a [[Definition:Field (Abstract Algebra)|field]].
Let $n \in \N_{>0}$.
Let $A = k \sqbrk {X_1, \ldots, X_n}$ be the [[Definition:Ring of Polynomials|ring of polynomials]] in $n$ [[Definition:Variable of Polynomial Ring|variables]] over $k$.
Let $S,T \subseteq A$.
Then:
:$\ds {\map V S} \cup \map V T = \m... | Let $x \in {\map V S} \cup \map V T$.
That is, either $x \in \map V S$ or $x \in \map V T$.
Then, if $x \in \map V S$:
:$\forall \paren {f,g} \in S \times T : \map f x \map g x = 0 \map g x = 0$
else:
:$\forall \paren {f,g} \in S \times T : \map f x \map g x = \map f x 0 = 0$
Therefore:
:$x \in \map V {S \cdot T}$... | Union of two Zero Loci is Zero Locus | https://proofwiki.org/wiki/Union_of_two_Zero_Loci_is_Zero_Locus | https://proofwiki.org/wiki/Union_of_two_Zero_Loci_is_Zero_Locus | [
"Algebraic Geometry"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Polynomial Ring",
"Definition:Polynomial Ring/Indeterminate",
"Definition:Zero Locus of Set of Polynomials"
] | [
"De Morgan's Laws (Set Theory)",
"Category:Algebraic Geometry"
] |
proofwiki-19818 | Zariski Topology is Topology | Let $k$ be a field.
Let $n \in \N_{>0}$.
Let $\tau$ be the Zariski topology on $k^n$.
Then $\tau$ is indeed a topology on $k^n$. | Let $A := k \sqbrk {X_1, \ldots, X_n}$ be the ring of polynomials.
Recall that by definition of Zariski topology:
:$\forall U \in \tau: \exists T_U \subseteq A: U = X \setminus \map V {T_U}$
where $\map V {T_U}$ denotes the zero locus of $T_U$.
Each of the open set axioms is examined in turn: | Let $k$ be a [[Definition:Field (Abstract Algebra)|field]].
Let $n \in \N_{>0}$.
Let $\tau$ be the [[Definition:Zariski Topology on Affine Space|Zariski topology]] on $k^n$.
Then $\tau$ is indeed a [[Definition:Topology|topology]] on $k^n$. | Let $A := k \sqbrk {X_1, \ldots, X_n}$ be the [[Definition:Ring of Polynomials|ring of polynomials]].
Recall that by definition of [[Definition:Zariski Topology on Affine Space|Zariski topology]]:
:$\forall U \in \tau: \exists T_U \subseteq A: U = X \setminus \map V {T_U}$
where $\map V {T_U}$ denotes the [[Definition... | Zariski Topology is Topology | https://proofwiki.org/wiki/Zariski_Topology_is_Topology | https://proofwiki.org/wiki/Zariski_Topology_is_Topology | [
"Affine Geometry",
"Zariski Topology"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Zariski Topology/Affine Space",
"Definition:Topology"
] | [
"Definition:Polynomial Ring",
"Definition:Zariski Topology/Affine Space",
"Definition:Zero Locus of Set of Polynomials",
"Axiom:Open Set Axioms",
"Axiom:Open Set Axioms"
] |
proofwiki-19819 | Pointwise Maximum of Finite Family of Seminorms is Seminorm | Let $\struct {K, \norm {\,\cdot\,}_K}$ be a normed division ring.
Let $X$ be a vector space over $K$.
Let $\II$ be a set of seminorms on $X$.
Define:
:$\ds q =: \max_{p \mathop \in \II} p$
where $\max$ denotes the pointwise maximum over $\II$. | === Proof of $(\text N 2)$ ===
Let $\lambda \in K$ and $x \in X$.
Then:
{{begin-eqn}}
{{eqn | l = \map q {\lambda x}
| r = \max_{p \mathop \in \II} \map p {\lambda x}
| c = Definition of $q$
}}
{{eqn | r = \max_{p \mathop \in \II} \norm \lambda_K \map p x
| c = $(\text N 2)$ in the definition of the semin... | Let $\struct {K, \norm {\,\cdot\,}_K}$ be a [[Definition:Normed Division Ring|normed division ring]].
Let $X$ be a [[Definition:Vector Space|vector space]] over $K$.
Let $\II$ be a [[Definition:Set|set]] of [[Definition:Seminorm|seminorms]] on $X$.
Define:
:$\ds q =: \max_{p \mathop \in \II} p$
where $\max$ den... | === Proof of $(\text N 2)$ ===
Let $\lambda \in K$ and $x \in X$.
Then:
{{begin-eqn}}
{{eqn | l = \map q {\lambda x}
| r = \max_{p \mathop \in \II} \map p {\lambda x}
| c = Definition of $q$
}}
{{eqn | r = \max_{p \mathop \in \II} \norm \lambda_K \map p x
| c = $(\text N 2)$ in the definition of the [[... | Pointwise Maximum of Finite Family of Seminorms is Seminorm | https://proofwiki.org/wiki/Pointwise_Maximum_of_Finite_Family_of_Seminorms_is_Seminorm | https://proofwiki.org/wiki/Pointwise_Maximum_of_Finite_Family_of_Seminorms_is_Seminorm | [
"Seminorms"
] | [
"Definition:Normed Division Ring",
"Definition:Vector Space",
"Definition:Set",
"Definition:Seminorm",
"Definition:Pointwise Maximum of Mappings/Real-Valued Functions"
] | [
"Definition:Seminorm",
"Definition:Seminorm"
] |
proofwiki-19820 | Linear Transformation between Topological Vector Spaces Continuous iff Continuous at Origin | Let $K$ be a topological field.
Let $X$ and $Y$ be topological vector spaces over $K$.
Let $T : X \to Y$ be a linear transformation.
Then $T$ is everywhere continuous {{iff}} it is continuous at ${\mathbf 0}_X$. | === Necessary Condition ===
Suppose that $T$ is everywhere continuous.
Then $T$ is continuous at every point.
In particular, $T$ is continuous at ${\mathbf 0}_X$.
{{qed|lemma}} | Let $K$ be a [[Definition:Topological Field|topological field]].
Let $X$ and $Y$ be [[Definition:Topological Vector Space|topological vector spaces]] over $K$.
Let $T : X \to Y$ be a [[Definition:Linear Transformation|linear transformation]].
Then $T$ is [[Definition:Everywhere Continuous Mapping Between Topologic... | === Necessary Condition ===
Suppose that $T$ is [[Definition:Everywhere Continuous Mapping Between Topological Spaces|everywhere continuous]].
Then $T$ is [[Definition:Continuous Mapping at Point (Topology)|continuous at every point]].
In particular, $T$ is [[Definition:Continuous Mapping at Point (Topology)|continu... | Linear Transformation between Topological Vector Spaces Continuous iff Continuous at Origin | https://proofwiki.org/wiki/Linear_Transformation_between_Topological_Vector_Spaces_Continuous_iff_Continuous_at_Origin | https://proofwiki.org/wiki/Linear_Transformation_between_Topological_Vector_Spaces_Continuous_iff_Continuous_at_Origin | [
"Topological Vector Spaces",
"Linear Transformations"
] | [
"Definition:Topological Field",
"Definition:Topological Vector Space",
"Definition:Linear Transformation",
"Definition:Continuous Mapping (Topology)/Everywhere",
"Definition:Continuous Mapping (Topology)/Point"
] | [
"Definition:Continuous Mapping (Topology)/Everywhere",
"Definition:Continuous Mapping (Topology)/Point",
"Definition:Continuous Mapping (Topology)/Point",
"Definition:Continuous Mapping (Topology)/Point",
"Definition:Continuous Mapping (Topology)/Point",
"Definition:Continuous Mapping (Topology)/Point"
] |
proofwiki-19821 | Topological Vector Space as Union of Dilations of Open Neighborhood of Origin | Let $\Bbb F \in \set {\R, \C}$.
Let $X$ be a topological vector space over $\Bbb F$.
Let $V$ be an open neighborhood of ${\mathbf 0}_X$.
Let $\sequence {r_n}_{n \mathop \in \N}$ be a sequence of positive real numbers such that $r_n \to \infty$.
Then:
:$\ds X = \bigcup_{n \mathop = 1}^\infty r_n V$
where $r_n V$ deno... | Clearly:
:$\ds \bigcup_{n \mathop = 1}^\infty r_n V \subseteq X$
We show that:
:$\ds X \subseteq \bigcup_{n \mathop = 1}^\infty r_n V$
Let $x \in X$.
From the definition of a topological vector space, the mapping $K \times X \to X$ defined by $\tuple {\lambda, x} \mapsto \lambda x$ is continuous.
From Horizontal Sec... | Let $\Bbb F \in \set {\R, \C}$.
Let $X$ be a [[Definition:Topological Vector Space|topological vector space]] over $\Bbb F$.
Let $V$ be an [[Definition:Open Neighborhood|open neighborhood]] of ${\mathbf 0}_X$.
Let $\sequence {r_n}_{n \mathop \in \N}$ be a [[Definition:Sequence|sequence]] of [[Definition:Positive Re... | Clearly:
:$\ds \bigcup_{n \mathop = 1}^\infty r_n V \subseteq X$
We show that:
:$\ds X \subseteq \bigcup_{n \mathop = 1}^\infty r_n V$
Let $x \in X$.
From the definition of a [[Definition:Topological Vector Space|topological vector space]], the [[Definition:Mapping|mapping]] $K \times X \to X$ defined by $\tupl... | Topological Vector Space as Union of Dilations of Open Neighborhood of Origin | https://proofwiki.org/wiki/Topological_Vector_Space_as_Union_of_Dilations_of_Open_Neighborhood_of_Origin | https://proofwiki.org/wiki/Topological_Vector_Space_as_Union_of_Dilations_of_Open_Neighborhood_of_Origin | [
"Topological Vector Spaces"
] | [
"Definition:Topological Vector Space",
"Definition:Open Neighborhood",
"Definition:Sequence",
"Definition:Positive/Real Number",
"Definition:Linear Combination of Subsets of Vector Space/Dilation"
] | [
"Definition:Topological Vector Space",
"Definition:Mapping",
"Definition:Continuous Mapping (Topology)",
"Horizontal Section of Continuous Function is Continuous",
"Definition:Horizontal Section of Function",
"Definition:Continuous Mapping (Topology)",
"Definition:Set",
"Definition:Open Set/Topology",... |
proofwiki-19822 | Condition for Point being in Closure/Topological Vector Space | Let $K$ be a topological field.
Let $X$ be a topological vector space over $K$.
Let $A \subseteq X$.
Let $A^-$ denote the closure of $A$ in $X$.
Let $x \in X$.
Then $x \in A^-$ {{iff}}:
:for each open neighborhood $V$ of ${\mathbf 0}_X$ we have $\paren {x + V} \cap A \ne \O$. | From the definition of the closure of $A$, $A^-$ is the set of adherent points of $A$.
So $x \in A^-$ {{iff}} for each open neighborhood $U$ of $x$ we have:
:$U \cap A \ne \O$
From Classification of Open Neighborhoods in Topological Vector Space, every such $U$ has the form $U = x + V$ for some open neighborhood $V$ ... | Let $K$ be a [[Definition:Topological Field|topological field]].
Let $X$ be a [[Definition:Topological Vector Space|topological vector space]] over $K$.
Let $A \subseteq X$.
Let $A^-$ denote the [[Definition:Closure (Topology)|closure]] of $A$ in $X$.
Let $x \in X$.
Then $x \in A^-$ {{iff}}:
:for each [[Defin... | From the definition of the [[Definition:Closure (Topology)|closure]] of $A$, $A^-$ is the set of [[Definition:Adherent Point of Set|adherent points]] of $A$.
So $x \in A^-$ {{iff}} for each [[Definition:Open Neighborhood|open neighborhood]] $U$ of $x$ we have:
:$U \cap A \ne \O$
From [[Classification of Open Neigh... | Condition for Point being in Closure/Topological Vector Space | https://proofwiki.org/wiki/Condition_for_Point_being_in_Closure/Topological_Vector_Space | https://proofwiki.org/wiki/Condition_for_Point_being_in_Closure/Topological_Vector_Space | [
"Condition for Point being in Closure",
"Topological Vector Spaces"
] | [
"Definition:Topological Field",
"Definition:Topological Vector Space",
"Definition:Closure (Topology)",
"Definition:Open Neighborhood"
] | [
"Definition:Closure (Topology)",
"Definition:Adherent Point of Set",
"Definition:Open Neighborhood",
"Classification of Open Neighborhoods in Topological Vector Space",
"Definition:Open Neighborhood",
"Definition:Open Neighborhood",
"Definition:Open Neighborhood",
"Category:Condition for Point being i... |
proofwiki-19823 | Expression for Closure of Set in Topological Vector Space | Let $K$ be a topological field.
Let $X$ be a topological vector space over $K$.
Let $A \subseteq X$.
Let $\mathcal V$ be the set of open neighborhoods of ${\mathbf 0}_X$.
Then:
:$\ds A^- = \bigcap_{V \in \mathcal V} \paren {A + V}$
where $A^-$ is the closure of $A$. | From Condition for Point being in Closure: Topological Vector Space, we have $x \in A^-$ {{iff}}:
:for each open neighborhood $V$ of ${\mathbf 0}_X$ we have $\paren {x + V} \cap A \ne \O$.
We have that $\paren {x + V} \cap A \ne \O$ {{iff}} there exists some $u \in X$ with $u \in A$ and $u \in x + V$.
This is equivale... | Let $K$ be a [[Definition:Topological Field|topological field]].
Let $X$ be a [[Definition:Topological Vector Space|topological vector space]] over $K$.
Let $A \subseteq X$.
Let $\mathcal V$ be the [[Definition:Set|set]] of [[Definition:Open Neighborhood|open neighborhoods]] of ${\mathbf 0}_X$.
Then:
:$\ds A^-... | From [[Condition for Point being in Closure/Topological Vector Space|Condition for Point being in Closure: Topological Vector Space]], we have $x \in A^-$ {{iff}}:
:for each [[Definition:Open Neighborhood|open neighborhood]] $V$ of ${\mathbf 0}_X$ we have $\paren {x + V} \cap A \ne \O$.
We have that $\paren {x + V} ... | Expression for Closure of Set in Topological Vector Space | https://proofwiki.org/wiki/Expression_for_Closure_of_Set_in_Topological_Vector_Space | https://proofwiki.org/wiki/Expression_for_Closure_of_Set_in_Topological_Vector_Space | [
"Topological Vector Spaces",
"Set Closures",
"Expression for Closure of Set in Topological Vector Space"
] | [
"Definition:Topological Field",
"Definition:Topological Vector Space",
"Definition:Set",
"Definition:Open Neighborhood",
"Definition:Closure (Topology)"
] | [
"Condition for Point being in Closure/Topological Vector Space",
"Definition:Open Neighborhood",
"Definition:Open Neighborhood",
"Dilation of Open Set in Topological Vector Space is Open",
"Definition:Open Set/Topology",
"Definition:Open Set/Topology"
] |
proofwiki-19824 | Union of Balanced Sets in Vector Space is Balanced | Let $\Bbb F \in \set {\R, \C}$.
Let $X$ be a vector space over $\Bbb F$.
Let $\sequence {E_\alpha}_{\alpha \mathop \in I}$ be an $I$-indexed family of balanced subsets of $X$.
Then:
:$\ds E = \bigcup_{\alpha \mathop \in I} E_\alpha$ is balanced. | Let $x \in E$.
Let $\lambda \in \Bbb F$ have $\cmod \lambda \le 1$.
We aim to show that $\lambda x \in E$.
Since $x \in E$, there exists $\alpha \in I$ with $x \in E_\alpha$.
Since $E_\alpha$ is balanced, we have $\lambda x \in E_\alpha$.
So:
:$\ds \lambda x \in \bigcup_{\alpha \mathop \in I} E_\alpha = E$
Since $x$... | Let $\Bbb F \in \set {\R, \C}$.
Let $X$ be a [[Definition:Vector Space|vector space]] over $\Bbb F$.
Let $\sequence {E_\alpha}_{\alpha \mathop \in I}$ be an [[Definition:Indexed Family|$I$-indexed family]] of [[Definition:Balanced Set|balanced subsets]] of $X$.
Then:
:$\ds E = \bigcup_{\alpha \mathop \in I} E_\al... | Let $x \in E$.
Let $\lambda \in \Bbb F$ have $\cmod \lambda \le 1$.
We aim to show that $\lambda x \in E$.
Since $x \in E$, there exists $\alpha \in I$ with $x \in E_\alpha$.
Since $E_\alpha$ is [[Definition:Balanced Set|balanced]], we have $\lambda x \in E_\alpha$.
So:
:$\ds \lambda x \in \bigcup_{\alpha \mat... | Union of Balanced Sets in Vector Space is Balanced | https://proofwiki.org/wiki/Union_of_Balanced_Sets_in_Vector_Space_is_Balanced | https://proofwiki.org/wiki/Union_of_Balanced_Sets_in_Vector_Space_is_Balanced | [
"Balanced Sets",
"Set Union"
] | [
"Definition:Vector Space",
"Definition:Indexing Set/Family",
"Definition:Balanced Set",
"Definition:Balanced Set"
] | [
"Definition:Balanced Set",
"Definition:Balanced Set",
"Category:Balanced Sets",
"Category:Set Union"
] |
proofwiki-19825 | Transfinite Recursion Theorem/Formulation 5 | Let $\On$ denote the class of all ordinals.
Let $h$ be a mapping.
Then there exists a '''unique''' mapping $F$ such that:
:$\forall \alpha \in \On: \map F \alpha = \map h {F \sqbrk \alpha}$ | This is a special case of the Transfinite Recursion Theorem: Formulation $2$.
Recall:
{{:Transfinite Recursion Theorem/Formulation 2}}{{qed|lemma}}
Consider $\map g x$.
If $x$ is a mapping, take $\map g x = \map h {\Img x}$.
If $x$ is not a mapping, take $\map g x = x$.
The result follows.
{{qed}} | Let $\On$ denote the [[Definition:Class of All Ordinals|class of all ordinals]].
Let $h$ be a [[Definition:Class Mapping|mapping]].
Then there exists a '''[[Definition:Unique|unique]]''' [[Definition:Class Mapping|mapping]] $F$ such that:
:$\forall \alpha \in \On: \map F \alpha = \map h {F \sqbrk \alpha}$ | This is a special case of the [[Transfinite Recursion Theorem/Formulation 2|Transfinite Recursion Theorem: Formulation $2$]].
Recall:
{{:Transfinite Recursion Theorem/Formulation 2}}{{qed|lemma}}
Consider $\map g x$.
If $x$ is a [[Definition:Class Mapping|mapping]], take $\map g x = \map h {\Img x}$.
If $x$ is not... | Transfinite Recursion Theorem/Formulation 5 | https://proofwiki.org/wiki/Transfinite_Recursion_Theorem/Formulation_5 | https://proofwiki.org/wiki/Transfinite_Recursion_Theorem/Formulation_5 | [
"Transfinite Recursion Theorem"
] | [
"Definition:Class of All Ordinals",
"Definition:Mapping/Class Theory",
"Definition:Unique",
"Definition:Mapping/Class Theory"
] | [
"Transfinite Recursion Theorem/Formulation 2",
"Definition:Mapping/Class Theory",
"Definition:Mapping/Class Theory"
] |
proofwiki-19826 | Class Mapping has Minimally Superinductive Class | Let $g$ be a (class) mapping.
Then there exists a class $M$ that is minimally superinductive under $g$. | {{ProofWanted|According to the rubric of the exercise, this is to be proved using one of the versions of the Transfinite Recursion Theorem that appears in S&F}} | Let $g$ be a [[Definition:Class Mapping|(class) mapping]].
Then there exists a [[Definition:Class (Class Theory)|class]] $M$ that is [[Definition:Minimally Superinductive Class|minimally superinductive]] under $g$. | {{ProofWanted|According to the rubric of the exercise, this is to be proved using one of the versions of the [[Transfinite Recursion Theorem]] that appears in S&F}} | Class Mapping has Minimally Superinductive Class | https://proofwiki.org/wiki/Class_Mapping_has_Minimally_Superinductive_Class | https://proofwiki.org/wiki/Class_Mapping_has_Minimally_Superinductive_Class | [
"Minimally Superinductive Classes"
] | [
"Definition:Mapping/Class Theory",
"Definition:Class (Class Theory)",
"Definition:Minimally Superinductive Class"
] | [
"Transfinite Recursion Theorem"
] |
proofwiki-19827 | Open Neighborhood of Origin in Topological Vector Space contains Balanced Open Neighborhood | Let $\Bbb F \in \set {\R, \C}$.
Let $X$ be a topological vector space over $\Bbb F$.
Let ${\mathbf 0}_X$ be the origin of $X$.
Let $U$ be an open neighborhood of ${\mathbf 0}_X$.
Then there exists a balanced open neighborhood $W$ of ${\mathbf 0}_X$ with $W \subseteq U$. | Equip $\Bbb F \times X$ with the product topology.
From Box Topology on Finite Product Space is Product Topology, this is precisely the box topology.
Define $M : \Bbb F \times X \to X$ by:
:$\map M {\lambda, x} = \lambda x$
for each $\tuple {\lambda, x} \in \Bbb F \times X$.
From the definition of a topological vector... | Let $\Bbb F \in \set {\R, \C}$.
Let $X$ be a [[Definition:Topological Vector Space|topological vector space]] over $\Bbb F$.
Let ${\mathbf 0}_X$ be the [[Definition:Origin of Vector Space|origin]] of $X$.
Let $U$ be an [[Definition:Open Neighborhood|open neighborhood]] of ${\mathbf 0}_X$.
Then there exists a [[D... | Equip $\Bbb F \times X$ with the [[Definition:Product Topology|product topology]].
From [[Box Topology on Finite Product Space is Product Topology]], this is precisely the [[Definition:Box Topology|box topology]].
Define $M : \Bbb F \times X \to X$ by:
:$\map M {\lambda, x} = \lambda x$
for each $\tuple {\lambda, ... | Open Neighborhood of Origin in Topological Vector Space contains Balanced Open Neighborhood | https://proofwiki.org/wiki/Open_Neighborhood_of_Origin_in_Topological_Vector_Space_contains_Balanced_Open_Neighborhood | https://proofwiki.org/wiki/Open_Neighborhood_of_Origin_in_Topological_Vector_Space_contains_Balanced_Open_Neighborhood | [
"Balanced Sets",
"Topological Vector Spaces"
] | [
"Definition:Topological Vector Space",
"Definition:Zero Vector",
"Definition:Open Neighborhood",
"Definition:Balanced Set",
"Definition:Open Neighborhood"
] | [
"Definition:Product Topology",
"Box Topology on Finite Product Space is Product Topology",
"Definition:Box Topology",
"Definition:Topological Vector Space",
"Definition:Continuous Mapping (Topology)",
"Definition:Continuous Mapping (Topology)/Point",
"Definition:Open Neighborhood",
"Basis for Box Topo... |
proofwiki-19828 | Finite Union of Dilations of Balanced Set | Let $\Bbb F \in \set {\R, \C}$.
Let $X$ be a vector space over $\Bbb F$.
Let $\lambda_1, \lambda_2, \ldots, \lambda_n$ be real numbers.
Let $W$ be a balanced set.
Then:
:$\ds \bigcup_{i \mathop = 1}^n \lambda_i W = \lambda W$
where $\lambda \in \set {\lambda_1, \lambda_2, \ldots, \lambda_n}$ is picked so that $\cmod... | If $\lambda = 0$, then $\lambda_i = 0$ for each $i \in \set {1, 2, \ldots, n}$, so that:
:$\ds \bigcup_{i \mathop = 1}^n \lambda_i W = \set {\mathbf 0_X} = \lambda W$
Now suppose that $\lambda \ne 0$.
Then for each $i$ we have:
:$\ds \cmod {\frac {\lambda_i} \lambda} = \frac {\cmod {\lambda_i} } {\cmod \lambda} \le 1$
... | Let $\Bbb F \in \set {\R, \C}$.
Let $X$ be a [[Definition:Vector Space|vector space]] over $\Bbb F$.
Let $\lambda_1, \lambda_2, \ldots, \lambda_n$ be [[Definition:Real Number|real numbers]].
Let $W$ be a [[Definition:Balanced Set|balanced set]].
Then:
:$\ds \bigcup_{i \mathop = 1}^n \lambda_i W = \lambda W$
w... | If $\lambda = 0$, then $\lambda_i = 0$ for each $i \in \set {1, 2, \ldots, n}$, so that:
:$\ds \bigcup_{i \mathop = 1}^n \lambda_i W = \set {\mathbf 0_X} = \lambda W$
Now suppose that $\lambda \ne 0$.
Then for each $i$ we have:
:$\ds \cmod {\frac {\lambda_i} \lambda} = \frac {\cmod {\lambda_i} } {\cmod \lambda} \le... | Finite Union of Dilations of Balanced Set | https://proofwiki.org/wiki/Finite_Union_of_Dilations_of_Balanced_Set | https://proofwiki.org/wiki/Finite_Union_of_Dilations_of_Balanced_Set | [
"Dilations of Subsets of Vector Spaces",
"Balanced Sets"
] | [
"Definition:Vector Space",
"Definition:Real Number",
"Definition:Balanced Set"
] | [
"Definition:Balanced Set",
"Union of Subsets is Subset",
"Set is Subset of Union",
"Definition:Set Equality/Definition 2",
"Category:Dilations of Subsets of Vector Spaces",
"Category:Balanced Sets"
] |
proofwiki-19829 | Compact Subspace of Topological Vector Space is von Neumann-Bounded | Let $\Bbb F \in \set {\R, \C}$.
Let $X$ be a topological vector space over $\Bbb F$.
Let $K$ be a compact subspace (where subspace is meant in the topological sense) of $X$.
Then $K$ is von Neumann-bounded. | Let $V$ be an open neighborhood of ${\mathbf 0}_X$.
We aim to find $s > 0$ such that $E \subseteq t V$ for $t > s$.
From Open Neighborhood of Origin in Topological Vector Space contains Balanced Open Neighborhood, there exists $W \subseteq V$ such that $W$ is a balanced open neighborhood of ${\mathbf 0}_X$.
From Topolo... | Let $\Bbb F \in \set {\R, \C}$.
Let $X$ be a [[Definition:Topological Vector Space|topological vector space]] over $\Bbb F$.
Let $K$ be a [[Definition:Compact Subspace|compact subspace]] (where [[Definition:Topological Subspace|subspace]] is meant in the topological sense) of $X$.
Then $K$ is [[Definition:Von Neum... | Let $V$ be an [[Definition:Open Neighborhood|open neighborhood]] of ${\mathbf 0}_X$.
We aim to find $s > 0$ such that $E \subseteq t V$ for $t > s$.
From [[Open Neighborhood of Origin in Topological Vector Space contains Balanced Open Neighborhood]], there exists $W \subseteq V$ such that $W$ is a [[Definition:Balanc... | Compact Subspace of Topological Vector Space is von Neumann-Bounded | https://proofwiki.org/wiki/Compact_Subspace_of_Topological_Vector_Space_is_von_Neumann-Bounded | https://proofwiki.org/wiki/Compact_Subspace_of_Topological_Vector_Space_is_von_Neumann-Bounded | [
"Von Neumann-Bounded Subsets of Topological Vector Spaces",
"Compact Topological Spaces"
] | [
"Definition:Topological Vector Space",
"Definition:Compact Topological Space/Subspace",
"Definition:Topological Subspace",
"Definition:Von Neumann-Bounded Subset of Topological Vector Space"
] | [
"Definition:Open Neighborhood",
"Open Neighborhood of Origin in Topological Vector Space contains Balanced Open Neighborhood",
"Definition:Balanced Set",
"Definition:Open Neighborhood",
"Topological Vector Space as Union of Dilations of Open Neighborhood of Origin",
"Definition:Compact Topological Space/S... |
proofwiki-19830 | Dilations of von Neumann-Bounded Neighborhood of Origin in Topological Vector Space form Local Basis for Origin | Let $\Bbb F \in \set {\R, \C}$.
Let $X$ be a topological vector space over $\Bbb F$.
Let $\sequence {\delta_n}_{n \mathop \in \N}$ be a strictly decreasing real sequence with $\delta_n \to 0$.
Let $V$ be a von Neumann-bounded open neighborhood of ${\mathbf 0}_X$.
Then:
:$\BB = \set {\delta_n V : n \in \N}$ is a local... | Let $U$ be an open neighborhood of ${\mathbf 0}_X$.
We show that there exists $N \in \N$ such that:
:$\delta_N V \subseteq U$
Since $V$ is von Neumann-bounded, there exists $s > 0$ such that:
:$V \subseteq t U$ for $t > s$.
Since $\delta_n \to 0$ and $\sequence {\delta_n}_{n \mathop \in \N}$ is a strictly decreasing... | Let $\Bbb F \in \set {\R, \C}$.
Let $X$ be a [[Definition:Topological Vector Space|topological vector space]] over $\Bbb F$.
Let $\sequence {\delta_n}_{n \mathop \in \N}$ be a [[Definition:Strictly Decreasing Real Sequence|strictly decreasing real sequence]] with $\delta_n \to 0$.
Let $V$ be a [[Definition:Von Neum... | Let $U$ be an [[Definition:Open Neighborhood|open neighborhood]] of ${\mathbf 0}_X$.
We show that there exists $N \in \N$ such that:
:$\delta_N V \subseteq U$
Since $V$ is [[Definition:Von Neumann-Bounded Subset of Topological Vector Space|von Neumann-bounded]], there exists $s > 0$ such that:
:$V \subseteq t U$... | Dilations of von Neumann-Bounded Neighborhood of Origin in Topological Vector Space form Local Basis for Origin | https://proofwiki.org/wiki/Dilations_of_von_Neumann-Bounded_Neighborhood_of_Origin_in_Topological_Vector_Space_form_Local_Basis_for_Origin | https://proofwiki.org/wiki/Dilations_of_von_Neumann-Bounded_Neighborhood_of_Origin_in_Topological_Vector_Space_form_Local_Basis_for_Origin | [
"Von Neumann-Bounded Subsets of Topological Vector Spaces",
"Dilations of Subsets of Vector Spaces",
"Local Bases"
] | [
"Definition:Topological Vector Space",
"Definition:Strictly Decreasing/Sequence/Real Sequence",
"Definition:Von Neumann-Bounded Subset of Topological Vector Space",
"Definition:Open Neighborhood",
"Definition:Local Basis"
] | [
"Definition:Open Neighborhood",
"Definition:Von Neumann-Bounded Subset of Topological Vector Space",
"Definition:Strictly Decreasing/Sequence/Real Sequence",
"Definition:Local Basis"
] |
proofwiki-19831 | Characterization of Continuous Linear Functionals on Topological Vector Space | Let $\Bbb F \in \set {\R, \C}$.
Let $X$ be a topological vector space over $\Bbb F$.
Let $f$ be a linear functional on $X$ such that $\map f x \ne 0$ for some $x \in X$.
{{TFAE}}
:$(1) \quad$ $f$ is continuous
:$(2) \quad$ $\ker f$ is closed, where $\ker f$ is the kernel of $f$
:$(3) \quad$ $\ker f$ is not everywhere... | === $(1)$ implies $(2)$ ===
Suppose that $f$ is continuous.
From the definition of the kernel, we have:
:$\ker f = f^{-1} \sqbrk {\set 0}$
Since $\set 0$ is closed in $\GF$, we have that:
:$f^{-1} \sqbrk {\set 0}$ is closed
from Continuity Defined from Closed Sets.
{{qed|lemma}} | Let $\Bbb F \in \set {\R, \C}$.
Let $X$ be a [[Definition:Topological Vector Space|topological vector space]] over $\Bbb F$.
Let $f$ be a [[Definition:Linear Functional|linear functional]] on $X$ such that $\map f x \ne 0$ for some $x \in X$.
{{TFAE}}
:$(1) \quad$ $f$ is [[Definition:Continuous Mapping (Topology... | === $(1)$ implies $(2)$ ===
Suppose that $f$ is [[Definition:Continuous Mapping (Topology)|continuous]].
From the definition of the [[Definition:Kernel of Linear Transformation|kernel]], we have:
:$\ker f = f^{-1} \sqbrk {\set 0}$
Since $\set 0$ is [[Definition:Closed Set (Topology)|closed]] in $\GF$, we have that... | Characterization of Continuous Linear Functionals on Topological Vector Space | https://proofwiki.org/wiki/Characterization_of_Continuous_Linear_Functionals_on_Topological_Vector_Space | https://proofwiki.org/wiki/Characterization_of_Continuous_Linear_Functionals_on_Topological_Vector_Space | [
"Topological Vector Spaces"
] | [
"Definition:Topological Vector Space",
"Definition:Linear Functional",
"Definition:Continuous Mapping (Topology)",
"Definition:Closed Set/Topology",
"Definition:Kernel of Linear Transformation",
"Definition:Everywhere Dense",
"Definition:Bounded Mapping",
"Definition:Open Neighborhood"
] | [
"Definition:Continuous Mapping (Topology)",
"Definition:Kernel of Linear Transformation",
"Definition:Closed Set/Topology",
"Definition:Closed Set/Topology",
"Continuity Defined from Closed Sets",
"Definition:Closed Set/Topology"
] |
proofwiki-19832 | Sum of Closures is Subset of Closure of Sum in Topological Vector Space | Let $K$ be a topological field.
Let $X$ be a topological vector space over $K$.
Let $A, B \subseteq X$.
Then:
:$A^- + B^- \subseteq \paren {A + B}^-$
where $A^-$, $B^-$ and $\paren {A + B}^-$ denote the closures of $A$, $B$ and $A + B$. | Let $x \in A^-$ and $y \in B^-$.
From Point in Set Closure iff Limit of Net, there exists directed sets $\struct {\Lambda_1, \preceq_1}$ and $\struct {\Lambda_2, \preceq_2}$ and nets $\sequence {x_\lambda}_{\lambda \mathop \in \Lambda_1}$ valued in $A$ and $\sequence {y_\lambda}_{\lambda \mathop \in \Lambda_2}$ valued ... | Let $K$ be a [[Definition:Topological Field|topological field]].
Let $X$ be a [[Definition:Topological Vector Space|topological vector space]] over $K$.
Let $A, B \subseteq X$.
Then:
:$A^- + B^- \subseteq \paren {A + B}^-$
where $A^-$, $B^-$ and $\paren {A + B}^-$ denote the [[Definition:Topological Closure|cl... | Let $x \in A^-$ and $y \in B^-$.
From [[Point in Set Closure iff Limit of Net]], there exists [[Definition:Directed Set|directed sets]] $\struct {\Lambda_1, \preceq_1}$ and $\struct {\Lambda_2, \preceq_2}$ and [[Definition:Net (Set Theory)|nets]] $\sequence {x_\lambda}_{\lambda \mathop \in \Lambda_1}$ valued in $A$ an... | Sum of Closures is Subset of Closure of Sum in Topological Vector Space/Proof 2 | https://proofwiki.org/wiki/Sum_of_Closures_is_Subset_of_Closure_of_Sum_in_Topological_Vector_Space | https://proofwiki.org/wiki/Sum_of_Closures_is_Subset_of_Closure_of_Sum_in_Topological_Vector_Space/Proof_2 | [
"Topological Vector Spaces",
"Set Closures",
"Sum of Closures is Subset of Closure of Sum in Topological Vector Space"
] | [
"Definition:Topological Field",
"Definition:Topological Vector Space",
"Definition:Closure (Topology)"
] | [
"Point in Set Closure iff Limit of Net",
"Definition:Directed Preordering",
"Definition:Net (Set Theory)",
"Definition:Convergent Net",
"Definition:Convergent Net",
"Definition:Relation",
"Product of Directed Sets is Directed Set",
"Definition:Directed Preordering",
"Definition:Convergent Net",
"D... |
proofwiki-19833 | Vanishing Ideal of Larger Subset of Affine Space is Smaller | Let $k$ be a field.
Let $n \ge 1$ be a natural number.
Let $\mathbb A^n_k$ be the standard affine space over $k$.
Let $S \subseteq T \subseteq \mathbb A^n_k$.
Then:
:$\map I S \supseteq \map I T$
where $\map I S$ and $\map I T$ denote the vanishing ideals of $S$ and $T$, respectively. | {{begin-eqn}}
{{eqn | l = f
| o = \in
| r = \map I T
}}
{{eqn | ll= \leadsto
| q = \forall x \in T
| l = \map f x
| r = 0
}}
{{eqn | ll= \leadsto
| q = \forall x \in S
| l = \map f x
| r = 0
| c = since $S \subseteq T$ by assumption
}}
{{eqn | ll= \leadsto
| l... | Let $k$ be a [[Definition:Field (Abstract Algebra)|field]].
Let $n \ge 1$ be a [[Definition:Natural Number|natural number]].
Let $\mathbb A^n_k$ be the [[Definition:Standard Affine Space|standard affine space]] over $k$.
Let $S \subseteq T \subseteq \mathbb A^n_k$.
Then:
:$\map I S \supseteq \map I T$
where $\map ... | {{begin-eqn}}
{{eqn | l = f
| o = \in
| r = \map I T
}}
{{eqn | ll= \leadsto
| q = \forall x \in T
| l = \map f x
| r = 0
}}
{{eqn | ll= \leadsto
| q = \forall x \in S
| l = \map f x
| r = 0
| c = since $S \subseteq T$ by assumption
}}
{{eqn | ll= \leadsto
| l... | Vanishing Ideal of Larger Subset of Affine Space is Smaller | https://proofwiki.org/wiki/Vanishing_Ideal_of_Larger_Subset_of_Affine_Space_is_Smaller | https://proofwiki.org/wiki/Vanishing_Ideal_of_Larger_Subset_of_Affine_Space_is_Smaller | [
"Algebraic Geometry"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Natural Numbers",
"Definition:Standard Affine Space",
"Definition:Vanishing Ideal of Subset of Affine Space"
] | [
"Category:Algebraic Geometry"
] |
proofwiki-19834 | Equivalence of Definitions of Convex Set in Vector Space | Let $\Bbb F \in \set {\R, \C}$.
Let $V$ be a vector space over $\Bbb F$.
Let $C \subseteq V$.
{{TFAE|def = Convex Set (Vector Space)|view = convexity}} | From the definition of the linear combination $t C + \paren {1 - t} C$, we have:
:$t C + \paren {1 - t} C = \set {t x + \paren {1 - t} y : x, y \in C}$
for $t \in \R$.
So we have:
:$t C + \paren {1 - t} C \subseteq C$
{{iff}}:
:$t x + \paren {1 - t} y \in C$ for all $x, y \in C$ and $t \in \closedint 0 1$.
Hence the... | Let $\Bbb F \in \set {\R, \C}$.
Let $V$ be a [[Definition:Vector Space|vector space]] over $\Bbb F$.
Let $C \subseteq V$.
{{TFAE|def = Convex Set (Vector Space)|view = convexity}} | From the definition of the [[Definition:Linear Combination of Subsets of Vector Space|linear combination]] $t C + \paren {1 - t} C$, we have:
:$t C + \paren {1 - t} C = \set {t x + \paren {1 - t} y : x, y \in C}$
for $t \in \R$.
So we have:
:$t C + \paren {1 - t} C \subseteq C$
{{iff}}:
:$t x + \paren {1 - t} ... | Equivalence of Definitions of Convex Set in Vector Space | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Convex_Set_in_Vector_Space | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Convex_Set_in_Vector_Space | [
"Convex Sets (Vector Spaces)"
] | [
"Definition:Vector Space"
] | [
"Definition:Linear Combination of Subsets of Vector Space",
"Category:Convex Sets (Vector Spaces)"
] |
proofwiki-19835 | Interior of Convex Set in Topological Vector Space is Convex | Let $\Bbb F \in \set {\R, \C}$.
Let $X$ be a topological vector space over $\Bbb F$.
Let $C \subseteq X$ be a convex set.
Then the interior of $C$, $C^\circ$, is convex. | Let $t \in \closedint 0 1$.
Since $C$ is convex, we have:
:$t C + \paren {1 - t} C \subseteq C$
Since $C^\circ \subseteq C$, we have:
{{begin-eqn}}
{{eqn | l = t C + \paren {1 - t} C
| r = \set {t x + \paren {1 - t} y : x, y \in C}
| c = {{Defof|Linear Combination of Subsets of Vector Space}}
}}
{{eqn | o = \sup... | Let $\Bbb F \in \set {\R, \C}$.
Let $X$ be a [[Definition:Topological Vector Space|topological vector space]] over $\Bbb F$.
Let $C \subseteq X$ be a [[Definition:Convex Set (Vector Space)|convex set]].
Then the [[Definition:Interior (Topology)|interior]] of $C$, $C^\circ$, is [[Definition:Convex Set (Vector Space)... | Let $t \in \closedint 0 1$.
Since $C$ is [[Definition:Convex Set (Vector Space)|convex]], we have:
:$t C + \paren {1 - t} C \subseteq C$
Since $C^\circ \subseteq C$, we have:
{{begin-eqn}}
{{eqn | l = t C + \paren {1 - t} C
| r = \set {t x + \paren {1 - t} y : x, y \in C}
| c = {{Defof|Linear Combination of ... | Interior of Convex Set in Topological Vector Space is Convex | https://proofwiki.org/wiki/Interior_of_Convex_Set_in_Topological_Vector_Space_is_Convex | https://proofwiki.org/wiki/Interior_of_Convex_Set_in_Topological_Vector_Space_is_Convex | [
"Set Interiors",
"Topological Vector Spaces",
"Convex Sets (Vector Spaces)"
] | [
"Definition:Topological Vector Space",
"Definition:Convex Set (Vector Space)",
"Definition:Interior (Topology)",
"Definition:Convex Set (Vector Space)"
] | [
"Definition:Convex Set (Vector Space)",
"Equivalence of Definitions of Interior (Topology)",
"Definition:Open Set/Topology",
"Dilation of Open Set in Topological Vector Space is Open",
"Definition:Open Set/Topology",
"Sum of Set and Open Set in Topological Vector Space is Open",
"Definition:Open Set/Top... |
proofwiki-19836 | Dilation of Closed Set in Topological Vector Space is Closed Set | Let $K$ be a topological field.
Let $X$ be a topological vector space over $K$.
Let $F$ be a closed set in $X$.
Let $\lambda \in K \setminus \set {0_K}$.
Then $\lambda F$ is a closed set in $X$. | We aim to show that $X \setminus \paren {\lambda F}$ is open.
Since $F$ is closed, $X \setminus F$ is open.
It follows from Dilation of Open Set in Topological Vector Space is Open that $\lambda \paren {X \setminus F}$ is open.
From Dilation of Complement of Set in Vector Space, we have:
:$X \setminus \paren {\lambda ... | Let $K$ be a [[Definition:Topological Field|topological field]].
Let $X$ be a [[Definition:Topological Vector Space|topological vector space]] over $K$.
Let $F$ be a [[Definition:Closed Set (Topology)|closed set]] in $X$.
Let $\lambda \in K \setminus \set {0_K}$.
Then $\lambda F$ is a [[Definition:Closed Set (To... | We aim to show that $X \setminus \paren {\lambda F}$ is [[Definition:Open Set (Topology)|open]].
Since $F$ is [[Definition:Closed Set (Topology)|closed]], $X \setminus F$ is [[Definition:Open Set (Topology)|open]].
It follows from [[Dilation of Open Set in Topological Vector Space is Open]] that $\lambda \paren {X \s... | Dilation of Closed Set in Topological Vector Space is Closed Set/Proof 1 | https://proofwiki.org/wiki/Dilation_of_Closed_Set_in_Topological_Vector_Space_is_Closed_Set | https://proofwiki.org/wiki/Dilation_of_Closed_Set_in_Topological_Vector_Space_is_Closed_Set/Proof_1 | [
"Dilation of Closed Set in Topological Vector Space is Closed Set",
"Dilations of Subsets of Vector Spaces",
"Topological Vector Spaces"
] | [
"Definition:Topological Field",
"Definition:Topological Vector Space",
"Definition:Closed Set/Topology",
"Definition:Closed Set/Topology"
] | [
"Definition:Open Set/Topology",
"Definition:Closed Set/Topology",
"Definition:Open Set/Topology",
"Dilation of Open Set in Topological Vector Space is Open",
"Definition:Open Set/Topology",
"Dilation of Complement of Set in Vector Space",
"Definition:Open Set/Topology",
"Definition:Open Set/Topology",... |
proofwiki-19837 | Dilation of Closure of Set in Topological Vector Space is Closure of Dilation | Let $F$ be a topological field.
Let $X$ be a topological vector space over $F$.
Let $A \subseteq X$.
Let $\lambda \in F \setminus \set {0_F}$.
Then we have:
:$\lambda A^- = \paren {\lambda A}^-$
where $A^-$ denotes the closure of $A$. | Let $x \in \lambda A^-$.
Then $\lambda^{-1} x \in A^-$.
From Point in Set Closure iff Limit of Net, there exists a directed set $\struct {\Lambda, \preceq}$ and a net $\family {y_\mu}_{\mu \mathop \in \Lambda}$ valued in $A$ converging to $\lambda^{-1} x$.
From Scalar Multiple of Convergent Net in Topological Vector Sp... | Let $F$ be a [[Definition:Topological Field|topological field]].
Let $X$ be a [[Definition:Topological Vector Space|topological vector space]] over $F$.
Let $A \subseteq X$.
Let $\lambda \in F \setminus \set {0_F}$.
Then we have:
:$\lambda A^- = \paren {\lambda A}^-$
where $A^-$ denotes the [[Definition:Topolo... | Let $x \in \lambda A^-$.
Then $\lambda^{-1} x \in A^-$.
From [[Point in Set Closure iff Limit of Net]], there exists a [[Definition:Directed Set|directed set]] $\struct {\Lambda, \preceq}$ and a [[Definition:Net (Set Theory)|net]] $\family {y_\mu}_{\mu \mathop \in \Lambda}$ valued in $A$ [[Definition:Convergent Net|c... | Dilation of Closure of Set in Topological Vector Space is Closure of Dilation/Proof 2 | https://proofwiki.org/wiki/Dilation_of_Closure_of_Set_in_Topological_Vector_Space_is_Closure_of_Dilation | https://proofwiki.org/wiki/Dilation_of_Closure_of_Set_in_Topological_Vector_Space_is_Closure_of_Dilation/Proof_2 | [
"Dilation of Closure of Set in Topological Vector Space is Closure of Dilation",
"Dilations of Subsets of Vector Spaces",
"Topological Vector Spaces",
"Set Closures"
] | [
"Definition:Topological Field",
"Definition:Topological Vector Space",
"Definition:Closure (Topology)"
] | [
"Point in Set Closure iff Limit of Net",
"Definition:Directed Preordering",
"Definition:Net (Set Theory)",
"Definition:Convergent Net",
"Scalar Multiple of Convergent Net in Topological Vector Space is Convergent",
"Definition:Convergent Net",
"Point in Set Closure iff Limit of Net",
"Point in Set Clo... |
proofwiki-19838 | Closure of Convex Set in Topological Vector Space is Convex | Let $\Bbb F \in \set {\R, \C}$.
Let $X$ be a topological vector space over $\Bbb F$.
Let $C \subseteq X$ be convex.
Then the closure $C^-$ of $C$ is convex. | Let $t \in \closedint 0 1$.
Since $C$ is convex, we have:
:$t C + \paren {1 - t} C \subseteq C$
We show that:
:$t C^- + \paren {1 - t} C^- \subseteq C^-$
We have:
{{begin-eqn}}
{{eqn | l = t C^- + \paren {1 - t} C^-
| r = \paren {t C}^- + \paren {\paren {1 - t} C}^-
| c = Dilation of Closure of Set in Topologic... | Let $\Bbb F \in \set {\R, \C}$.
Let $X$ be a [[Definition:Topological Vector Space|topological vector space]] over $\Bbb F$.
Let $C \subseteq X$ be [[Definition:Convex Set (Vector Space)|convex]].
Then the [[Definition:Topological Closure|closure]] $C^-$ of $C$ is [[Definition:Convex Set (Vector Space)|convex]]. | Let $t \in \closedint 0 1$.
Since $C$ is [[Definition:Convex Set (Vector Space)|convex]], we have:
:$t C + \paren {1 - t} C \subseteq C$
We show that:
:$t C^- + \paren {1 - t} C^- \subseteq C^-$
We have:
{{begin-eqn}}
{{eqn | l = t C^- + \paren {1 - t} C^-
| r = \paren {t C}^- + \paren {\paren {1 - t} C}^-
... | Closure of Convex Set in Topological Vector Space is Convex | https://proofwiki.org/wiki/Closure_of_Convex_Set_in_Topological_Vector_Space_is_Convex | https://proofwiki.org/wiki/Closure_of_Convex_Set_in_Topological_Vector_Space_is_Convex | [
"Convex Sets (Vector Spaces)",
"Topological Vector Spaces",
"Set Closures"
] | [
"Definition:Topological Vector Space",
"Definition:Convex Set (Vector Space)",
"Definition:Closure (Topology)",
"Definition:Convex Set (Vector Space)"
] | [
"Definition:Convex Set (Vector Space)",
"Dilation of Closure of Set in Topological Vector Space is Closure of Dilation",
"Sum of Closures is Subset of Closure of Sum in Topological Vector Space",
"Topological Closure of Subset is Subset of Topological Closure"
] |
proofwiki-19839 | Closure of Linear Subspace of Topological Vector Space is Linear Subspace | Let $K$ be a topological field.
Let $X$ be a topological vector space over $K$.
Let $Y$ be a linear subspace of $X$.
Then the closure $Y^-$ of $Y$ is a linear subspace of $X$. | Since $Y \subseteq Y^-$, we have that $Y^-$ is non-empty.
We now use the One-Step Vector Subspace Test.
We show that for each $\lambda \in K$ and $u, v \in Y^-$ we have:
:$u + \lambda v \in Y^-$
That is, we want to show that:
:$Y^- + \lambda Y^- \subseteq Y^-$
for each $\lambda \in K$.
Let $\lambda \in K$.
Then:
{{... | Let $K$ be a [[Definition:Topological Field|topological field]].
Let $X$ be a [[Definition:Topological Vector Space|topological vector space]] over $K$.
Let $Y$ be a [[Definition:Linear Subspace|linear subspace]] of $X$.
Then the [[Definition:Topological Closure|closure]] $Y^-$ of $Y$ is a [[Definition:Linear Subsp... | Since $Y \subseteq Y^-$, we have that $Y^-$ is [[Definition:Non-Empty Set|non-empty]].
We now use the [[One-Step Vector Subspace Test]].
We show that for each $\lambda \in K$ and $u, v \in Y^-$ we have:
:$u + \lambda v \in Y^-$
That is, we want to show that:
:$Y^- + \lambda Y^- \subseteq Y^-$
for each $\lambda ... | Closure of Linear Subspace of Topological Vector Space is Linear Subspace | https://proofwiki.org/wiki/Closure_of_Linear_Subspace_of_Topological_Vector_Space_is_Linear_Subspace | https://proofwiki.org/wiki/Closure_of_Linear_Subspace_of_Topological_Vector_Space_is_Linear_Subspace | [
"Topological Vector Spaces",
"Set Closures"
] | [
"Definition:Topological Field",
"Definition:Topological Vector Space",
"Definition:Linear Subspace",
"Definition:Closure (Topology)",
"Definition:Linear Subspace"
] | [
"Definition:Non-Empty Set",
"One-Step Vector Subspace Test",
"Dilation of Closure of Set in Topological Vector Space is Closure of Dilation",
"Sum of Closures is Subset of Closure of Sum in Topological Vector Space",
"Definition:Linear Subspace",
"Topological Closure of Subset is Subset of Topological Clo... |
proofwiki-19840 | Closure of Balanced Set in Topological Vector Space is Balanced | Let $\Bbb F \in \set {\R, \C}$.
Let $X$ be a topological vector space over $\Bbb F$.
Let $B \subseteq X$ be a balanced set.
Then the closure $B^-$ of $B$ is balanced. | Let $\lambda \in \Bbb F$ have $\cmod \lambda \le 1$.
Then, we have:
:$\lambda B \subseteq B$
So, from Topological Closure of Subset is Subset of Topological Closure we have:
:$\paren {\lambda B}^- \subseteq B^-$
From Dilation of Closure of Set in Topological Vector Space is Closure of Dilation we have $\paren {\lam... | Let $\Bbb F \in \set {\R, \C}$.
Let $X$ be a [[Definition:Topological Vector Space|topological vector space]] over $\Bbb F$.
Let $B \subseteq X$ be a [[Definition:Balanced Set|balanced set]].
Then the [[Definition:Topological Closure|closure]] $B^-$ of $B$ is [[Definition:Balanced Set|balanced]]. | Let $\lambda \in \Bbb F$ have $\cmod \lambda \le 1$.
Then, we have:
:$\lambda B \subseteq B$
So, from [[Topological Closure of Subset is Subset of Topological Closure]] we have:
:$\paren {\lambda B}^- \subseteq B^-$
From [[Dilation of Closure of Set in Topological Vector Space is Closure of Dilation]] we have ... | Closure of Balanced Set in Topological Vector Space is Balanced | https://proofwiki.org/wiki/Closure_of_Balanced_Set_in_Topological_Vector_Space_is_Balanced | https://proofwiki.org/wiki/Closure_of_Balanced_Set_in_Topological_Vector_Space_is_Balanced | [
"Set Closures",
"Balanced Sets",
"Topological Vector Spaces"
] | [
"Definition:Topological Vector Space",
"Definition:Balanced Set",
"Definition:Closure (Topology)",
"Definition:Balanced Set"
] | [
"Topological Closure of Subset is Subset of Topological Closure",
"Dilation of Closure of Set in Topological Vector Space is Closure of Dilation",
"Definition:Balanced Set"
] |
proofwiki-19841 | Vanishing Ideal of Zero Locus of Ideal is Radical | Let $k$ be an algebraically closed field.
Let $n \ge 0$ be a natural number.
Let $k \sqbrk {X_1, \ldots, X_n}$ be the polynomial ring in $n$ variables over $k$.
Let $\mathfrak a \subseteq k \sqbrk {X_1, \ldots, X_n}$ be an ideal.
Then:
:$\map I {\map V {\mathfrak a} } = \map \Rad {\mathfrak a}$
where:
:$\map V \cdot$ d... | This is exactly Hilbert's Nullstellensatz.
{{qed}}
Category:Algebraic Geometry
9z1byf2yecuy18dmzg9bfp6okifwdxi | Let $k$ be an [[Definition:Algebraically Closed Field|algebraically closed field]].
Let $n \ge 0$ be a [[Definition:Natural Number|natural number]].
Let $k \sqbrk {X_1, \ldots, X_n}$ be the [[Definition:Polynomial Ring in Multiple Variables|polynomial ring]] in $n$ variables over $k$.
Let $\mathfrak a \subseteq k \s... | This is exactly [[Hilbert's Nullstellensatz]].
{{qed}}
[[Category:Algebraic Geometry]]
9z1byf2yecuy18dmzg9bfp6okifwdxi | Vanishing Ideal of Zero Locus of Ideal is Radical | https://proofwiki.org/wiki/Vanishing_Ideal_of_Zero_Locus_of_Ideal_is_Radical | https://proofwiki.org/wiki/Vanishing_Ideal_of_Zero_Locus_of_Ideal_is_Radical | [
"Algebraic Geometry"
] | [
"Definition:Algebraically Closed Field",
"Definition:Natural Numbers",
"Definition:Polynomial Ring",
"Definition:Ideal of Ring",
"Definition:Zero Locus of Set of Polynomials",
"Definition:Vanishing Ideal of Subset of Affine Space",
"Definition:Radical of Ideal of Ring"
] | [
"Hilbert's Nullstellensatz",
"Category:Algebraic Geometry"
] |
proofwiki-19842 | Zero Locus of Vanishing Ideal is Closure | Let $k$ be a field.
Let $n \ge 1$ be a natural number.
Let $k \sqbrk {x_1, \ldots, x_n}$ be the polynomial ring in $n$ variables over $k$.
Let $\mathbb A^n_k$ be the standard affine space over $k$.
Let $\mathbb A^n_k$ be equipped with the Zariski topology.
Let $S \subseteq \mathbb A^n_k$.
Then:
:$\map V {\map I S} = S^... | {{MissingLinks|still plenty}}
{{explain|Explanatory words needed to bridge some of the gaps still}}
Let $P := k \sqbrk {x_1, \ldots, x_n}$.
By definition of closure, the set $S^-$ is closed in the Zariski topology, therefore $S^-$ is an affine algebraic set.
By definition of an affine algebraic set, there exists some s... | Let $k$ be a [[Definition:Field (Abstract Algebra)|field]].
Let $n \ge 1$ be a [[Definition:Natural Number|natural number]].
Let $k \sqbrk {x_1, \ldots, x_n}$ be the [[Definition:Polynomial Ring in Multiple Variables|polynomial ring]] in $n$ variables over $k$.
Let $\mathbb A^n_k$ be the [[Definition:Standard Affine... | {{MissingLinks|still plenty}}
{{explain|Explanatory words needed to bridge some of the gaps still}}
Let $P := k \sqbrk {x_1, \ldots, x_n}$.
By definition of [[Definition:Closure (Topology)|closure]], the [[Definition:Set|set]] $S^-$ is [[Definition:Closed Set (Topology)|closed]] in the [[Definition:Zariski Topology ... | Zero Locus of Vanishing Ideal is Closure | https://proofwiki.org/wiki/Zero_Locus_of_Vanishing_Ideal_is_Closure | https://proofwiki.org/wiki/Zero_Locus_of_Vanishing_Ideal_is_Closure | [
"Algebraic Geometry"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Natural Numbers",
"Definition:Polynomial Ring",
"Definition:Standard Affine Space",
"Definition:Zariski Topology/Affine Space",
"Definition:Vanishing Ideal of Subset of Affine Space",
"Definition:Zero Locus of Set of Polynomials",
"Definition:Closure ... | [
"Definition:Closure (Topology)",
"Definition:Set",
"Definition:Closed Set/Topology",
"Definition:Zariski Topology/Affine Space",
"Definition:Affine Algebraic Set",
"Definition:Affine Algebraic Set",
"Definition:Set",
"Definition:Generated Ideal of Ring",
"Zero Locus of Set is Zero Locus of Generated... |
proofwiki-19843 | Existence of Disjoint Well-Ordered Sets Isomorphic to Ordinals | Let $\alpha$ and $\beta$ be ordinals.
Then there exist two well-ordered sets $a$ and $b$ such that:
:$a$ and $b$ are order isomorphic to $\alpha$ and $\beta$ respectively
:$a$ and $b$ are disjoint. | Let $a$ and $b$ be defined as:
{{begin-eqn}}
{{eqn | l = a
| o = :=
| r = \alpha \times \set 0
}}
{{eqn | l = b
| o = :=
| r = \beta \times \set 1
}}
{{end-eqn}}
Then let $a$ and $b$ be ordered by their first coordinate.
The result follows.
{{qed}} | Let $\alpha$ and $\beta$ be [[Definition:Ordinal|ordinals]].
Then there exist two [[Definition:Well-Ordered Set|well-ordered sets]] $a$ and $b$ such that:
:$a$ and $b$ are [[Definition:Order Isomorphism|order isomorphic]] to $\alpha$ and $\beta$ respectively
:$a$ and $b$ are [[Definition:Disjoint Sets|disjoint]]. | Let $a$ and $b$ be defined as:
{{begin-eqn}}
{{eqn | l = a
| o = :=
| r = \alpha \times \set 0
}}
{{eqn | l = b
| o = :=
| r = \beta \times \set 1
}}
{{end-eqn}}
Then let $a$ and $b$ be [[Definition:Ordering|ordered]] by their [[Definition:Coordinate of Ordered Tuple|first coordinate]].
The r... | Existence of Disjoint Well-Ordered Sets Isomorphic to Ordinals | https://proofwiki.org/wiki/Existence_of_Disjoint_Well-Ordered_Sets_Isomorphic_to_Ordinals | https://proofwiki.org/wiki/Existence_of_Disjoint_Well-Ordered_Sets_Isomorphic_to_Ordinals | [
"Ordinals",
"Well-Orderings"
] | [
"Definition:Ordinal",
"Definition:Well-Ordered Set",
"Definition:Order Isomorphism",
"Definition:Disjoint Sets"
] | [
"Definition:Ordering",
"Definition:Cartesian Product/Coordinate"
] |
proofwiki-19844 | Dilation of Interior of Set in Topological Vector Space is Interior of Dilation | Let $K$ be a topological field.
Let $X$ be a topological vector space over $K$.
Let $A \subseteq X$.
Let $\lambda \in K \setminus \set 0$.
Then we have:
:$\lambda A^\circ = \paren {\lambda A}^\circ$
where $A^\circ$ denotes the interior of $A$. | From the definition of interior we have:
:$\ds A^\circ = \bigcup \leftset {U \subseteq A: U}$ is open in $\rightset X$
and:
:$\ds \paren {\lambda A}^\circ = \bigcup \leftset {U \subseteq \lambda A: U}$ is open in $\rightset X$
For brevity write:
:$\ds \SS_1 = \leftset {U \subseteq A: U}$ is open in $\rightset X$
and:
... | Let $K$ be a [[Definition:Topological Field|topological field]].
Let $X$ be a [[Definition:Topological Vector Space|topological vector space]] over $K$.
Let $A \subseteq X$.
Let $\lambda \in K \setminus \set 0$.
Then we have:
:$\lambda A^\circ = \paren {\lambda A}^\circ$
where $A^\circ$ denotes the [[Definitio... | From the definition of [[Definition:Interior (Topology)|interior]] we have:
:$\ds A^\circ = \bigcup \leftset {U \subseteq A: U}$ is [[Definition:Open Set (Topology)|open]] in $\rightset X$
and:
:$\ds \paren {\lambda A}^\circ = \bigcup \leftset {U \subseteq \lambda A: U}$ is [[Definition:Open Set (Topology)|open]] in... | Dilation of Interior of Set in Topological Vector Space is Interior of Dilation | https://proofwiki.org/wiki/Dilation_of_Interior_of_Set_in_Topological_Vector_Space_is_Interior_of_Dilation | https://proofwiki.org/wiki/Dilation_of_Interior_of_Set_in_Topological_Vector_Space_is_Interior_of_Dilation | [
"Dilations of Subsets of Vector Spaces",
"Set Interiors",
"Topological Vector Spaces"
] | [
"Definition:Topological Field",
"Definition:Topological Vector Space",
"Definition:Interior (Topology)"
] | [
"Definition:Interior (Topology)",
"Definition:Open Set/Topology",
"Definition:Open Set/Topology",
"Definition:Open Set/Topology",
"Definition:Open Set/Topology",
"Dilation of Union of Subsets of Vector Space",
"Definition:Open Set/Topology",
"Dilation of Open Set in Topological Vector Space is Open",
... |
proofwiki-19845 | Interior of Balanced Set containing Origin in Topological Vector Space is Balanced | Let $\Bbb F \in \set {\R, \C}$.
Let $X$ be a topological vector space over $\Bbb F$.
Let $B \subseteq X$ be a balanced set such that:
:${\mathbf 0}_X \in B^\circ$
where $B^\circ$ is the interior of $B$.
Then $B^\circ$ is balanced. | Let $\lambda \in \Bbb F$ have $0 < \cmod \lambda \le 1$.
Then, we have:
:$\lambda B \subseteq B$
From Interior of Subset, we have:
:$\paren {\lambda B}^\circ \subseteq B^\circ$
Then from Dilation of Interior of Set in Topological Vector Space is Interior of Dilation we have:
:$\lambda B^\circ \subseteq B^\circ$
fo... | Let $\Bbb F \in \set {\R, \C}$.
Let $X$ be a [[Definition:Topological Vector Space|topological vector space]] over $\Bbb F$.
Let $B \subseteq X$ be a [[Definition:Balanced Set|balanced set]] such that:
:${\mathbf 0}_X \in B^\circ$
where $B^\circ$ is the [[Definition:Interior (Topology)|interior]] of $B$.
Then $B... | Let $\lambda \in \Bbb F$ have $0 < \cmod \lambda \le 1$.
Then, we have:
:$\lambda B \subseteq B$
From [[Interior of Subset]], we have:
:$\paren {\lambda B}^\circ \subseteq B^\circ$
Then from [[Dilation of Interior of Set in Topological Vector Space is Interior of Dilation]] we have:
:$\lambda B^\circ \subset... | Interior of Balanced Set containing Origin in Topological Vector Space is Balanced | https://proofwiki.org/wiki/Interior_of_Balanced_Set_containing_Origin_in_Topological_Vector_Space_is_Balanced | https://proofwiki.org/wiki/Interior_of_Balanced_Set_containing_Origin_in_Topological_Vector_Space_is_Balanced | [
"Balanced Sets",
"Set Interiors",
"Topological Vector Spaces"
] | [
"Definition:Topological Vector Space",
"Definition:Balanced Set",
"Definition:Interior (Topology)",
"Definition:Balanced Set"
] | [
"Interior of Subset",
"Dilation of Interior of Set in Topological Vector Space is Interior of Dilation",
"Definition:Balanced Set"
] |
proofwiki-19846 | Disjoint Compact Set and Closed Set in Topological Vector Space separated by Open Neighborhood | Let $F$ be a topological field.
Let $X$ be a topological vector space over $F$.
Let $K$ be a compact subspace of $X$.
Let $C \subseteq X$ be a closed set such that:
:$K \cap C = \O$
Then there exists an open neighborhood $V$ of ${\mathbf 0}_X$ such that:
:$\paren {K + V} \cap \paren {C + V} = \O$ | If $K = \O$, we have $K + V = \O$, so we have:
:$\paren {K + V} \cap \paren {C + V} = \O$
from Intersection with Empty Set.
Now take $K \ne \O$.
Let $x \in K$.
Since $K$ and $C$ are disjoint, it follows that $x \in X \setminus C$, so ${\mathbf 0}_X \in \paren {X \setminus C} - x$.
Since $C$ is closed, $X \setminus C... | Let $F$ be a [[Definition:Topological Field|topological field]].
Let $X$ be a [[Definition:Topological Vector Space|topological vector space]] over $F$.
Let $K$ be a [[Definition:Compact Subspace|compact subspace]] of $X$.
Let $C \subseteq X$ be a [[Definition:Closed Set (Topology)|closed set]] such that:
:$K \cap... | If $K = \O$, we have $K + V = \O$, so we have:
:$\paren {K + V} \cap \paren {C + V} = \O$
from [[Intersection with Empty Set]].
Now take $K \ne \O$.
Let $x \in K$.
Since $K$ and $C$ are [[Definition:Disjoint Sets|disjoint]], it follows that $x \in X \setminus C$, so ${\mathbf 0}_X \in \paren {X \setminus C} - x$... | Disjoint Compact Set and Closed Set in Topological Vector Space separated by Open Neighborhood | https://proofwiki.org/wiki/Disjoint_Compact_Set_and_Closed_Set_in_Topological_Vector_Space_separated_by_Open_Neighborhood | https://proofwiki.org/wiki/Disjoint_Compact_Set_and_Closed_Set_in_Topological_Vector_Space_separated_by_Open_Neighborhood | [
"Topological Vector Spaces",
"Disjoint Compact Set and Closed Set in Topological Vector Space separated by Open Neighborhood"
] | [
"Definition:Topological Field",
"Definition:Topological Vector Space",
"Definition:Compact Topological Space/Subspace",
"Definition:Closed Set/Topology",
"Definition:Open Neighborhood"
] | [
"Intersection with Empty Set",
"Definition:Disjoint Sets",
"Definition:Closed Set/Topology",
"Definition:Open Set/Topology",
"Translation of Open Set in Topological Vector Space is Open",
"Definition:Open Neighborhood",
"Definition:Symmetric Set/Vector Space",
"Definition:Open Neighborhood",
"Defini... |
proofwiki-19847 | Open Neighborhood of Point in Topological Vector Space contains Sum of Open Neighborhoods | Let $K$ be a topological field.
Let $X$ be a topological vector space over $K$.
Let $a, b \in X$.
Let $W$ be an open neighborhood of $a + b$.
Then there exists an open neighborhood $W_1$ of $a$ and an open neighborhood $W_2$ of $b$ such that:
:$W_1 + W_2 \subseteq W$
where $W_1 + W_2$ denotes the sum of $W_1$ and $W_... | Equip $X \times X$ with the product topology.
From Box Topology on Finite Product Space is Product Topology, this is precisely the box topology.
Define $S : X \times X \to X$ by:
:$\map S {x, y} = x + y$
for each $\tuple {x, y} \in X \times X$.
From the definition of a topological vector space, $S$ is continuous.
In ... | Let $K$ be a [[Definition:Topological Field|topological field]].
Let $X$ be a [[Definition:Topological Vector Space|topological vector space]] over $K$.
Let $a, b \in X$.
Let $W$ be an [[Definition:Open Neighborhood|open neighborhood]] of $a + b$.
Then there exists an [[Definition:Open Neighborhood|open neighborh... | Equip $X \times X$ with the [[Definition:Product Topology|product topology]].
From [[Box Topology on Finite Product Space is Product Topology]], this is precisely the [[Definition:Box Topology|box topology]].
Define $S : X \times X \to X$ by:
:$\map S {x, y} = x + y$
for each $\tuple {x, y} \in X \times X$.
From... | Open Neighborhood of Point in Topological Vector Space contains Sum of Open Neighborhoods | https://proofwiki.org/wiki/Open_Neighborhood_of_Point_in_Topological_Vector_Space_contains_Sum_of_Open_Neighborhoods | https://proofwiki.org/wiki/Open_Neighborhood_of_Point_in_Topological_Vector_Space_contains_Sum_of_Open_Neighborhoods | [
"Topological Vector Spaces",
"Open Neighborhood of Point in Topological Vector Space contains Sum of Open Neighborhoods"
] | [
"Definition:Topological Field",
"Definition:Topological Vector Space",
"Definition:Open Neighborhood",
"Definition:Open Neighborhood",
"Definition:Open Neighborhood",
"Definition:Linear Combination of Subsets of Vector Space"
] | [
"Definition:Product Topology",
"Box Topology on Finite Product Space is Product Topology",
"Definition:Box Topology",
"Definition:Topological Vector Space",
"Definition:Continuous Mapping (Topology)",
"Definition:Continuous Mapping (Topology)/Point",
"Definition:Open Neighborhood",
"Basis for Box Topo... |
proofwiki-19848 | Open Neighborhood of Point in Topological Vector Space contains Sum of Open Neighborhoods/Corollary 1 | Let $W$ be an open neighborhood of ${\mathbf 0}_X$.
Then there exists a symmetric open neighborhood $U$ of ${\mathbf 0}_X$ such that:
:$U + U \subseteq W$ | Since ${\mathbf 0}_X + {\mathbf 0}_X = {\mathbf 0}_X$, we can apply Open Neighborhood of Point in Topological Vector Space contains Sum of Open Neighborhoods to obtain an open neighborhood $V_1$ of ${\mathbf 0}_X$ and an open neighborhood $V_2$ of ${\mathbf 0}_X$ such that:
:$V_1 + V_2 \subseteq W$
Note at this point ... | Let $W$ be an [[Definition:Open Neighborhood|open neighborhood]] of ${\mathbf 0}_X$.
Then there exists a [[Definition:Symmetric Subset of Vector Space|symmetric]] [[Definition:Open Neighborhood|open neighborhood]] $U$ of ${\mathbf 0}_X$ such that:
:$U + U \subseteq W$ | Since ${\mathbf 0}_X + {\mathbf 0}_X = {\mathbf 0}_X$, we can apply [[Open Neighborhood of Point in Topological Vector Space contains Sum of Open Neighborhoods]] to obtain an [[Definition:Open Neighborhood|open neighborhood]] $V_1$ of ${\mathbf 0}_X$ and an [[Definition:Open Neighborhood|open neighborhood]] $V_2$ of ${... | Open Neighborhood of Point in Topological Vector Space contains Sum of Open Neighborhoods/Corollary 1 | https://proofwiki.org/wiki/Open_Neighborhood_of_Point_in_Topological_Vector_Space_contains_Sum_of_Open_Neighborhoods/Corollary_1 | https://proofwiki.org/wiki/Open_Neighborhood_of_Point_in_Topological_Vector_Space_contains_Sum_of_Open_Neighborhoods/Corollary_1 | [
"Open Neighborhood of Point in Topological Vector Space contains Sum of Open Neighborhoods"
] | [
"Definition:Open Neighborhood",
"Definition:Symmetric Set/Vector Space",
"Definition:Open Neighborhood"
] | [
"Open Neighborhood of Point in Topological Vector Space contains Sum of Open Neighborhoods",
"Definition:Open Neighborhood",
"Definition:Open Neighborhood",
"Dilation of Open Set in Topological Vector Space is Open",
"Definition:Open Neighborhood",
"Definition:Set Intersection/Finite Intersection",
"Def... |
proofwiki-19849 | Sum of Union of Subsets of Vector Space and Subset | Let $K$ be a field.
Let $X$ be a vector space over $K$.
Let $\sequence {E_\alpha}_{\alpha \mathop \in I}$ be a $I$-indexed family of subsets of $X$.
Let $C \subseteq X$.
Then:
:$\ds C + \bigcup_{\alpha \mathop \in I} E_\alpha = \bigcup_{\alpha \mathop \in I} \paren {C + E_\alpha}$ | We have:
:$\ds v \in \bigcup_{\alpha \mathop \in I} \paren {C + E_\alpha}$
{{iff}} there exists $c \in C$, $\alpha \in I$, $x \in E_\alpha$ such that:
:$v = c + x$
That is, {{iff}} $v = c + x$ for $c \in C$ and:
:$\ds x \in \bigcup_{\alpha \mathop \in I} E_\alpha$
This is equivalent to:
:$\ds v \in C + \bigcup_{\al... | Let $K$ be a [[Definition:Field (Abstract Algebra)|field]].
Let $X$ be a [[Definition:Vector Space|vector space]] over $K$.
Let $\sequence {E_\alpha}_{\alpha \mathop \in I}$ be a [[Definition:Indexed Family|$I$-indexed family]] of [[Definition:Subset|subsets]] of $X$.
Let $C \subseteq X$.
Then:
:$\ds C + \bigc... | We have:
:$\ds v \in \bigcup_{\alpha \mathop \in I} \paren {C + E_\alpha}$
{{iff}} there exists $c \in C$, $\alpha \in I$, $x \in E_\alpha$ such that:
:$v = c + x$
That is, {{iff}} $v = c + x$ for $c \in C$ and:
:$\ds x \in \bigcup_{\alpha \mathop \in I} E_\alpha$
This is equivalent to:
:$\ds v \in C + \bigc... | Sum of Union of Subsets of Vector Space and Subset | https://proofwiki.org/wiki/Sum_of_Union_of_Subsets_of_Vector_Space_and_Subset | https://proofwiki.org/wiki/Sum_of_Union_of_Subsets_of_Vector_Space_and_Subset | [
"Vector Spaces",
"Set Union"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Vector Space",
"Definition:Indexing Set/Family",
"Definition:Subset"
] | [
"Definition:Set Equality",
"Category:Vector Spaces",
"Category:Set Union"
] |
proofwiki-19850 | Closure of von Neumann-Bounded Subset of Topological Vector Space is von Neumann-Bounded | Let $\Bbb F \in \set {\R, \C}$.
Let $X$ be a topological vector space over $\Bbb F$.
Let $E$ be a von Neumann-bounded subset of $X$.
Then the closure $E^-$ of $E$ is von Neumann-bounded. | Let $V$ be an open neighborhood of ${\mathbf 0}_X$.
From Disjoint Compact Set and Closed Set in Topological Vector Space separated by Open Neighborhood: Corollary, there exists an open neighborhood $W$ of ${\mathbf 0}_X$ such that:
:$W^- \subseteq V$
Since $E$ is von Neumann-bounded, there exists $s > 0$ such that:
:... | Let $\Bbb F \in \set {\R, \C}$.
Let $X$ be a [[Definition:Topological Vector Space|topological vector space]] over $\Bbb F$.
Let $E$ be a [[Definition:Von Neumann-Bounded Subset of Topological Vector Space|von Neumann-bounded]] [[Definition:Subset|subset]] of $X$.
Then the [[Definition:Topological Closure|closure]]... | Let $V$ be an [[Definition:Open Neighborhood|open neighborhood]] of ${\mathbf 0}_X$.
From [[Disjoint Compact Set and Closed Set in Topological Vector Space separated by Open Neighborhood/Corollary|Disjoint Compact Set and Closed Set in Topological Vector Space separated by Open Neighborhood: Corollary]], there exists ... | Closure of von Neumann-Bounded Subset of Topological Vector Space is von Neumann-Bounded | https://proofwiki.org/wiki/Closure_of_von_Neumann-Bounded_Subset_of_Topological_Vector_Space_is_von_Neumann-Bounded | https://proofwiki.org/wiki/Closure_of_von_Neumann-Bounded_Subset_of_Topological_Vector_Space_is_von_Neumann-Bounded | [
"Set Closures",
"Von Neumann-Bounded Subsets of Topological Vector Spaces"
] | [
"Definition:Topological Vector Space",
"Definition:Von Neumann-Bounded Subset of Topological Vector Space",
"Definition:Subset",
"Definition:Closure (Topology)",
"Definition:Von Neumann-Bounded Subset of Topological Vector Space"
] | [
"Definition:Open Neighborhood",
"Disjoint Compact Set and Closed Set in Topological Vector Space separated by Open Neighborhood/Corollary",
"Definition:Open Neighborhood",
"Definition:Von Neumann-Bounded Subset of Topological Vector Space",
"Topological Closure of Subset is Subset of Topological Closure",
... |
proofwiki-19851 | Convex Open Neighborhood of Origin in Topological Vector Space contains Balanced Convex Open Neighborhood | Let $\Bbb F \in \set {\R, \C}$.
Let $X$ be a topological vector space over $\Bbb F$.
Let $U$ be a convex open neighborhood of ${\mathbf 0}_X$, where ${\mathbf 0}_X$ is the zero vector of $X$.
Then there exists a balanced convex open neighborhood $V$ of ${\mathbf 0}_X$ with $V \subseteq U$. | Let:
:$\ds A = \bigcap_{\cmod \alpha = 1} \alpha U$
From Open Neighborhood of Origin in Topological Vector Space contains Balanced Open Neighborhood, there exists an balanced open neighborhood $W$ of ${\mathbf 0}_X$ with $W \subseteq U$.
Since $W$ is balanced, we have:
:$\alpha W \subseteq W$
for all $\alpha \in \Bbb... | Let $\Bbb F \in \set {\R, \C}$.
Let $X$ be a [[Definition:Topological Vector Space|topological vector space]] over $\Bbb F$.
Let $U$ be a [[Definition:Convex Set (Vector Space)|convex]] [[Definition:Open Neighborhood|open neighborhood]] of ${\mathbf 0}_X$, where ${\mathbf 0}_X$ is the [[Definition:Zero Vector|zero ... | Let:
:$\ds A = \bigcap_{\cmod \alpha = 1} \alpha U$
From [[Open Neighborhood of Origin in Topological Vector Space contains Balanced Open Neighborhood]], there exists an [[Definition:Balanced Set|balanced]] [[Definition:Open Neighborhood|open neighborhood]] $W$ of ${\mathbf 0}_X$ with $W \subseteq U$.
Since $W$ is ... | Convex Open Neighborhood of Origin in Topological Vector Space contains Balanced Convex Open Neighborhood | https://proofwiki.org/wiki/Convex_Open_Neighborhood_of_Origin_in_Topological_Vector_Space_contains_Balanced_Convex_Open_Neighborhood | https://proofwiki.org/wiki/Convex_Open_Neighborhood_of_Origin_in_Topological_Vector_Space_contains_Balanced_Convex_Open_Neighborhood | [
"Balanced Sets",
"Convex Sets (Vector Spaces)",
"Topological Vector Spaces"
] | [
"Definition:Topological Vector Space",
"Definition:Convex Set (Vector Space)",
"Definition:Open Neighborhood",
"Definition:Zero Vector",
"Definition:Balanced Set",
"Definition:Convex Set (Vector Space)",
"Definition:Open Neighborhood"
] | [
"Open Neighborhood of Origin in Topological Vector Space contains Balanced Open Neighborhood",
"Definition:Balanced Set",
"Definition:Open Neighborhood",
"Definition:Balanced Set",
"Set Intersection Preserves Subsets",
"Definition:Open Set/Topology",
"Definition:Interior (Topology)",
"Definition:Open ... |
proofwiki-19852 | Function is Convex iff Epigraph is Convex | Let $\GF \in \set {\R, \C}$.
Let $X$ be a vector space over $\GF$.
Let $C \subseteq X$ be a convex set.
Consider $X \times \GF$ as the direct product of $X$ and $\GF$.
Let $f : C \to \overline \R$ be a function.
Let $\map {\operatorname {epi} } f$ be the epigraph of $f$.
Then $f$ is convex {{iff}} $\map {\operatorname ... | === Necessary Condition ===
Suppose that $f$ is convex.
Let $t \in \openint 0 1$.
Let $\tuple {x_1, \alpha_1}, \tuple {x_2, \alpha_2} \in \map {\operatorname {epi} } f$.
Then:
:$\map f {x_1} \le \alpha_1 < \infty$
and:
:$\map f {x_2} \le \alpha_2 < \infty$
We therefore have, since $t > 0$:
:$t \map f {x_1} + \paren {... | Let $\GF \in \set {\R, \C}$.
Let $X$ be a [[Definition:Vector Space|vector space]] over $\GF$.
Let $C \subseteq X$ be a [[Definition:Convex Set (Vector Space)|convex set]].
Consider $X \times \GF$ as the [[Definition:Direct Product of Vector Spaces|direct product]] of $X$ and $\GF$.
Let $f : C \to \overline \R$ be ... | === Necessary Condition ===
Suppose that $f$ is [[Definition:Convex Real Function/Vector Space|convex]].
Let $t \in \openint 0 1$.
Let $\tuple {x_1, \alpha_1}, \tuple {x_2, \alpha_2} \in \map {\operatorname {epi} } f$.
Then:
:$\map f {x_1} \le \alpha_1 < \infty$
and:
:$\map f {x_2} \le \alpha_2 < \infty$
We ther... | Function is Convex iff Epigraph is Convex | https://proofwiki.org/wiki/Function_is_Convex_iff_Epigraph_is_Convex | https://proofwiki.org/wiki/Function_is_Convex_iff_Epigraph_is_Convex | [
"Convex Real Functions",
"Convex Sets (Vector Spaces)",
"Epigraphs"
] | [
"Definition:Vector Space",
"Definition:Convex Set (Vector Space)",
"Definition:Direct Product of Vector Spaces",
"Definition:Function",
"Definition:Epigraph",
"Definition:Convex Real Function/Vector Space",
"Definition:Convex Set (Vector Space)"
] | [
"Definition:Convex Real Function/Vector Space",
"Definition:Convex Real Function/Vector Space",
"Definition:Convex Set (Vector Space)",
"Definition:Convex Set (Vector Space)",
"Definition:Convex Set (Vector Space)",
"Definition:Convex Set (Vector Space)",
"Definition:Convex Set (Vector Space)",
"Defin... |
proofwiki-19853 | Supremum Operator Norm is Well-Defined | Let $K$ be a field.
Let $X, Y$ be normed vector spaces over $K$.
Let $\map {CL} {X, Y}$ be the continuous linear transformation space.
Let $\norm {\, \cdot \,}$ be an operator norm on $\map {CL} {X, Y}$ defined as:
:$\norm T := \map \sup {\norm {Tx} : x \in X : \norm x \le 1}$
Then $\norm {\, \cdot \,}$ is well-defined... | Let $S = \set {\norm {T x} : x \in X : \norm x \le 1}$
By definition of the norm:
:$S \subseteq \R$ | Let $K$ be a [[Definition:Field (Abstract Algebra)|field]].
Let $X, Y$ be [[Definition:Normed Vector Space|normed vector spaces]] over $K$.
Let $\map {CL} {X, Y}$ be the [[Definition:Continuous Linear Transformation Space|continuous linear transformation space]].
Let $\norm {\, \cdot \,}$ be an [[Definition:Operator... | Let $S = \set {\norm {T x} : x \in X : \norm x \le 1}$
By definition of the [[Definition:Norm|norm]]:
:$S \subseteq \R$ | Supremum Operator Norm is Well-Defined | https://proofwiki.org/wiki/Supremum_Operator_Norm_is_Well-Defined | https://proofwiki.org/wiki/Supremum_Operator_Norm_is_Well-Defined | [
"Supremum Norm",
"Linear Transformations",
"Continuous Linear Transformations",
"Continuous Mappings",
"Examples of Well-Defined Mappings"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Normed Vector Space",
"Definition:Continuous Linear Transformation Space",
"Definition:Operator Norm",
"Definition:Well-Defined/Mapping"
] | [
"Definition:Norm"
] |
proofwiki-19854 | Equivalence of Defintions of Ergodic Measure-Preserving Transformation | Let $\struct {X, \BB, \mu}$ be a probability space.
Let $T: X \to X$ be a measure-preserving transformation.
{{TFAE|def = Ergodic Measure-Preserving Transformation}} | === Definition 1 implies Definition 2 ===
First, we claim that for each $j \in \N$:
:$\map \mu {T^{-j} \sqbrk A \symdif A} = 0$.
Indeed, applying Symmetric Difference is Subset of Union of Symmetric Differences, inductively:
{{begin-eqn}}
{{eqn | l = T^{-j} \sqbrk A \symdif A
| o = \subseteq
| r = \bigcup _... | Let $\struct {X, \BB, \mu}$ be a [[Definition:Probability Space|probability space]].
Let $T: X \to X$ be a [[Definition:Measure-Preserving Transformation|measure-preserving transformation]].
{{TFAE|def = Ergodic Measure-Preserving Transformation}} | === Definition 1 implies Definition 2 ===
First, we claim that for each $j \in \N$:
:$\map \mu {T^{-j} \sqbrk A \symdif A} = 0$.
Indeed, applying [[Symmetric Difference is Subset of Union of Symmetric Differences]], inductively:
{{begin-eqn}}
{{eqn | l = T^{-j} \sqbrk A \symdif A
| o = \subseteq
| r = \big... | Equivalence of Defintions of Ergodic Measure-Preserving Transformation | https://proofwiki.org/wiki/Equivalence_of_Defintions_of_Ergodic_Measure-Preserving_Transformation | https://proofwiki.org/wiki/Equivalence_of_Defintions_of_Ergodic_Measure-Preserving_Transformation | [
"Ergodic Theory",
"Ergodic Measure-Preserving Transformations"
] | [
"Definition:Probability Space",
"Definition:Measure-Preserving Transformation"
] | [
"Symmetric Difference is Subset of Union of Symmetric Differences"
] |
proofwiki-19855 | Quasilinear Differential Equation/Examples/x + y y' = 0 | The first order quasilinear ordinary differential equation over the real numbers $\R$:
:$x + y y' = 0$
has the general solution:
:$x^2 + y^2 = C$
where:
:$C > 0$
:$y \ne 0$
:$x < \size {\sqrt C}$
with the singular point:
:$x = y = 0$ | Let us rearrange the equation in question:
{{begin-eqn}}
{{eqn | l = x + y y'
| r = 0
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = y \dfrac {\d y} {\d x}
| r = -x
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = y \rd y
| r = \paren {-1} x \rd x
| c =
}}
{{end-eqn}}
This is in... | The [[Definition:First Order ODE|first order]] [[Definition:Quasilinear Differential Equation|quasilinear]] [[Definition:Ordinary Differential Equation|ordinary differential equation]] over the [[Definition:Real Number|real numbers]] $\R$:
:$x + y y' = 0$
has the [[Definition:General Solution to Differential Equation|... | Let us rearrange the equation in question:
{{begin-eqn}}
{{eqn | l = x + y y'
| r = 0
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = y \dfrac {\d y} {\d x}
| r = -x
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = y \rd y
| r = \paren {-1} x \rd x
| c =
}}
{{end-eqn}}
This is ... | Quasilinear Differential Equation/Examples/x + y y' = 0/Proof 1 | https://proofwiki.org/wiki/Quasilinear_Differential_Equation/Examples/x_+_y_y'_=_0 | https://proofwiki.org/wiki/Quasilinear_Differential_Equation/Examples/x_+_y_y'_=_0/Proof_1 | [
"Examples of First Order ODEs",
"Examples of Quasilinear Differential Equations",
"Quasilinear Differential Equation/Examples/x + y y' = 0"
] | [
"Definition:First Order Ordinary Differential Equation",
"Definition:Differential Equation/Degree/First",
"Definition:Differential Equation/Ordinary",
"Definition:Real Number",
"Definition:Differential Equation/Solution/General Solution",
"Definition:Singular Point/Real"
] | [
"First Order ODE/y dy = k x dx"
] |
proofwiki-19856 | Quasilinear Differential Equation/Examples/x + y y' = 0 | The first order quasilinear ordinary differential equation over the real numbers $\R$:
:$x + y y' = 0$
has the general solution:
:$x^2 + y^2 = C$
where:
:$C > 0$
:$y \ne 0$
:$x < \size {\sqrt C}$
with the singular point:
:$x = y = 0$ | Observe that from Derivative of Power and Chain Rule for Derivatives:
:$\dfrac {\map \d {x^2 + y^2} } {\d y} = 2 x + 2 y \dfrac {\d y} {\d x}$
Hence:
:$\dfrac {\map \d {x^2 + y^2} } {\d y} = 0$
So from Derivative of Constant:
:$x^2 + y^2 = C$
where $C$ is abitrary.
{{qed}} | The [[Definition:First Order ODE|first order]] [[Definition:Quasilinear Differential Equation|quasilinear]] [[Definition:Ordinary Differential Equation|ordinary differential equation]] over the [[Definition:Real Number|real numbers]] $\R$:
:$x + y y' = 0$
has the [[Definition:General Solution to Differential Equation|... | Observe that from [[Derivative of Power]] and [[Chain Rule for Derivatives]]:
:$\dfrac {\map \d {x^2 + y^2} } {\d y} = 2 x + 2 y \dfrac {\d y} {\d x}$
Hence:
:$\dfrac {\map \d {x^2 + y^2} } {\d y} = 0$
So from [[Derivative of Constant]]:
:$x^2 + y^2 = C$
where $C$ is abitrary.
{{qed}} | Quasilinear Differential Equation/Examples/x + y y' = 0/Proof 2 | https://proofwiki.org/wiki/Quasilinear_Differential_Equation/Examples/x_+_y_y'_=_0 | https://proofwiki.org/wiki/Quasilinear_Differential_Equation/Examples/x_+_y_y'_=_0/Proof_2 | [
"Examples of First Order ODEs",
"Examples of Quasilinear Differential Equations",
"Quasilinear Differential Equation/Examples/x + y y' = 0"
] | [
"Definition:First Order Ordinary Differential Equation",
"Definition:Differential Equation/Degree/First",
"Definition:Differential Equation/Ordinary",
"Definition:Real Number",
"Definition:Differential Equation/Solution/General Solution",
"Definition:Singular Point/Real"
] | [
"Power Rule for Derivatives",
"Derivative of Composite Function",
"Derivative of Constant"
] |
proofwiki-19857 | Quasilinear Differential Equation/Examples/x + y y' = 0/Explicit Solution | The first order quasilinear ordinary differential equation:
:$x + y y' = 0$
has a general solution which can be expressed explicitly as:
:$y = \pm \sqrt {C - x^2}$
over the domain:
:$-\sqrt C < x < \sqrt C$ | The general solution of $x + y y' = 0$ is implicitly over the real numbers.
We have:
{{begin-eqn}}
{{eqn | l = x^2 + y^2
| r = y
| c = Quasilinear Differential Equation: $x + y y' = 0$
}}
{{eqn | ll= \leadsto
| l = y^2
| r = C - x^2
| c = where $C - x^2 \ge 0$
}}
{{eqn | ll= \leadsto
... | The [[Definition:First Order ODE|first order]] [[Definition:Quasilinear Differential Equation|quasilinear]] [[Definition:Ordinary Differential Equation|ordinary differential equation]]:
:$x + y y' = 0$
has a [[Definition:General Solution to Differential Equation|general solution]] which can be expressed explicitly as:... | The [[Definition:General Solution to Differential Equation|general solution]] of $x + y y' = 0$ is implicitly over the [[Definition:Real Number|real numbers]].
We have:
{{begin-eqn}}
{{eqn | l = x^2 + y^2
| r = y
| c = [[Quasilinear Differential Equation/Examples/x + y y' = 0|Quasilinear Differential Equa... | Quasilinear Differential Equation/Examples/x + y y' = 0/Explicit Solution | https://proofwiki.org/wiki/Quasilinear_Differential_Equation/Examples/x_+_y_y'_=_0/Explicit_Solution | https://proofwiki.org/wiki/Quasilinear_Differential_Equation/Examples/x_+_y_y'_=_0/Explicit_Solution | [
"Quasilinear Differential Equation/Examples/x + y y' = 0"
] | [
"Definition:First Order Ordinary Differential Equation",
"Definition:Differential Equation/Degree/First",
"Definition:Differential Equation/Ordinary",
"Definition:Differential Equation/Solution/General Solution",
"Definition:Domain (Set Theory)/Mapping"
] | [
"Definition:Differential Equation/Solution/General Solution",
"Definition:Real Number",
"Quasilinear Differential Equation/Examples/x + y y' = 0",
"Definition:Derivative",
"Definition:Differential Equation/Ordinary",
"Definition:Infinity",
"Definition:Zero (Number)"
] |
proofwiki-19858 | Universal Upper Bound greater than Supremum Operator Norm | Let $\map {CL} {X, Y}$ be the continuous linear transformation space.
Let $T \in \map {CL} {X, Y}$.
Let $\norm {\, \cdot \,}$ be an operator norm on $\map {CL} {X, Y}$ defined as:
:$\norm T := \sup \set {\norm {Tx}_Y : x \in X : \norm x_X \le 1}$
Suppose:
:$\exists M \in \R_{> 0} : \forall x \in X : \norm {T x}_Y \le M... | Let $x \in X$ be such that $\norm x_X \le 1$.
Then:
{{begin-eqn}}
{{eqn | l = \norm {T x}_Y
| o = \le
| r = M \norm x_X
}}
{{eqn | o = \le
| r = M \cdot 1
}}
{{eqn | r = M
}}
{{end-eqn}}
Hence, $M$ is an upper bound for $S = \set {\norm {T x}_Y : x \in X : \norm x_X \le 1}$.
By definition of the supr... | Let $\map {CL} {X, Y}$ be the [[Definition:Continuous Linear Transformation Space|continuous linear transformation space]].
Let $T \in \map {CL} {X, Y}$.
Let $\norm {\, \cdot \,}$ be an [[Definition:Operator Norm|operator norm]] on $\map {CL} {X, Y}$ defined as:
:$\norm T := \sup \set {\norm {Tx}_Y : x \in X : \norm... | Let $x \in X$ be such that $\norm x_X \le 1$.
Then:
{{begin-eqn}}
{{eqn | l = \norm {T x}_Y
| o = \le
| r = M \norm x_X
}}
{{eqn | o = \le
| r = M \cdot 1
}}
{{eqn | r = M
}}
{{end-eqn}}
Hence, $M$ is an [[Definition:Upper Bound of Subset of Real Numbers|upper bound]] for $S = \set {\norm {T x}_Y ... | Universal Upper Bound greater than Supremum Operator Norm | https://proofwiki.org/wiki/Universal_Upper_Bound_greater_than_Supremum_Operator_Norm | https://proofwiki.org/wiki/Universal_Upper_Bound_greater_than_Supremum_Operator_Norm | [
"Linear Transformations",
"Continuous Linear Transformations",
"Continuous Mappings",
"Supremum Norm"
] | [
"Definition:Continuous Linear Transformation Space",
"Definition:Operator Norm"
] | [
"Definition:Upper Bound of Set/Real Numbers",
"Definition:Supremum of Set/Real Numbers"
] |
proofwiki-19859 | Supremum Operator Norm as Universal Upper Bound | Let $\map {CL} {X, Y}$ be the continuous linear transformation space.
Let $T \in \map {CL} {X, Y}$.
Let $\norm {\, \cdot \,}$ be the supremum operator norm on $\map {CL} {X, Y}$.
Then:
:$\forall x \in X : \norm {Tx}_Y \le \norm T \norm x_X$ | Let $S = \set {\norm {Tx}_Y : x \in X : \norm x_X \le 1}$.
Suppose $x = \mathbf 0$.
Then:
{{begin-eqn}}
{{eqn | l = \norm {T x}_Y
| r = \norm {T \mathbf 0}_Y
}}
{{eqn | r = \norm {\mathbf 0}_Y
}}
{{eqn | r = 0
}}
{{eqn | r = \norm T 0
}}
{{eqn | r = \norm T \norm {\mathbf 0}_X
}}
{{eqn | r = \norm T \norm x_X
}}
... | Let $\map {CL} {X, Y}$ be the [[Definition:Continuous Linear Transformation Space|continuous linear transformation space]].
Let $T \in \map {CL} {X, Y}$.
Let $\norm {\, \cdot \,}$ be the [[Definition:Supremum Operator Norm|supremum operator norm]] on $\map {CL} {X, Y}$.
Then:
:$\forall x \in X : \norm {Tx}_Y \le \... | Let $S = \set {\norm {Tx}_Y : x \in X : \norm x_X \le 1}$.
Suppose $x = \mathbf 0$.
Then:
{{begin-eqn}}
{{eqn | l = \norm {T x}_Y
| r = \norm {T \mathbf 0}_Y
}}
{{eqn | r = \norm {\mathbf 0}_Y
}}
{{eqn | r = 0
}}
{{eqn | r = \norm T 0
}}
{{eqn | r = \norm T \norm {\mathbf 0}_X
}}
{{eqn | r = \norm T \norm x_X
... | Supremum Operator Norm as Universal Upper Bound | https://proofwiki.org/wiki/Supremum_Operator_Norm_as_Universal_Upper_Bound | https://proofwiki.org/wiki/Supremum_Operator_Norm_as_Universal_Upper_Bound | [
"Linear Transformations",
"Continuous Linear Transformations",
"Continuous Mappings",
"Supremum Norm"
] | [
"Definition:Continuous Linear Transformation Space",
"Definition:Supremum Operator Norm"
] | [] |
proofwiki-19860 | Reversed Complex Contour is Contour | Let $C$ be a contour in the complex plane $\C$ that is defined as a concatenation of a finite sequence $\sequence{ C_1, \ldots, C_n }$ of directed smooth curves in $\C$.
Then the finite sequence of reversed directed smooth curves:
:$\sequence{ -C_n, -C_{n - 1}, \ldots, -C_1 }$
defines a contour that is independent of t... | Let $C_k$ be parameterized by the smooth path $\gamma_k: \closedint {a_k} {b_k} \to \C$ for all $k \in \set {1, \ldots, n}$.
From Reversed Directed Smooth Curve is Directed Smooth Curve, it follows that $-C_k$ is independent of the parameterization $\gamma_k$ of $C_k$.
We now prove that the end point of $-C_k$ is equal... | Let $C$ be a [[Definition:Contour (Complex Plane)|contour]] in the [[Definition:Complex Plane|complex plane]] $\C$ that is defined as a [[Definition:Concatenation of Contours (Complex Plane)|concatenation]] of a [[Definition:Finite Sequence|finite sequence]] $\sequence{ C_1, \ldots, C_n }$ of [[Definition:Directed Smoo... | Let $C_k$ be [[Definition:Parameterization of Directed Smooth Curve (Complex Plane)|parameterized]] by the [[Definition:Smooth Path (Complex Analysis)|smooth path]] $\gamma_k: \closedint {a_k} {b_k} \to \C$ for all $k \in \set {1, \ldots, n}$.
From [[Reversed Directed Smooth Curve is Directed Smooth Curve]], it follow... | Reversed Complex Contour is Contour | https://proofwiki.org/wiki/Reversed_Complex_Contour_is_Contour | https://proofwiki.org/wiki/Reversed_Complex_Contour_is_Contour | [
"Complex Contour Integrals"
] | [
"Definition:Contour/Complex Plane",
"Definition:Complex Number/Complex Plane",
"Definition:Concatenation of Contours/Complex Plane",
"Definition:Finite Sequence",
"Definition:Directed Smooth Curve/Complex Plane",
"Definition:Reversed Directed Smooth Curve/Complex Plane",
"Definition:Contour/Complex Plan... | [
"Definition:Directed Smooth Curve/Parameterization/Complex Plane",
"Definition:Smooth Path/Complex",
"Reversed Directed Smooth Curve is Directed Smooth Curve",
"Definition:Directed Smooth Curve/Parameterization/Complex Plane",
"Definition:Contour/Endpoints/Complex Plane",
"Definition:Contour/Endpoints/Com... |
proofwiki-19861 | Composite of Strictly Increasing Mappings is Strictly Increasing | Let $\struct {S_1, \preceq_1}$, $\struct {S_2, \preceq_2}$ and $\struct {S_3, \preceq_3}$ be ordered sets.
Let:
:$\phi: \struct {S_1, \preceq_1} \to \struct {S_2, \preceq_2}$
and:
:$\psi: \struct {S_2, \preceq_2} \to \struct {S_3, \preceq_3}$
be strictly increasing mappings.
Then $\psi \circ \phi: \struct {S_1, \preceq... | By definition of composition of mappings:
:$\map {\paren {\psi \circ \phi} } x = \map \psi {\map \phi x}$
As $\phi$ is a strictly increasing mapping, we have:
:$\forall x_1, y_1 \in S_1: x_1 \prec_1 y_1 \implies \map \phi {x_1} \prec_2 \map \phi {y_1}$
where $a \prec_1 b$ denotes that $a \preceq_1 b$ and $a \ne b$.
As ... | Let $\struct {S_1, \preceq_1}$, $\struct {S_2, \preceq_2}$ and $\struct {S_3, \preceq_3}$ be [[Definition:Ordered Set|ordered sets]].
Let:
:$\phi: \struct {S_1, \preceq_1} \to \struct {S_2, \preceq_2}$
and:
:$\psi: \struct {S_2, \preceq_2} \to \struct {S_3, \preceq_3}$
be [[Definition:Strictly Increasing Mapping|stric... | By definition of [[Definition:Composition of Mappings|composition of mappings]]:
:$\map {\paren {\psi \circ \phi} } x = \map \psi {\map \phi x}$
As $\phi$ is a [[Definition:Strictly Increasing Mapping|strictly increasing mapping]], we have:
:$\forall x_1, y_1 \in S_1: x_1 \prec_1 y_1 \implies \map \phi {x_1} \prec_2 ... | Composite of Strictly Increasing Mappings is Strictly Increasing | https://proofwiki.org/wiki/Composite_of_Strictly_Increasing_Mappings_is_Strictly_Increasing | https://proofwiki.org/wiki/Composite_of_Strictly_Increasing_Mappings_is_Strictly_Increasing | [
"Increasing Mappings"
] | [
"Definition:Ordered Set",
"Definition:Strictly Increasing/Mapping",
"Definition:Strictly Increasing/Mapping"
] | [
"Definition:Composition of Mappings",
"Definition:Strictly Increasing/Mapping",
"Definition:Strictly Increasing/Mapping",
"Category:Increasing Mappings"
] |
proofwiki-19862 | Primitive of Reciprocal of 1 plus x squared/Arctangent Form | :$\ds \int \frac {\d x} {1 + x^2} = \arctan x + C$ | {{begin-eqn}}
{{eqn | l = \map {\dfrac \d {\d x} } {\arctan x}
| r = \dfrac 1 {1 + x^2}
| c = Derivative of Arctangent Function
}}
{{eqn | ll= \leadsto
| l = \int \dfrac {\d x} {1 + x^2}
| r = \arctan x + C
| c = {{Defof|Primitive (Calculus)}}
}}
{{end-eqn}}
{{qed}} | :$\ds \int \frac {\d x} {1 + x^2} = \arctan x + C$ | {{begin-eqn}}
{{eqn | l = \map {\dfrac \d {\d x} } {\arctan x}
| r = \dfrac 1 {1 + x^2}
| c = [[Derivative of Arctangent Function]]
}}
{{eqn | ll= \leadsto
| l = \int \dfrac {\d x} {1 + x^2}
| r = \arctan x + C
| c = {{Defof|Primitive (Calculus)}}
}}
{{end-eqn}}
{{qed}} | Primitive of Reciprocal of 1 plus x squared/Arctangent Form/Proof 2 | https://proofwiki.org/wiki/Primitive_of_Reciprocal_of_1_plus_x_squared/Arctangent_Form | https://proofwiki.org/wiki/Primitive_of_Reciprocal_of_1_plus_x_squared/Arctangent_Form/Proof_2 | [
"Primitive of Reciprocal of x squared plus a squared"
] | [] | [
"Derivative of Arctangent Function"
] |
proofwiki-19863 | Primitive of Reciprocal of Root of 1 minus x squared/Arcsine Form | :$\ds \int \frac {\d x} {\sqrt {1 - x^2} } = \arcsin x + C$ | From Primitive of $\dfrac 1 {\sqrt {a^2 - x^2} }$: Arcsine Form:
:$\ds \int \frac {\d x} {\sqrt {a^2 - x^2} } = \arcsin \frac x a + C$
The result follows by setting $a = 1$.
{{Qed}} | :$\ds \int \frac {\d x} {\sqrt {1 - x^2} } = \arcsin x + C$ | From [[Primitive of Reciprocal of Root of a squared minus x squared/Arcsine Form|Primitive of $\dfrac 1 {\sqrt {a^2 - x^2} }$: Arcsine Form]]:
:$\ds \int \frac {\d x} {\sqrt {a^2 - x^2} } = \arcsin \frac x a + C$
The result follows by setting $a = 1$.
{{Qed}} | Primitive of Reciprocal of Root of 1 minus x squared/Arcsine Form/Proof 1 | https://proofwiki.org/wiki/Primitive_of_Reciprocal_of_Root_of_1_minus_x_squared/Arcsine_Form | https://proofwiki.org/wiki/Primitive_of_Reciprocal_of_Root_of_1_minus_x_squared/Arcsine_Form/Proof_1 | [
"Primitive of Reciprocal of Root of a squared minus x squared"
] | [] | [
"Primitive of Reciprocal of Root of a squared minus x squared/Arcsine Form"
] |
proofwiki-19864 | Primitive of Reciprocal of Root of 1 minus x squared/Arcsine Form | :$\ds \int \frac {\d x} {\sqrt {1 - x^2} } = \arcsin x + C$ | {{begin-eqn}}
{{eqn | l = \map {\dfrac \d {\d x} } {\arcsin x}
| r = \dfrac 1 {\sqrt {1 - x^2} }
| c = Derivative of Arcsine Function
}}
{{eqn | ll= \leadsto
| l = \int \dfrac {\d x} {\sqrt {1 - x^2} }
| r = \arcsin x + C
| c = {{Defof|Primitive (Calculus)}}
}}
{{end-eqn}}
{{qed}} | :$\ds \int \frac {\d x} {\sqrt {1 - x^2} } = \arcsin x + C$ | {{begin-eqn}}
{{eqn | l = \map {\dfrac \d {\d x} } {\arcsin x}
| r = \dfrac 1 {\sqrt {1 - x^2} }
| c = [[Derivative of Arcsine Function]]
}}
{{eqn | ll= \leadsto
| l = \int \dfrac {\d x} {\sqrt {1 - x^2} }
| r = \arcsin x + C
| c = {{Defof|Primitive (Calculus)}}
}}
{{end-eqn}}
{{qed}} | Primitive of Reciprocal of Root of 1 minus x squared/Arcsine Form/Proof 2 | https://proofwiki.org/wiki/Primitive_of_Reciprocal_of_Root_of_1_minus_x_squared/Arcsine_Form | https://proofwiki.org/wiki/Primitive_of_Reciprocal_of_Root_of_1_minus_x_squared/Arcsine_Form/Proof_2 | [
"Primitive of Reciprocal of Root of a squared minus x squared"
] | [] | [
"Derivative of Arcsine Function"
] |
proofwiki-19865 | Sine Function is Absolutely Convergent/Complex Case | The complex sine function $\sin: \C \to \C$ is absolutely convergent. | The definition of the complex sine function is:
:$\ds \forall z \in \C: \sin z = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {z^{2 n + 1} } {\paren {2 n + 1}!}$
By definition of absolutely convergent complex series, we must show that the power series
:$\ds \sum_{n \mathop = 0}^\infty \size {\paren {-1}^n \frac {z^{... | The [[Definition:Complex Sine Function|complex sine function]] $\sin: \C \to \C$ is [[Definition:Absolutely Convergent Complex Series|absolutely convergent]]. | The definition of the [[Definition:Complex Sine Function|complex sine function]] is:
:$\ds \forall z \in \C: \sin z = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {z^{2 n + 1} } {\paren {2 n + 1}!}$
By [[Definition:Absolutely Convergent Complex Series|definition of absolutely convergent complex series]], we must s... | Sine Function is Absolutely Convergent/Complex Case/Proof 1 | https://proofwiki.org/wiki/Sine_Function_is_Absolutely_Convergent/Complex_Case | https://proofwiki.org/wiki/Sine_Function_is_Absolutely_Convergent/Complex_Case/Proof_1 | [
"Complex Sine Function is Absolutely Convergent",
"Complex Sine Function",
"Absolute Convergence"
] | [
"Definition:Sine/Complex Function",
"Definition:Absolutely Convergent Series/Complex Numbers"
] | [
"Definition:Sine/Complex Function",
"Definition:Absolutely Convergent Series/Complex Numbers",
"Definition:Power Series/Complex Domain",
"Squeeze Theorem/Sequences/Complex Numbers",
"Power Series Expansion for Exponential Function",
"Squeeze Theorem/Sequences/Complex Numbers"
] |
proofwiki-19866 | Sine Function is Absolutely Convergent/Complex Case | The complex sine function $\sin: \C \to \C$ is absolutely convergent. | Radius of Convergence of Power Series Expansion for Sine Function shows that the radius of convergence of the complex sine function is infinite.
Then Existence of Radius of Convergence of Complex Power Series shows that the complex sine function is absolutely convergent.
{{qed}} | The [[Definition:Complex Sine Function|complex sine function]] $\sin: \C \to \C$ is [[Definition:Absolutely Convergent Complex Series|absolutely convergent]]. | [[Radius of Convergence of Power Series Expansion for Sine Function]] shows that the [[Definition:Radius of Convergence of Complex Power Series|radius of convergence]] of the [[Definition:Complex Sine Function|complex sine function]] is infinite.
Then [[Existence of Radius of Convergence of Complex Power Series/Absolu... | Sine Function is Absolutely Convergent/Complex Case/Proof 2 | https://proofwiki.org/wiki/Sine_Function_is_Absolutely_Convergent/Complex_Case | https://proofwiki.org/wiki/Sine_Function_is_Absolutely_Convergent/Complex_Case/Proof_2 | [
"Complex Sine Function is Absolutely Convergent",
"Complex Sine Function",
"Absolute Convergence"
] | [
"Definition:Sine/Complex Function",
"Definition:Absolutely Convergent Series/Complex Numbers"
] | [
"Radius of Convergence of Power Series Expansion for Sine Function",
"Definition:Radius of Convergence/Complex Domain",
"Definition:Sine/Complex Function",
"Existence of Radius of Convergence of Complex Power Series/Absolute Convergence",
"Definition:Sine/Complex Function",
"Definition:Absolutely Converge... |
proofwiki-19867 | Cosine Function is Absolutely Convergent/Complex Case | The complex cosine function $\cos: \C \to \C$ is absolutely convergent. | The definition of the complex cosine function is:
:$\ds \cos z = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {z^{2 n} } {\paren {2 n}!}$
By definition of absolutely convergent complex series, we must show that the power series
:$\ds \sum_{n \mathop = 0}^\infty \size {\paren {-1}^n \frac {z^{2 n} } {\paren {2 n}!} }... | The [[Definition:Complex Cosine Function|complex cosine function]] $\cos: \C \to \C$ is [[Definition:Absolutely Convergent Complex Series|absolutely convergent]]. | The definition of the [[Definition:Complex Cosine Function|complex cosine function]] is:
:$\ds \cos z = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {z^{2 n} } {\paren {2 n}!}$
By definition of [[Definition:Absolutely Convergent Complex Series|absolutely convergent complex series]], we must show that the [[Definit... | Cosine Function is Absolutely Convergent/Complex Case/Proof 1 | https://proofwiki.org/wiki/Cosine_Function_is_Absolutely_Convergent/Complex_Case | https://proofwiki.org/wiki/Cosine_Function_is_Absolutely_Convergent/Complex_Case/Proof_1 | [
"Complex Cosine Function is Absolutely Convergent",
"Cosine Function is Absolutely Convergent",
"Complex Cosine Function",
"Absolute Convergence"
] | [
"Definition:Cosine/Complex Function",
"Definition:Absolutely Convergent Series/Complex Numbers"
] | [
"Definition:Cosine/Complex Function",
"Definition:Absolutely Convergent Series/Complex Numbers",
"Definition:Power Series/Complex Domain",
"Squeeze Theorem/Sequences/Complex Numbers",
"Power Series Expansion for Exponential Function",
"Squeeze Theorem/Sequences/Complex Numbers"
] |
proofwiki-19868 | Cosine Function is Absolutely Convergent/Complex Case | The complex cosine function $\cos: \C \to \C$ is absolutely convergent. | Radius of Convergence of Power Series Expansion for Cosine Function shows that the radius of convergence of the complex cosine function is infinite.
Then Existence of Radius of Convergence of Complex Power Series shows that the complex cosine function is absolutely convergent.
{{qed}} | The [[Definition:Complex Cosine Function|complex cosine function]] $\cos: \C \to \C$ is [[Definition:Absolutely Convergent Complex Series|absolutely convergent]]. | [[Radius of Convergence of Power Series Expansion for Cosine Function]] shows that the [[Definition:Radius of Convergence of Complex Power Series|radius of convergence]] of the [[Definition:Complex Cosine Function|complex cosine function]] is infinite.
Then [[Existence of Radius of Convergence of Complex Power Series/... | Cosine Function is Absolutely Convergent/Complex Case/Proof 2 | https://proofwiki.org/wiki/Cosine_Function_is_Absolutely_Convergent/Complex_Case | https://proofwiki.org/wiki/Cosine_Function_is_Absolutely_Convergent/Complex_Case/Proof_2 | [
"Complex Cosine Function is Absolutely Convergent",
"Cosine Function is Absolutely Convergent",
"Complex Cosine Function",
"Absolute Convergence"
] | [
"Definition:Cosine/Complex Function",
"Definition:Absolutely Convergent Series/Complex Numbers"
] | [
"Radius of Convergence of Power Series Expansion for Cosine Function",
"Definition:Radius of Convergence/Complex Domain",
"Definition:Cosine/Complex Function",
"Existence of Radius of Convergence of Complex Power Series/Absolute Convergence",
"Definition:Absolutely Convergent Series/Complex Numbers"
] |
proofwiki-19869 | Contour Integral of Closed Contour Split into Two Contours | Let $f: D \to \C$ be a continuous complex function, where $D$ is a connected domain.
Let $C = \sequence {C_1, \ldots, C_n}$ be a closed contour in $D$.
Let $C'$ be a contour in $D$ with start point $z_1$ and end point $z_2$.
Let $-C'$ denote the reversed contour of $C'$.
Suppose:
:$z_1$ is equal to the end point of $C_... | By definition of closed contour, $C^{\paren 1}$ and $C^{\paren 2}$ are closed contours.
Then:
{{begin-eqn}}
{{eqn | l = \oint_C \map f z \rd z
| r = \sum_{j \mathop = 1}^n \int_{C_j} \map f z \rd z
| c = Contour Integral of Concatenation of Contours
}}
{{eqn | r = \sum_{j \mathop = 1}^n \int_{C_j} \map f z ... | Let $f: D \to \C$ be a [[Definition:Continuous Complex Function|continuous complex function]], where $D$ is a [[Definition:Connected Domain (Complex Analysis)|connected domain]].
Let $C = \sequence {C_1, \ldots, C_n}$ be a [[Definition:Closed Contour (Complex Plane)|closed contour]] in $D$.
Let $C'$ be a contour in $... | By definition of [[Definition:Closed Contour (Complex Plane)|closed contour]], $C^{\paren 1}$ and $C^{\paren 2}$ are [[Definition:Closed Contour (Complex Plane)|closed contours]].
Then:
{{begin-eqn}}
{{eqn | l = \oint_C \map f z \rd z
| r = \sum_{j \mathop = 1}^n \int_{C_j} \map f z \rd z
| c = [[Contour ... | Contour Integral of Closed Contour Split into Two Contours | https://proofwiki.org/wiki/Contour_Integral_of_Closed_Contour_Split_into_Two_Contours | https://proofwiki.org/wiki/Contour_Integral_of_Closed_Contour_Split_into_Two_Contours | [
"Complex Contour Integrals"
] | [
"Definition:Continuous Complex Function",
"Definition:Connected Domain (Complex Analysis)",
"Definition:Contour/Closed/Complex Plane",
"Definition:Directed Smooth Curve/Endpoints/Complex Plane",
"Definition:Directed Smooth Curve/Endpoints/Complex Plane",
"Definition:Reversed Contour/Complex Plane",
"Def... | [
"Definition:Contour/Closed/Complex Plane",
"Definition:Contour/Closed/Complex Plane",
"Contour Integral of Concatenation of Contours",
"Contour Integral along Reversed Contour",
"Contour Integral of Concatenation of Contours"
] |
proofwiki-19870 | Power Series Expansion for Exponential Integral Function/Formulation 2 | Let $\Ei: \R_{>0} \to \R$ denote the exponential integral function:
:$\map \Ei x = \ds \int_{t \mathop \to -\infty}^{t \mathop = x} \frac {e^t} t \rd t$
Then:
{{begin-eqn}}
{{eqn | l = \map \Ei x
| r = \gamma + \ln x + \sum_{n \mathop = 1}^\infty \frac {x^n} {n \times n!}
| c =
}}
{{eqn | r = \gamma + \ln ... | {{begin-eqn}}
{{eqn | l = \map \Ei x
| r = \gamma + \ln x + \int_0^x \frac {e^u - 1} u \rd u
| c = Characterization of Exponential Integral Function
}}
{{eqn | r = \gamma + \ln x + \int_0^x \frac 1 u \paren {\sum_{n \mathop = 0}^\infty \frac {u^n} {n!} - 1} \rd u
| c = {{Defof|Real Exponential Functio... | Let $\Ei: \R_{>0} \to \R$ denote the [[Definition:Exponential Integral Function/Formulation 2|exponential integral function]]:
:$\map \Ei x = \ds \int_{t \mathop \to -\infty}^{t \mathop = x} \frac {e^t} t \rd t$
Then:
{{begin-eqn}}
{{eqn | l = \map \Ei x
| r = \gamma + \ln x + \sum_{n \mathop = 1}^\infty \frac ... | {{begin-eqn}}
{{eqn | l = \map \Ei x
| r = \gamma + \ln x + \int_0^x \frac {e^u - 1} u \rd u
| c = [[Characterization of Exponential Integral Function/Formulation 2|Characterization of Exponential Integral Function]]
}}
{{eqn | r = \gamma + \ln x + \int_0^x \frac 1 u \paren {\sum_{n \mathop = 0}^\infty \fra... | Power Series Expansion for Exponential Integral Function/Formulation 2 | https://proofwiki.org/wiki/Power_Series_Expansion_for_Exponential_Integral_Function/Formulation_2 | https://proofwiki.org/wiki/Power_Series_Expansion_for_Exponential_Integral_Function/Formulation_2 | [
"Power Series Expansion for Exponential Integral Function"
] | [
"Definition:Exponential Integral Function/Formulation 2"
] | [
"Characterization of Exponential Integral Function/Formulation 2",
"Power Series is Termwise Integrable within Radius of Convergence",
"Primitive of Power"
] |
proofwiki-19871 | Complex Logarithm Function is Transcendental | Let $a \in \R_{>0}$ be a strictly positive real number such that $a \ne 1$.
Let $\log_a: \C \to \C$ denote the complex general logarithm base $a$.
$\log_a$ is a transcendental function. | {{ProofWanted}}
Category:Logarithms
Category:Examples of Transcendental Functions
r8kquxl6v4xuy1lgbml72nv66z0epn1 | Let $a \in \R_{>0}$ be a [[Definition:Strictly Positive Real Number|strictly positive real number]] such that $a \ne 1$.
Let $\log_a: \C \to \C$ denote the [[Definition:Complex General Logarithm|complex general logarithm base $a$]].
$\log_a$ is a [[Definition:Transcendental Function|transcendental function]]. | {{ProofWanted}}
[[Category:Logarithms]]
[[Category:Examples of Transcendental Functions]]
r8kquxl6v4xuy1lgbml72nv66z0epn1 | Complex Logarithm Function is Transcendental | https://proofwiki.org/wiki/Complex_Logarithm_Function_is_Transcendental | https://proofwiki.org/wiki/Complex_Logarithm_Function_is_Transcendental | [
"Logarithms",
"Examples of Transcendental Functions"
] | [
"Definition:Strictly Positive/Real Number",
"Definition:General Logarithm/Complex",
"Definition:Transcendental Function"
] | [
"Category:Logarithms",
"Category:Examples of Transcendental Functions"
] |
proofwiki-19872 | Epsilon-Function Differentiability Condition/Complex Case | Let $f: D \to \C$ be a continuous function, where $D \subseteq \C$ is an open set.
Let $z \in \C$.
Then $f$ is differentiable at $z$ {{iff}} there exist $\alpha \in \C$ and $r \in \R_{>0}$ such that for all $h \in \map {B_r} 0 \setminus \set 0$:
:$\map f {z + h} = \map f z + h \paren {\alpha + \map \epsilon h}$
where:
... | === Necessary Condition ===
Assume that $f$ is differentiable in $z$.
By definition of open set, there exists $r \in \R_{>0}$ such that the open ball $\map {B_r} z \subseteq D$.
Define $\epsilon: \map {B_r} 0 \setminus \set 0 \to \C$ by:
:$\map \epsilon h = \dfrac {\map f {z + h} - \map f z} h - \map {f'} z$
Let $h \in... | Let $f: D \to \C$ be a [[Definition:Continuous Complex Function|continuous function]], where $D \subseteq \C$ is an [[Definition:Open Set (Complex Analysis)|open set]].
Let $z \in \C$.
Then $f$ is [[Definition:Complex-Differentiable Function|differentiable]] at $z$ {{iff}} there exist $\alpha \in \C$ and $r \in \R_{... | === Necessary Condition ===
Assume that $f$ is [[Definition:Complex-Differentiable Function|differentiable]] in $z$.
By definition of [[Definition:Open Set (Complex Analysis)|open set]], there exists $r \in \R_{>0}$ such that the [[Definition:Open Ball|open ball]] $\map {B_r} z \subseteq D$.
Define $\epsilon: \map {... | Epsilon-Function Differentiability Condition/Complex Case | https://proofwiki.org/wiki/Epsilon-Function_Differentiability_Condition/Complex_Case | https://proofwiki.org/wiki/Epsilon-Function_Differentiability_Condition/Complex_Case | [
"Complex Differential Calculus",
"Epsilon-Function Differentiability Condition"
] | [
"Definition:Continuous Complex Function",
"Definition:Open Set/Complex Analysis",
"Definition:Differentiable Mapping/Complex Function",
"Definition:Open Ball",
"Definition:Complex Function"
] | [
"Definition:Differentiable Mapping/Complex Function",
"Definition:Open Set/Complex Analysis",
"Definition:Open Ball",
"Definition:Well-Defined/Mapping",
"Definition:Differentiable Mapping/Complex Function",
"Definition:Differentiable Mapping/Complex Function"
] |
proofwiki-19873 | Supremum Operator Norm is Norm | Let $\map {CL} {X, Y}$ be the continuous linear transformation space.
Let $\norm {\, \cdot \,} : \map {CL} {X, Y} \to \R$ be the supremum operator norm such that:
:$\forall T \in \map {CL} {X, Y} : \norm T := \sup \set {\norm {Tx}_Y : x \in X : \norm x_X \le 1}$
Then $\norm {\, \cdot \,}$ is a norm. | === Positive definiteness ===
Let $T \in \map {CL} {X, Y}$.
By definition of norm:
:$\norm {Tx}_Y \ge 0$
Then:
:$\ds \norm T = \sup_{\begin{split} x \mathop \in X \\ \norm x \mathop \le 1 \end{split} } \norm {T x}_Y \ge 0$
Suppose $\norm T = 0$.
Then:
{{begin-eqn}}
{{eqn | l = \norm {T x}_Y
| o = \le
| r ... | Let $\map {CL} {X, Y}$ be the [[Definition:Continuous Linear Transformation Space|continuous linear transformation space]].
Let $\norm {\, \cdot \,} : \map {CL} {X, Y} \to \R$ be the [[Definition:Supremum Operator Norm|supremum operator norm]] such that:
:$\forall T \in \map {CL} {X, Y} : \norm T := \sup \set {\norm ... | === Positive definiteness ===
Let $T \in \map {CL} {X, Y}$.
By definition of [[Definition:Norm on Vector Space|norm]]:
:$\norm {Tx}_Y \ge 0$
Then:
:$\ds \norm T = \sup_{\begin{split} x \mathop \in X \\ \norm x \mathop \le 1 \end{split} } \norm {T x}_Y \ge 0$
Suppose $\norm T = 0$.
Then:
{{begin-eqn}}
{{eqn | l ... | Supremum Operator Norm is Norm | https://proofwiki.org/wiki/Supremum_Operator_Norm_is_Norm | https://proofwiki.org/wiki/Supremum_Operator_Norm_is_Norm | [
"Linear Transformations",
"Continuous Linear Transformations",
"Continuous Mappings",
"Supremum Norm",
"Examples of Norms"
] | [
"Definition:Continuous Linear Transformation Space",
"Definition:Supremum Operator Norm",
"Definition:Norm"
] | [
"Definition:Norm/Vector Space",
"Supremum Operator Norm as Universal Upper Bound",
"Definition:Norm/Vector Space",
"Definition:Zero Vector",
"Definition:Normed Vector Space",
"Supremum Operator Norm as Universal Upper Bound"
] |
proofwiki-19874 | Combination Theorem for Complex Derivatives/Product Rule/Proof 1 | Let $D$ be an open subset of the set of complex numbers.
Let $f, g: D \to \C$ be complex-differentiable functions on $D$
Let $f g$ denote the pointwise product of the functions $f$ and $g$.
Then $f g$ is complex-differentiable in $D$, and its derivative $\paren {f g}'$ is defined by:
:$\map {\paren {f g}'} z = \map {f'... | Define $k: D \to \C$ as the pointwise product of $f$ and $g$, so $k = fg$.
Let $z_0 \in D$ be a point in $D$.
{{begin-eqn}}
{{eqn | l = \map {k'} {z_0}
| r = \lim_{h \mathop \to 0} \frac {\map k {z_0 + h} - \map k {z_0} } h
| c =
}}
{{eqn | r = \lim_{h \mathop \to 0} \frac {\map f {z_0 + h} \, \map g {z_0 +... | Let $D$ be an [[Definition:Open Set (Complex Analysis)|open]] [[Definition:Subset|subset]] of the [[Definition:Set|set]] of [[Definition:Complex Number|complex numbers]].
Let $f, g: D \to \C$ be [[Definition:Complex-Differentiable Function|complex-differentiable functions]] on $D$
Let $f g$ denote the [[Definition:P... | Define $k: D \to \C$ as the [[Definition:Pointwise Multiplication of Complex-Valued Functions|pointwise product]] of $f$ and $g$, so $k = fg$.
Let $z_0 \in D$ be a point in $D$.
{{begin-eqn}}
{{eqn | l = \map {k'} {z_0}
| r = \lim_{h \mathop \to 0} \frac {\map k {z_0 + h} - \map k {z_0} } h
| c =
}}
{{eqn... | Combination Theorem for Complex Derivatives/Product Rule/Proof 1 | https://proofwiki.org/wiki/Combination_Theorem_for_Complex_Derivatives/Product_Rule/Proof_1 | https://proofwiki.org/wiki/Combination_Theorem_for_Complex_Derivatives/Product_Rule/Proof_1 | [
"Combination Theorem for Complex Derivatives"
] | [
"Definition:Open Set/Complex Analysis",
"Definition:Subset",
"Definition:Set",
"Definition:Complex Number",
"Definition:Differentiable Mapping/Complex Function",
"Definition:Pointwise Multiplication of Complex-Valued Functions",
"Definition:Complex Function",
"Definition:Differentiable Mapping/Complex... | [
"Definition:Pointwise Multiplication of Complex-Valued Functions"
] |
proofwiki-19875 | Combination Theorem for Complex Derivatives/Product Rule/Proof 2 | Let $D$ be an open subset of the set of complex numbers.
Let $f, g: D \to \C$ be complex-differentiable functions on $D$
Let $f g$ denote the pointwise product of the functions $f$ and $g$.
Then $f g$ is complex-differentiable in $D$, and its derivative $\paren {f g}'$ is defined by:
:$\map {\paren {f g}'} z = \map {f'... | Denote the open ball of $0$ by radius $r \in \R_{>0}$ as $\map {B_r} 0$.
Let $z \in D$.
By the Epsilon-Function Complex Differentiability Condition, it follows that there exists $r \in \R_{>0}$ such that for all $h \in \map {B_r} 0 \setminus \set 0$:
:$\map f {z + h} = \map f z + \map h {\map {f'} z + \map {\epsilon_f}... | Let $D$ be an [[Definition:Open Set (Complex Analysis)|open]] [[Definition:Subset|subset]] of the [[Definition:Set|set]] of [[Definition:Complex Number|complex numbers]].
Let $f, g: D \to \C$ be [[Definition:Complex-Differentiable Function|complex-differentiable functions]] on $D$
Let $f g$ denote the [[Definition:P... | Denote the [[Definition:Open Ball|open ball]] of $0$ by [[Definition:Radius of Open Ball|radius]] $r \in \R_{>0}$ as $\map {B_r} 0$.
Let $z \in D$.
By the [[Epsilon-Function Complex Differentiability Condition]], it follows that there exists $r \in \R_{>0}$ such that for all $h \in \map {B_r} 0 \setminus \set 0$:
:$... | Combination Theorem for Complex Derivatives/Product Rule/Proof 2 | https://proofwiki.org/wiki/Combination_Theorem_for_Complex_Derivatives/Product_Rule/Proof_2 | https://proofwiki.org/wiki/Combination_Theorem_for_Complex_Derivatives/Product_Rule/Proof_2 | [
"Combination Theorem for Complex Derivatives"
] | [
"Definition:Open Set/Complex Analysis",
"Definition:Subset",
"Definition:Set",
"Definition:Complex Number",
"Definition:Differentiable Mapping/Complex Function",
"Definition:Pointwise Multiplication of Complex-Valued Functions",
"Definition:Complex Function",
"Definition:Differentiable Mapping/Complex... | [
"Definition:Open Ball",
"Definition:Open Ball/Radius",
"Epsilon-Function Differentiability Condition/Complex Case",
"Definition:Complex Function",
"Definition:Convergent Mapping/Complex Function",
"Combination Theorem for Limits of Functions/Complex/Product Rule",
"Combination Theorem for Limits of Func... |
proofwiki-19876 | Prime Gap Size is Unbounded | Let $n \in \N$ be a natural number, arbitrarily large.
Then there exist consecutive prime numbers $p_1$ and $p_2$ such that:
:$p_2 - p_1 > n$
That is, the size of prime gaps is unbounded.
That is, there are blocks of consecutive composite numbers whose length exceeds any given $n \in \N$. | Let $p$ be a prime number greater than $n + 1$.
Consider the primorial:
:$q = 2 \times 3 \times 5 \times \cdots \times p$
All of the $p - 1$ numbers:
:$q + 2, q + 3, q + 4, \ldots, q + p$
are composite.
Hence the result.
{{qed}} | Let $n \in \N$ be a [[Definition:Natural Number|natural number]], arbitrarily large.
Then there exist consecutive [[Definition:Prime Number|prime numbers]] $p_1$ and $p_2$ such that:
:$p_2 - p_1 > n$
That is, the size of [[Definition:Prime Gap|prime gaps]] is [[Definition:Unbounded|unbounded]].
That is, there are bl... | Let $p$ be a [[Definition:Prime Number|prime number]] greater than $n + 1$.
Consider the [[Definition:Primorial|primorial]]:
:$q = 2 \times 3 \times 5 \times \cdots \times p$
All of the $p - 1$ numbers:
:$q + 2, q + 3, q + 4, \ldots, q + p$
are [[Definition:Composite Number|composite]].
Hence the result.
{{qed}} | Prime Gap Size is Unbounded | https://proofwiki.org/wiki/Prime_Gap_Size_is_Unbounded | https://proofwiki.org/wiki/Prime_Gap_Size_is_Unbounded | [
"Prime Gaps"
] | [
"Definition:Natural Numbers",
"Definition:Prime Number",
"Definition:Prime Gap",
"Definition:Unbounded",
"Definition:Composite Number"
] | [
"Definition:Prime Number",
"Definition:Primorial",
"Definition:Composite Number"
] |
proofwiki-19877 | Linear Bound between Complex Function and Derivative | Let $f: D \to \C$ be a complex-differentiable function, where $D \subseteq \C$ is an open set.
Let $z_0 \in D$.
Let $\epsilon \in \R_{>0}$.
Then there exists $r \in \R_{>0}$ such that for all $z \in \map {B_r} {z_0}$:
:$\size {\map f z - \map f {z_0} - \map {f'} {z_0} \paren {z - z_0} } < \epsilon \size {z - z_0}$
wher... | {{begin-eqn}}
{{eqn | l = \map {f'} {z_0}
| r = \ds \lim_{z \mathop \to z_0} \dfrac{ \map f z - \map f {z_0} } {z - z_0}
| c = {{Defof|Complex-Differentiable at Point}}
}}
{{eqn | ll = \leadsto
| l = 0
| r = \lim_{z_0 \mathop \to z} \dfrac {\map f z - \map f {z_0} - \map {f'} {z_0} \paren{ z - ... | Let $f: D \to \C$ be a [[Definition:Complex-Differentiable Function|complex-differentiable function]], where $D \subseteq \C$ is an [[Definition:Open Set (Complex Analysis)|open set]].
Let $z_0 \in D$.
Let $\epsilon \in \R_{>0}$.
Then there exists $r \in \R_{>0}$ such that for all $z \in \map {B_r} {z_0}$:
:$\size... | {{begin-eqn}}
{{eqn | l = \map {f'} {z_0}
| r = \ds \lim_{z \mathop \to z_0} \dfrac{ \map f z - \map f {z_0} } {z - z_0}
| c = {{Defof|Complex-Differentiable at Point}}
}}
{{eqn | ll = \leadsto
| l = 0
| r = \lim_{z_0 \mathop \to z} \dfrac {\map f z - \map f {z_0} - \map {f'} {z_0} \paren{ z - ... | Linear Bound between Complex Function and Derivative | https://proofwiki.org/wiki/Linear_Bound_between_Complex_Function_and_Derivative | https://proofwiki.org/wiki/Linear_Bound_between_Complex_Function_and_Derivative | [
"Complex Differential Calculus"
] | [
"Definition:Differentiable Mapping/Complex Function",
"Definition:Open Set/Complex Analysis",
"Definition:Open Ball",
"Definition:Open Ball/Center",
"Definition:Radius of Open Ball "
] | [
"Combination Theorem for Limits of Functions/Complex",
"Modulus of Limit",
"Complex Modulus of Quotient of Complex Numbers",
"Definition:Limit of Complex Function"
] |
proofwiki-19878 | Supremum Operator Norm of Linear Transformation is Bounded Above by Hilbert-Schmidt Norm | Let $T_A : \R^n \to \R^m$ be the linear transformation such that:
:$\forall \mathbf x \in \R^n : T_A \mathbf x := A \mathbf x$
Let $\norm {\, \cdot \,}$ be the supremum operator norm.
Let $\norm {\, \cdot \,}_{HS}$ be the Hilbert Schmidt norm.
Then:
:$\norm {T_A} \le \norm A_{HS}$ | We have that Norms on Finite-Dimensional Real Vector Space are Equivalent.
Choose the Euclidean norm.
Let $X = \struct {\R^n, \norm {\, \cdot \,}_2}$ and $Y = \struct {\R^m, \norm {\, \cdot \,}_2}$ be normed vector spaces.
Let the matrix $A \in \R^{m \times n}$ be given by:
:$A = \begin {bmatrix}
a_{1 1} & \cdots & a_{... | Let $T_A : \R^n \to \R^m$ be the [[Definition:Linear Transformation on Vector Space|linear transformation]] such that:
:$\forall \mathbf x \in \R^n : T_A \mathbf x := A \mathbf x$
Let $\norm {\, \cdot \,}$ be the [[Definition:Supremum Operator Norm|supremum operator norm]].
Let $\norm {\, \cdot \,}_{HS}$ be the [[De... | We have that [[Norms on Finite-Dimensional Real Vector Space are Equivalent]].
Choose the [[Definition:Euclidean Norm|Euclidean norm]].
Let $X = \struct {\R^n, \norm {\, \cdot \,}_2}$ and $Y = \struct {\R^m, \norm {\, \cdot \,}_2}$ be [[Definition:Normed Vector Space|normed vector spaces]].
Let the [[Definition:Matr... | Supremum Operator Norm of Linear Transformation is Bounded Above by Hilbert-Schmidt Norm | https://proofwiki.org/wiki/Supremum_Operator_Norm_of_Linear_Transformation_is_Bounded_Above_by_Hilbert-Schmidt_Norm | https://proofwiki.org/wiki/Supremum_Operator_Norm_of_Linear_Transformation_is_Bounded_Above_by_Hilbert-Schmidt_Norm | [
"Inequalities"
] | [
"Definition:Linear Transformation/Vector Space",
"Definition:Supremum Operator Norm",
"Definition:Hilbert-Schmidt Norm"
] | [
"Norms on Finite-Dimensional Real Vector Space are Equivalent",
"Definition:Euclidean Norm",
"Definition:Normed Vector Space",
"Definition:Matrix",
"Set of Linear Transformations is Isomorphic to Matrix Space",
"Definition:Linear Transformation/Vector Space",
"Linear Transformations between Finite-Dimen... |
proofwiki-19879 | Goursat's Integral Lemma | Let $f: D \to \C$ be a holomorphic function in $D$, where $D \subseteq \C$ is a connected domain.
Let $\triangle$ be a triangle embedded in the complex plane $\C$ with boundary $\partial \triangle$.
Let $\partial \triangle \cup \Int \triangle \subseteq D$, where $\Int \triangle$ is the interior of $\partial \triangle$.... | {{tidy|Sorry, still stuff to do to improve / tidy this further. Please leave this template in place til I've had a chance to give it a workover.}}
We will create a sequence of triangles $\sequence {\triangle_n}_{n \mathop \in \N}$ by an inductive process.
Put $\triangle_0 = \triangle$ as the first term of the sequence.... | Let $f: D \to \C$ be a [[Definition:Holomorphic Complex Function|holomorphic function]] in $D$, where $D \subseteq \C$ is a [[Definition:Connected Domain (Complex Analysis)|connected domain]].
Let $\triangle$ be a [[Definition:Triangle (Geometry)|triangle]] embedded in the [[Definition:Complex Plane|complex plane]] $\... | {{tidy|Sorry, still stuff to do to improve / tidy this further. Please leave this template in place til I've had a chance to give it a workover.}}
We will create a [[Definition:Sequence|sequence]] of [[Definition:Triangle (Geometry)|triangles]] $\sequence {\triangle_n}_{n \mathop \in \N}$ by an inductive process.
Put... | Goursat's Integral Lemma | https://proofwiki.org/wiki/Goursat's_Integral_Lemma | https://proofwiki.org/wiki/Goursat's_Integral_Lemma | [
"Complex Contour Integrals"
] | [
"Definition:Holomorphic Function/Complex Plane",
"Definition:Connected Domain (Complex Analysis)",
"Definition:Triangle (Geometry)",
"Definition:Complex Number/Complex Plane",
"Definition:Boundary (Topology)",
"Definition:Interior of Simple Closed Contour",
"Definition:Contour Integral/Complex/Closed",
... | [
"Definition:Sequence",
"Definition:Triangle (Geometry)",
"Definition:Term of Sequence",
"Definition:Sequence",
"Definition:Polygon/Vertex",
"Definition:Convex Set (Vector Space)/Line Segment",
"Definition:Line/Endpoint",
"Definition:Line/Midpoint",
"Definition:Convex Set (Vector Space)/Line Segment"... |
proofwiki-19880 | Continuous Image of Path-Connected Set is Path-Connected/Metric Space | Let $M_1, M_2$ be metric spaces whose metrics are $d_1, d_2$ respectively.
Let $f: M_1 \to M_2$ be a continuous mapping.
Let $S \subseteq M_1$ be a path-connected subspace of $M_1$.
Then $f \sqbrk S$ is a path-connected subspace of $M_2$. | Let $\map f s, \map f {s'} \in f \sqbrk S$, for some $s, s' \in S$.
Let $\mathbb I$ denote the closed unit interval:
:$\mathbb I = \closedint 0 1$
Let $p: \mathbb I \to S$ be a continuous mapping such that:
:$\map p 0 = s, \map p 1 = s'$
Such a $p$ exists since $S$ is a path-connected subspace of $M_1$.
Now define $q: ... | Let $M_1, M_2$ be [[Definition:Metric Space|metric spaces]] whose [[Definition:Metric|metrics]] are $d_1, d_2$ respectively.
Let $f: M_1 \to M_2$ be a [[Definition:Continuous Mapping (Metric Spaces)|continuous mapping]].
Let $S \subseteq M_1$ be a [[Definition:Path-Connected Metric Subspace|path-connected subspace]] ... | Let $\map f s, \map f {s'} \in f \sqbrk S$, for some $s, s' \in S$.
Let $\mathbb I$ denote the [[Definition:Closed Unit Interval|closed unit interval]]:
:$\mathbb I = \closedint 0 1$
Let $p: \mathbb I \to S$ be a [[Definition:Continuous Mapping (Metric Spaces)|continuous mapping]] such that:
:$\map p 0 = s, \map p 1... | Continuous Image of Path-Connected Set is Path-Connected/Metric Space | https://proofwiki.org/wiki/Continuous_Image_of_Path-Connected_Set_is_Path-Connected/Metric_Space | https://proofwiki.org/wiki/Continuous_Image_of_Path-Connected_Set_is_Path-Connected/Metric_Space | [
"Path-Connected Metric Spaces",
"Metric Spaces"
] | [
"Definition:Metric Space",
"Definition:Metric Space/Metric",
"Definition:Continuous Mapping (Metric Space)",
"Definition:Path-Connected/Metric Space/Subset",
"Definition:Path-Connected/Metric Space/Subset"
] | [
"Definition:Real Interval/Unit Interval/Closed",
"Definition:Continuous Mapping (Metric Space)",
"Definition:Path-Connected/Metric Space/Subset",
"Definition:Continuous Mapping (Metric Space)",
"Definition:Composition of Mappings",
"Definition:Path (Topology)"
] |
proofwiki-19881 | Path-Connectedness is Preserved under Homeomorphism | Let $\struct {T_1, \tau_1}, \struct {T_2, \tau_2}$ be topological spaces.
Let $\phi: T_1 \to T_2$ be a homeomorphism.
Let $S \subseteq T_1$ be a subset of $T_1$.
Let $S$ be path-connected in $\struct {T_1, \tau_1}$.
Then $\phi \sqbrk S$ is path-connected in $\struct {T_2, \tau_2}$.
That is, path-connectedness is a topo... | By definition of homeomorphism, $\phi$ is a continuous mapping.
The result now follows from Continuous Image of Path-Connected Set is Path-Connected.
{{qed}}
Category:Topological Properties
Category:Path-Connected Sets
7susxuru3awz4i4gh6u86ag6wha5h0r | Let $\struct {T_1, \tau_1}, \struct {T_2, \tau_2}$ be [[Definition:Topological Space|topological spaces]].
Let $\phi: T_1 \to T_2$ be a [[Definition:Homeomorphism (Topological Spaces)|homeomorphism]].
Let $S \subseteq T_1$ be a [[Definition:Subset|subset]] of $T_1$.
Let $S$ be [[Definition:Path-Connected Set|path-co... | By [[Definition:Homeomorphism/Topological Spaces/Definition 4|definition of homeomorphism]], $\phi$ is a [[Definition:Everywhere Continuous Mapping (Topology)|continuous mapping]].
The result now follows from [[Continuous Image of Path-Connected Set is Path-Connected]].
{{qed}}
[[Category:Topological Properties]]
[[C... | Path-Connectedness is Preserved under Homeomorphism | https://proofwiki.org/wiki/Path-Connectedness_is_Preserved_under_Homeomorphism | https://proofwiki.org/wiki/Path-Connectedness_is_Preserved_under_Homeomorphism | [
"Topological Properties",
"Path-Connected Sets"
] | [
"Definition:Topological Space",
"Definition:Homeomorphism/Topological Spaces",
"Definition:Subset",
"Definition:Path-Connected/Set",
"Definition:Path-Connected/Set",
"Definition:Path-Connected/Set",
"Definition:Topological Property"
] | [
"Definition:Homeomorphism/Topological Spaces/Definition 4",
"Definition:Continuous Mapping (Topology)/Everywhere",
"Continuous Image of Path-Connected Set is Path-Connected",
"Category:Topological Properties",
"Category:Path-Connected Sets"
] |
proofwiki-19882 | Holomorphic Function is Continuously Differentiable | Let $U \subseteq \C$ be an open set.
Let $f : U \to \C$ be a holomorphic function.
Then $f$ is continuously differentiable in $U$. | Let $z \in U$.
The derivatives $\map {f'} z$ and $\map {f' '} z$ exist by Cauchy's Integral Formula for Derivatives.
Therefore $f'$ is continuous at $z$ by Complex-Differentiable Function is Continuous.
{{qed}} | Let $U \subseteq \C$ be an [[Definition:Open Subset of Complex Plane|open set]].
Let $f : U \to \C$ be a [[Definition:Holomorphic Complex Function|holomorphic function]].
Then $f$ is [[Definition:Continuously Differentiable|continuously differentiable]] in $U$. | Let $z \in U$.
The [[Definition:Derivative of Complex Function|derivatives]] $\map {f'} z$ and $\map {f' '} z$ exist by [[Cauchy's Integral Formula for Derivatives]].
Therefore $f'$ is [[Definition:Continuous Complex Function|continuous]] at $z$ by [[Complex-Differentiable Function is Continuous]].
{{qed}} | Holomorphic Function is Continuously Differentiable | https://proofwiki.org/wiki/Holomorphic_Function_is_Continuously_Differentiable | https://proofwiki.org/wiki/Holomorphic_Function_is_Continuously_Differentiable | [
"Holomorphic Functions"
] | [
"Definition:Open Set/Complex Analysis",
"Definition:Holomorphic Function/Complex Plane",
"Definition:Continuously Differentiable"
] | [
"Definition:Derivative/Complex Function",
"Cauchy's Integral Formula/General Result",
"Definition:Continuous Complex Function",
"Complex-Differentiable Function is Continuous"
] |
proofwiki-19883 | Simple Connectedness is Preserved under Homeomorphism | Let $\struct {T_1, \tau_1}, \struct {T_2, \tau_2}$ be topological spaces.
Let $\phi: T_1 \to T_2$ be a homeomorphism.
Let $S \subseteq T_1$ be a subset of $T_1$.
Let $S$ be simply connected in $\struct {T_1, \tau_1}$.
Then $\phi \sqbrk S$ is simply connected in $\struct {T_2, \tau_2}$.
That is, simple connectedness is ... | By {{Defof|Simply Connected}}, $S$ is path-connected.
Path-Connectedness is Preserved under Homeomorphism shows that $\phi \sqbrk S$ is path-connected.
Let $x \in S$.
Let:
:$\gamma_1 : \closedint 0 1 \to \phi \sqbrk S$
:$\gamma_2 : \closedint 0 1 \to \phi \sqbrk S$
be two loops in $\phi \sqbrk S$.
Let $\gamma_1, \gamma... | Let $\struct {T_1, \tau_1}, \struct {T_2, \tau_2}$ be [[Definition:Topological Space|topological spaces]].
Let $\phi: T_1 \to T_2$ be a [[Definition:Homeomorphism (Topological Spaces)|homeomorphism]].
Let $S \subseteq T_1$ be a [[Definition:Subset|subset]] of $T_1$.
Let $S$ be [[Definition:Simply Connected|simply co... | By {{Defof|Simply Connected}}, $S$ is [[Definition:Path-Connected Space|path-connected]].
[[Path-Connectedness is Preserved under Homeomorphism]] shows that $\phi \sqbrk S$ is [[Definition:Path-Connected Space|path-connected]].
Let $x \in S$.
Let:
:$\gamma_1 : \closedint 0 1 \to \phi \sqbrk S$
:$\gamma_2 : \closed... | Simple Connectedness is Preserved under Homeomorphism | https://proofwiki.org/wiki/Simple_Connectedness_is_Preserved_under_Homeomorphism | https://proofwiki.org/wiki/Simple_Connectedness_is_Preserved_under_Homeomorphism | [
"Simply Connected Spaces",
"Topological Properties"
] | [
"Definition:Topological Space",
"Definition:Homeomorphism/Topological Spaces",
"Definition:Subset",
"Definition:Simply Connected",
"Definition:Simply Connected",
"Definition:Simply Connected",
"Definition:Topological Property"
] | [
"Definition:Path-Connected/Topological Space",
"Path-Connectedness is Preserved under Homeomorphism",
"Definition:Path-Connected/Topological Space",
"Definition:Loop (Topology)",
"Definition:Loop (Topology)/Base Point",
"Composite of Continuous Mappings is Continuous",
"Definition:Continuous Mapping (To... |
proofwiki-19884 | Equivalence of Definitions of Continuous Complex Function | {{TFAE|def = Continuous Complex Function}}
Let $A_1, A_2 \subseteq \C$ be subsets of the complex plane.
Let $f: A_1 \to A_2$ be a complex function from $A_1$ to $A_2$.
Let $a \in A_1$. | === Definition using Limit implies Epsilon-Delta Definition ===
Let $\epsilon > 0$.
By the Epsilon-Delta condition of limit, there exists $\delta > 0$, so for all $z \in A_1$:
:$0 < \cmod {z - a} < \delta \implies \cmod {\map f z - \map f a} < \epsilon$
In the case of $0 = \cmod {z - a}$, we must have $z = a$, so
:$\cm... | {{TFAE|def = Continuous Complex Function}}
Let $A_1, A_2 \subseteq \C$ be [[Definition:Subset|subsets]] of the [[Definition:Complex Plane|complex plane]].
Let $f: A_1 \to A_2$ be a [[Definition:Complex Function|complex function]] from $A_1$ to $A_2$.
Let $a \in A_1$. | === [[Definition:Continuous Complex Function/Using Limit|Definition using Limit]] implies [[Definition:Continuous Complex Function/Epsilon-Delta|Epsilon-Delta Definition]] ===
Let $\epsilon > 0$.
By the Epsilon-Delta condition of [[Definition:Limit of Complex Function|limit]], there exists $\delta > 0$, so for all $z... | Equivalence of Definitions of Continuous Complex Function | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Continuous_Complex_Function | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Continuous_Complex_Function | [
"Continuous Complex Functions"
] | [
"Definition:Subset",
"Definition:Complex Number/Complex Plane",
"Definition:Complex Function"
] | [
"Definition:Continuous Complex Function/Using Limit",
"Definition:Continuous Complex Function/Epsilon-Delta",
"Definition:Limit of Complex Function",
"Definition:Continuous Complex Function/Epsilon-Delta",
"Definition:Continuous Complex Function/Using Limit",
"Definition:Continuous Complex Function/Epsilo... |
proofwiki-19885 | Function f is Big-O of g iff g is Big-Omega of f | Let $f: \N \to \R, g: \N \to \R$ be two real sequences, expressed here as real-valued functions on the set of natural numbers $\N$.
Then:
:$\map f n \in \map \OO {\map g n}$
{{iff}}:
:$\map g n \in \map \Omega {\map f n}$
where:
:$\OO$ denotes big-$\OO$ notation
:$\Omega$ denotes big-$\Omega$ notation. | Recall the definitions:
:$\map \Omega f = \set {g: \N \to \R: \exists c_1 \in \R_{>0}: \exists n_1 \in \N: \forall n > n_1: 0 \le c_1 \cdot \size {\map f n} \le \size {\map g n} }$
:$\map \OO g = \set {f: \N \to \R: \exists c_2 \in \R_{>0}: \exists n_2 \in \N: \forall n > n_2: 0 \le \size {\map f n} \le c_2 \cdot \size... | Let $f: \N \to \R, g: \N \to \R$ be two [[Definition:Real Sequence|real sequences]], expressed here as [[Definition:Real-Valued Function|real-valued functions]] on the [[Definition:Natural Number|set of natural numbers]] $\N$.
Then:
:$\map f n \in \map \OO {\map g n}$
{{iff}}:
:$\map g n \in \map \Omega {\map f n}$
... | Recall the definitions:
:$\map \Omega f = \set {g: \N \to \R: \exists c_1 \in \R_{>0}: \exists n_1 \in \N: \forall n > n_1: 0 \le c_1 \cdot \size {\map f n} \le \size {\map g n} }$
:$\map \OO g = \set {f: \N \to \R: \exists c_2 \in \R_{>0}: \exists n_2 \in \N: \forall n > n_2: 0 \le \size {\map f n} \le c_2 \cdot \si... | Function f is Big-O of g iff g is Big-Omega of f | https://proofwiki.org/wiki/Function_f_is_Big-O_of_g_iff_g_is_Big-Omega_of_f | https://proofwiki.org/wiki/Function_f_is_Big-O_of_g_iff_g_is_Big-Omega_of_f | [
"Big-O Notation",
"Big-Omega Notation"
] | [
"Definition:Real Sequence",
"Definition:Real-Valued Function",
"Definition:Natural Numbers",
"Definition:Big-O Notation",
"Definition:Big-Omega Notation"
] | [
"Definition:Sign of Number"
] |
proofwiki-19886 | Noetherian Hausdorff Space is Finite Discrete Space | Let $\struct {S, \tau}$ be a Noetherian Hausdorff space.
Let $S \ne \O$.
Then $\struct {S, \tau}$ is a finite discrete space. | {{ProofWanted}}
Category:Hausdorff Spaces
Category:Noetherian Topological Spaces
Category:Discrete Topologies
Category:Finite Topological Spaces
3hu53kqjr3kn79vdbik404icicva5uq | Let $\struct {S, \tau}$ be a [[Definition:Noetherian Topological Space|Noetherian]] [[Definition:Hausdorff Space|Hausdorff space]].
Let $S \ne \O$.
Then $\struct {S, \tau}$ is a [[Definition:Finite Discrete Topology|finite discrete space]]. | {{ProofWanted}}
[[Category:Hausdorff Spaces]]
[[Category:Noetherian Topological Spaces]]
[[Category:Discrete Topologies]]
[[Category:Finite Topological Spaces]]
3hu53kqjr3kn79vdbik404icicva5uq | Noetherian Hausdorff Space is Finite Discrete Space | https://proofwiki.org/wiki/Noetherian_Hausdorff_Space_is_Finite_Discrete_Space | https://proofwiki.org/wiki/Noetherian_Hausdorff_Space_is_Finite_Discrete_Space | [
"Hausdorff Spaces",
"Noetherian Topological Spaces",
"Discrete Topologies",
"Finite Topological Spaces"
] | [
"Definition:Noetherian Topological Space",
"Definition:T2 Space",
"Definition:Discrete Topology/Finite"
] | [
"Category:Hausdorff Spaces",
"Category:Noetherian Topological Spaces",
"Category:Discrete Topologies",
"Category:Finite Topological Spaces"
] |
proofwiki-19887 | Sum of Additive Function Values is Well-Defined | Let $\SS$ be an algebra of sets.
Let $f: \SS \to \overline \R$ be an additive function on $\SS$.
Let $A, B \in \SS$.
Then the sum
:$\map f A + \map f B$
is well-defined in the extended real numbers $\overline \R$. | Suppose the sum $\map f A + \map f B$ is void.
By {{Defof|Extended Real Addition}}, this happens when the sum is $\paren{ +\infty } + \paren{ -\infty }$, or $\paren{ -\infty } + \paren{ +\infty }$.
{{WLOG}}, assume that $\map f A = +\infty$, and $\map f B = -\infty$.
Set Difference and Intersection are Disjoint shows t... | Let $\SS$ be an [[Definition:Algebra of Sets|algebra of sets]].
Let $f: \SS \to \overline \R$ be an [[Definition:Additive Function (Measure Theory)|additive function]] on $\SS$.
Let $A, B \in \SS$.
Then the [[Definition:Extended Real Addition|sum]]
:$\map f A + \map f B$
is well-defined in the [[Definition:Extend... | Suppose the [[Definition:Extended Real Addition|sum]] $\map f A + \map f B$ is void.
By {{Defof|Extended Real Addition}}, this happens when the sum is $\paren{ +\infty } + \paren{ -\infty }$, or $\paren{ -\infty } + \paren{ +\infty }$.
{{WLOG}}, assume that $\map f A = +\infty$, and $\map f B = -\infty$.
[[Set Diffe... | Sum of Additive Function Values is Well-Defined | https://proofwiki.org/wiki/Sum_of_Additive_Function_Values_is_Well-Defined | https://proofwiki.org/wiki/Sum_of_Additive_Function_Values_is_Well-Defined | [
"Set Systems",
"Additive Functions"
] | [
"Definition:Algebra of Sets",
"Definition:Additive Function (Measure Theory)",
"Definition:Extended Real Addition",
"Definition:Extended Real Number Line"
] | [
"Definition:Extended Real Addition",
"Set Difference and Intersection are Disjoint",
"Definition:Disjoint Sets",
"Set Difference and Intersection are Disjoint",
"Definition:Disjoint Sets",
"Definition:Additive Function (Measure Theory)",
"Set Difference Union Intersection",
"Definition:Additive Functi... |
proofwiki-19888 | Path-Homotopic Paths have Path Homotopy | Let $X$ be a topological space.
Let $f, g: \closedint 0 1 \to X$ be paths.
Then $f$ and $g$ are path-homotopic, {{iff}} there exists a path homotopy between $f$ and $g$. | === Necessary Condition ===
By {{Defof|Path-Homotopic}}, there exists a homotopy $H : \closedint 0 1 \times \closedint 0 1 \to X$ such that:
:$\forall s \in \closedint 0 1: \map H {s, 0} = \map f s$
:$\forall s \in \closedint 0 1: \map H {s, 1} = \map g s$
Homotopic Paths have Same Endpoints shows that:
:$\map f 0 = \m... | Let $X$ be a [[Definition:Topological Space|topological space]].
Let $f, g: \closedint 0 1 \to X$ be [[Definition:Path (Topology)|paths]].
Then $f$ and $g$ are [[Definition:Path-Homotopic|path-homotopic]], {{iff}} there exists a [[Definition:Path Homotopy|path homotopy]] between $f$ and $g$. | === Necessary Condition ===
By {{Defof|Path-Homotopic}}, there exists a [[Definition:Free Homotopy|homotopy]] $H : \closedint 0 1 \times \closedint 0 1 \to X$ such that:
:$\forall s \in \closedint 0 1: \map H {s, 0} = \map f s$
:$\forall s \in \closedint 0 1: \map H {s, 1} = \map g s$
[[Homotopic Paths have Same E... | Path-Homotopic Paths have Path Homotopy | https://proofwiki.org/wiki/Path-Homotopic_Paths_have_Path_Homotopy | https://proofwiki.org/wiki/Path-Homotopic_Paths_have_Path_Homotopy | [
"Homotopy Theory"
] | [
"Definition:Topological Space",
"Definition:Path (Topology)",
"Definition:Homotopy/Path",
"Definition:Homotopy/Path/Path Homotopy"
] | [
"Definition:Homotopy/Free",
"Homotopic Paths have Same Endpoints",
"Definition:Homotopy/Relative",
"Definition:Homotopy/Relative"
] |
proofwiki-19889 | Diagonal Operator over 2-Sequence Space is Continuous Linear Transformation | Let $\Bbb K = \set {\R, \C}$.
Let $\sequence {\lambda_n}_{n \mathop \in \N_{> 0} }$ be a bounded sequence in $\Bbb K$.
Let $\struct {\ell^2, \norm {\, \cdot \,}_2}$ be the normed $2$-sequence space.
Let $\mathbf x = \tuple {a_1, a_2, a_3, \ldots} \in \ell^2$.
Suppose $\Lambda : \ell^2 \to \ell^2$ is a diagonal operator... | === Linearity ===
Let $\sequence {a_n}_{n \mathop \in \N_{> 0} }, \sequence {b_n}_{n \mathop \in \N_{> 0} } \in \ell^2$.
{{begin-eqn}}
{{eqn | l = \map \Lambda {k \sequence {a_n} + \sequence {b_n} }
| r = \map \Lambda {\sequence {k a_n + b_n} }
| c = $P$-sequence space is a vector space
}}
{{eqn | r = \sequ... | Let $\Bbb K = \set {\R, \C}$.
Let $\sequence {\lambda_n}_{n \mathop \in \N_{> 0} }$ be a [[Definition:Bounded Sequence|bounded sequence]] in $\Bbb K$.
Let $\struct {\ell^2, \norm {\, \cdot \,}_2}$ be the [[P-Sequence Space with P-Norm forms Normed Vector Space|normed $2$-sequence space]].
Let $\mathbf x = \tuple {a_... | === Linearity ===
Let $\sequence {a_n}_{n \mathop \in \N_{> 0} }, \sequence {b_n}_{n \mathop \in \N_{> 0} } \in \ell^2$.
{{begin-eqn}}
{{eqn | l = \map \Lambda {k \sequence {a_n} + \sequence {b_n} }
| r = \map \Lambda {\sequence {k a_n + b_n} }
| c = [[P-Sequence Space with Pointwise Addition and Pointwis... | Diagonal Operator over 2-Sequence Space is Continuous Linear Transformation | https://proofwiki.org/wiki/Diagonal_Operator_over_2-Sequence_Space_is_Continuous_Linear_Transformation | https://proofwiki.org/wiki/Diagonal_Operator_over_2-Sequence_Space_is_Continuous_Linear_Transformation | [
"Linear Transformations",
"Continuous Linear Transformations",
"Continuous Mappings"
] | [
"Definition:Bounded Sequence",
"P-Sequence Space with P-Norm forms Normed Vector Space",
"Definition:Diagonal Operator"
] | [
"P-Sequence Space with Pointwise Addition and Pointwise Scalar Multiplication on Ring of Sequences forms Vector Space",
"P-Sequence Space with Pointwise Addition and Pointwise Scalar Multiplication on Ring of Sequences forms Vector Space",
"Definition:Linear Transformation/Vector Space"
] |
proofwiki-19890 | Supremum Operator Norm of Diagonal Operator over 2-Sequence Space | Let $\Bbb K = \set {\R, \C}$.
Let $\sequence {\lambda_n}_{n \mathop \in \N_{> 0} }$ be a bounded sequence in $\Bbb K$.
Let $\struct {\ell^2, \norm {\, \cdot \,}_2}$ be the normed $2$-sequence space.
Let $\mathbf x = \tuple {a_1, a_2, a_3, \ldots} \in \ell^2$.
Suppose $\Lambda : \ell^2 \to \ell^2$ be a diagonal operator... | From Diagonal Operator over 2-Sequence Space is Continuous Linear Transformation:
:$\ds \norm {\Lambda \mathbf x }_2^2 \le \paren {\sup_{n \mathop \in \N_{> 0} } \size{\lambda_n} }^2 \norm {\mathbf x}_2^2$
Therefore:
:$\ds \norm {\Lambda \mathbf x}_2 \le \sup_{n \mathop \in \N_{> 0} } \size{\lambda_n} \norm {\mathbf x}... | Let $\Bbb K = \set {\R, \C}$.
Let $\sequence {\lambda_n}_{n \mathop \in \N_{> 0} }$ be a [[Definition:Bounded Sequence|bounded sequence]] in $\Bbb K$.
Let $\struct {\ell^2, \norm {\, \cdot \,}_2}$ be the [[P-Sequence Space with P-Norm forms Normed Vector Space|normed $2$-sequence space]].
Let $\mathbf x = \tuple {a_... | From [[Diagonal Operator over 2-Sequence Space is Continuous Linear Transformation]]:
:$\ds \norm {\Lambda \mathbf x }_2^2 \le \paren {\sup_{n \mathop \in \N_{> 0} } \size{\lambda_n} }^2 \norm {\mathbf x}_2^2$
Therefore:
:$\ds \norm {\Lambda \mathbf x}_2 \le \sup_{n \mathop \in \N_{> 0} } \size{\lambda_n} \norm {\ma... | Supremum Operator Norm of Diagonal Operator over 2-Sequence Space | https://proofwiki.org/wiki/Supremum_Operator_Norm_of_Diagonal_Operator_over_2-Sequence_Space | https://proofwiki.org/wiki/Supremum_Operator_Norm_of_Diagonal_Operator_over_2-Sequence_Space | [
"Supremum Norm",
"Examples of Norms"
] | [
"Definition:Bounded Sequence",
"P-Sequence Space with P-Norm forms Normed Vector Space",
"Definition:Diagonal Operator",
"Definition:Supremum Operator Norm"
] | [
"Diagonal Operator over 2-Sequence Space is Continuous Linear Transformation",
"Definition:Supremum Operator Norm",
"Supremum Operator Norm as Universal Upper Bound",
"Definition:Supremum of Real Sequence",
"Definition:Left Hand Side",
"Definition:Real Number",
"Definition:Right Hand Side"
] |
proofwiki-19891 | Reparameterization of Directed Smooth Curve with Given Domain | Let $\gamma: \closedint a b \to \C$ be a smooth path in the complex plane.
Let $C$ be a directed smooth curve with parameterization $\gamma$.
Let $\closedint { a_0 }{ b_0 }$ be a closed real interval, where $a_0 < b_0$.
Then there exists a smooth path
:$\gamma_0 : \closedint { a_0 }{ b_0 } \to \C$
that is a reparameter... | Define $\phi : \closedint a b \to \closedint { a_0 }{ b_0 }$ by:
:$\map{ \phi }{ t } = \dfrac{ b_0 - a_0 }{ b - a } \paren{ t - a } + a_0$
Power Rule for Derivatives shows that
:$\map{ \phi' }{ t } = \dfrac{ b_0 - a_0 }{ b - a }$
Real Function with Strictly Positive Derivative is Strictly Increasing shows that $\phi$ i... | Let $\gamma: \closedint a b \to \C$ be a [[Definition:Smooth Path (Complex Analysis)|smooth path]] in the [[Definition:Complex Plane|complex plane]].
Let $C$ be a [[Definition:Directed Smooth Curve (Complex Plane)|directed smooth curve]] with [[Definition:Parameterization of Directed Smooth Curve (Complex Plane)|param... | Define $\phi : \closedint a b \to \closedint { a_0 }{ b_0 }$ by:
:$\map{ \phi }{ t } = \dfrac{ b_0 - a_0 }{ b - a } \paren{ t - a } + a_0$
[[Power Rule for Derivatives]] shows that
:$\map{ \phi' }{ t } = \dfrac{ b_0 - a_0 }{ b - a }$
[[Real Function with Strictly Positive Derivative is Strictly Increasing]] shows... | Reparameterization of Directed Smooth Curve with Given Domain | https://proofwiki.org/wiki/Reparameterization_of_Directed_Smooth_Curve_with_Given_Domain | https://proofwiki.org/wiki/Reparameterization_of_Directed_Smooth_Curve_with_Given_Domain | [
"Directed Smooth Curves (Complex Plane)"
] | [
"Definition:Smooth Path/Complex",
"Definition:Complex Number/Complex Plane",
"Definition:Directed Smooth Curve/Complex Plane",
"Definition:Directed Smooth Curve/Parameterization/Complex Plane",
"Definition:Real Interval/Closed",
"Definition:Smooth Path/Complex",
"Definition:Directed Smooth Curve/Paramet... | [
"Power Rule for Derivatives",
"Real Function with Strictly Positive Derivative is Strictly Increasing",
"Definition:Strictly Increasing/Real Function",
"Strictly Monotone Real Function is Bijective",
"Definition:Bijection",
"Definition:Directed Smooth Curve/Complex Plane",
"Definition:Directed Smooth Cu... |
proofwiki-19892 | Interior of Simple Closed Contour is Well-Defined | Let $C$ be a simple closed contour in the complex plane.
Let $f : \closedint 0 1 \to \R^2$ be a Jordan curve.
Let $\phi : \R^2 \to \C$ be defined by:
:$\map \phi {x, y} = x + i y$
Let $\Img C = \map \phi {\Img f}$, where $\Img C$ denotes the image of $C$, and $\Img f$ denotes the image of $f$.
Then the interior of $C$:... | Let $C$ be defined as a concatenation of a (finite) sequence of directed smooth curves $\sequence {C_1, \ldots, C_n}$.
We show existence of a Jordan curve $f : \closedint 0 1 \to \R^2$ that fulfills the criteria $\map \phi {\Img f} = \Img C$.
Reparameterization of Directed Smooth Curve with Given Domain shows that $C_k... | Let $C$ be a [[Definition:Simple Contour (Complex Plane)|simple]] [[Definition:Closed Contour (Complex Plane)|closed contour]] in the [[Definition:Complex Plane|complex plane]].
Let $f : \closedint 0 1 \to \R^2$ be a [[Definition:Jordan Curve|Jordan curve]].
Let $\phi : \R^2 \to \C$ be defined by:
:$\map \phi {x, y}... | Let $C$ be defined as a [[Definition:Concatenation of Contours (Complex Plane)|concatenation]] of a [[Definition:Finite Sequence|(finite) sequence]] of [[Definition:Directed Smooth Curve (Complex Plane)|directed smooth curves]] $\sequence {C_1, \ldots, C_n}$.
We show existence of a [[Definition:Jordan Curve|Jordan cur... | Interior of Simple Closed Contour is Well-Defined | https://proofwiki.org/wiki/Interior_of_Simple_Closed_Contour_is_Well-Defined | https://proofwiki.org/wiki/Interior_of_Simple_Closed_Contour_is_Well-Defined | [
"Complex Contour Integrals"
] | [
"Definition:Contour/Simple/Complex Plane",
"Definition:Contour/Closed/Complex Plane",
"Definition:Complex Number/Complex Plane",
"Definition:Jordan Curve",
"Definition:Contour/Image/Complex Plane",
"Definition:Image (Set Theory)/Mapping/Mapping",
"Definition:Interior of Simple Closed Contour",
"Defini... | [
"Definition:Concatenation of Contours/Complex Plane",
"Definition:Finite Sequence",
"Definition:Directed Smooth Curve/Complex Plane",
"Definition:Jordan Curve",
"Reparameterization of Directed Smooth Curve with Given Domain",
"Definition:Directed Smooth Curve/Parameterization/Complex Plane",
"Definition... |
proofwiki-19893 | Jordan Curve Characterization of Simply Connected Set | Let $D \subseteq \R^2$ be an open path-connected subset of the Euclidean plane.
Then $D$ is simply connected, {{Iff}} the following condition holds:
:For all Jordan curves $\gamma : \closedint 0 1 \to \R^2$ with $\Img \gamma \subseteq D$, we have:
:$\Int \gamma \subseteq D$
Here $\Img \gamma$ denotes the image of $\gam... | === Sufficient condition ===
Suppose that $D$ is simply connected.
Let $\gamma : \closedint 0 1 \to \R^2$ be a Jordan curve with $\Img \gamma \subseteq D$.
By definition of simple connectedness, there exists a constant loop $c : \closedint 0 1 \to D$ such that $\gamma$ and $c$ are path-homotopic in $D$.
Let $H : \close... | Let $D \subseteq \R^2$ be an [[Definition:Open Set (Metric Space)|open]] [[Definition:Path-Connected Metric Subspace|path-connected subset]] of the [[Definition:Real Number Plane with Euclidean Topology |Euclidean plane]].
Then $D$ is [[Definition:Simply Connected|simply connected]], {{Iff}} the following condition h... | === Sufficient condition ===
Suppose that $D$ is [[Definition:Simply Connected|simply connected]].
Let $\gamma : \closedint 0 1 \to \R^2$ be a [[Definition:Jordan Curve|Jordan curve]] with $\Img \gamma \subseteq D$.
By definition of [[Definition:Simply Connected/Definition 4|simple connectedness]], there exists a [[... | Jordan Curve Characterization of Simply Connected Set | https://proofwiki.org/wiki/Jordan_Curve_Characterization_of_Simply_Connected_Set | https://proofwiki.org/wiki/Jordan_Curve_Characterization_of_Simply_Connected_Set | [
"Simply Connected Spaces"
] | [
"Definition:Open Set/Metric Space",
"Definition:Path-Connected/Metric Space/Subset",
"Definition:Real Number Plane with Euclidean Topology ",
"Definition:Simply Connected",
"Definition:Jordan Curve",
"Definition:Image (Set Theory)/Mapping/Mapping",
"Definition:Jordan Curve/Interior"
] | [
"Definition:Simply Connected",
"Definition:Jordan Curve",
"Definition:Simply Connected/Definition 4",
"Definition:Loop (Topology)/Constant Loop",
"Definition:Homotopy/Path",
"Definition:Homotopy/Path/Path Homotopy",
"Interior of Jordan Curve is Subset of Image of Null-Homotopy",
"Definition:Jordan Cur... |
proofwiki-19894 | Coprime Integers cannot Both be Zero | Let $a$ and $b$ be integers.
Let $a$ and $b$ be coprime.
Then it cannot be the case that $a = b = 0$. | Let $a$ and $b$ be coprime.
Then by definition:
:$\gcd \set {a, b} = 1$
{{AimForCont}} $a = b = 0$.
Then $\gcd \set {a, b}$ is undefined.
But it is not possible both:
:for $\gcd \set {a, b}$ to be undefined
:for $\gcd \set {a, b} = 1$.
Hence by Proof by Contradiction it follows that it cannot be the case that $a = b = ... | Let $a$ and $b$ be [[Definition:Integer|integers]].
Let $a$ and $b$ be [[Definition:Coprime Integers|coprime]].
Then it cannot be the case that $a = b = 0$. | Let $a$ and $b$ be [[Definition:Coprime Integers|coprime]].
Then by definition:
:$\gcd \set {a, b} = 1$
{{AimForCont}} $a = b = 0$.
Then $\gcd \set {a, b}$ is [[Definition:Undefined|undefined]].
But it is not possible both:
:for $\gcd \set {a, b}$ to be [[Definition:Undefined|undefined]]
:for $\gcd \set {a, b} = 1$... | Coprime Integers cannot Both be Zero | https://proofwiki.org/wiki/Coprime_Integers_cannot_Both_be_Zero | https://proofwiki.org/wiki/Coprime_Integers_cannot_Both_be_Zero | [
"Coprime Integers",
"Zero"
] | [
"Definition:Integer",
"Definition:Coprime/Integers"
] | [
"Definition:Coprime/Integers",
"Definition:Undefined",
"Definition:Undefined",
"Proof by Contradiction",
"Category:Coprime Integers",
"Category:Zero"
] |
proofwiki-19895 | Complex Plane is Homeomorphic to Real Plane | Let $\C$ denote the complex plane.
Let $\R^2$ denote the Euclidean plane.
Then the function $\phi : \R^2 \to \C$ defined by:
:$\map \phi {x, y} = x + i y$
is a homeomorphism between $\R^2$ and $\C$. | Define $\phi^{-1} : \C \to \R^2$ by:
:$\map {\phi^{-1} } z = \tuple {\map \Re z, \map \Im z}$
Then $\phi^{-1}$ is the inverse of $\phi$, as:
:$\map \phi {\map { \phi^{-1} } z} = \map \Re z + i \map \Im z = z$
:$\map {\phi^{-1} } {\map \phi {x, y} } = \tuple {\map \Re {x + i y}, \map \Im {x + i y} } = \tuple {x, y}$
By ... | Let $\C$ denote the [[Definition:Complex Plane|complex plane]].
Let $\R^2$ denote the [[Definition:Real Number Plane with Euclidean Topology|Euclidean plane]].
Then the [[Definition:Complex-Valued Function|function]] $\phi : \R^2 \to \C$ defined by:
:$\map \phi {x, y} = x + i y$
is a [[Definition:Homeomorphic Metr... | Define $\phi^{-1} : \C \to \R^2$ by:
:$\map {\phi^{-1} } z = \tuple {\map \Re z, \map \Im z}$
Then $\phi^{-1}$ is the [[Definition:Inverse Mapping|inverse]] of $\phi$, as:
:$\map \phi {\map { \phi^{-1} } z} = \map \Re z + i \map \Im z = z$
:$\map {\phi^{-1} } {\map \phi {x, y} } = \tuple {\map \Re {x + i y}, \map ... | Complex Plane is Homeomorphic to Real Plane | https://proofwiki.org/wiki/Complex_Plane_is_Homeomorphic_to_Real_Plane | https://proofwiki.org/wiki/Complex_Plane_is_Homeomorphic_to_Real_Plane | [
"Complex Analysis",
"Examples of Homeomorphisms"
] | [
"Definition:Complex Number/Complex Plane",
"Definition:Euclidean Space/Euclidean Topology/Real Number Plane",
"Definition:Complex-Valued Function",
"Definition:Homeomorphism/Metric Spaces"
] | [
"Definition:Inverse Mapping",
"Definition:Bijection/Definition 2",
"Definition:Bijection",
"Definition:Continuous Mapping (Metric Space)/Space",
"Complex Plane is Metric Space",
"Definition:Euclidean Metric/Real Number Plane",
"Definition:Real Number Plane",
"Definition:Continuous Mapping (Metric Spac... |
proofwiki-19896 | Polygon not Equal to Triangle has Chord | Let $P$ be a polygon that is not a triangle.
Then $P$ has a chord that lies completely in the interior of $P$. | From Polygon has Salient Angle, it follows that $P$ has a convex internal angle.
Let $A, B, C$ be vertices of $P$ such that $\angle BAC$ is a convex internal angle of the vertex $A$.
Let $BC$ subtend the angle $\angle BAC$.
Suppose $BC$ is equal to a side of $P$.
Then $P = \triangle ABC$, which would contradict our ass... | Let $P$ be a [[Definition:Polygon|polygon]] that is not a [[Definition:Triangle (Geometry)|triangle]].
Then $P$ has a [[Definition:Chord of Polygon|chord]] that lies completely in the [[Definition:Interior of Region|interior]] of $P$. | From [[Polygon has Salient Angle]], it follows that $P$ has a [[Definition:Convex Angle|convex]] [[Definition:Internal Angle|internal angle]].
Let $A, B, C$ be [[Definition:Vertex of Polygon|vertices]] of $P$ such that $\angle BAC$ is a [[Definition:Convex Angle|convex]] [[Definition:Internal Angle|internal angle]] of... | Polygon not Equal to Triangle has Chord | https://proofwiki.org/wiki/Polygon_not_Equal_to_Triangle_has_Chord | https://proofwiki.org/wiki/Polygon_not_Equal_to_Triangle_has_Chord | [
"Chords",
"Polygons"
] | [
"Definition:Polygon",
"Definition:Triangle (Geometry)",
"Definition:Polygon/Chord",
"Definition:Region"
] | [
"Polygon has Salient Angle",
"Definition:Convex Angle",
"Definition:Polygon/Internal Angle",
"Definition:Polygon/Vertex",
"Definition:Convex Angle",
"Definition:Polygon/Internal Angle",
"Definition:Polygon/Vertex",
"Definition:Angle/Subtend",
"Definition:Angle",
"Definition:Polygon/Side",
"Defin... |
proofwiki-19897 | Reductio ad Absurdum for Hilbert Proof System Instance 1 for Predicate Logic | Let $\LL$ be the language of predicate logic.
Let $\mathscr H$ be instance 1 of a Hilbert proof system for predicate logic.
Then Reductio ad Absurdum is a derived rule of $\mathscr H$:
{{:Reductio ad Absurdum/Proof Rule}} | {{MissingLinks|To the propositional tautologies}}
Suppose that $\Sigma, \neg \phi \vdash_{\mathscr H} \bot$.
By Contradictory Antecedent, $\bot \implies \phi$ is a tautology.
Therefore, $\bot \implies \phi$ is an axiom of $\mathscr H$, so that by Modus Ponendo Ponens:
:$\Sigma, \neg \phi \vdash_{\mathscr H} \phi$
By De... | Let $\LL$ be the [[Definition:Language of Predicate Logic|language of predicate logic]].
Let $\mathscr H$ be [[Definition:Hilbert Proof System Instance 1 for Predicate Logic|instance 1 of a Hilbert proof system for predicate logic]].
Then [[Reductio ad Absurdum/Proof Rule|Reductio ad Absurdum]] is a [[Definition:Der... | {{MissingLinks|To the propositional tautologies}}
Suppose that $\Sigma, \neg \phi \vdash_{\mathscr H} \bot$.
By [[Contradictory Antecedent/Proof by Truth Table|Contradictory Antecedent]], $\bot \implies \phi$ is a [[Definition:Propositional Tautology|tautology]].
Therefore, $\bot \implies \phi$ is an [[Definition:Axi... | Reductio ad Absurdum for Hilbert Proof System Instance 1 for Predicate Logic | https://proofwiki.org/wiki/Reductio_ad_Absurdum_for_Hilbert_Proof_System_Instance_1_for_Predicate_Logic | https://proofwiki.org/wiki/Reductio_ad_Absurdum_for_Hilbert_Proof_System_Instance_1_for_Predicate_Logic | [
"Reductio ad Absurdum",
"Hilbert Proof System Instance 1 for Predicate Logic"
] | [
"Definition:Language of Predicate Logic",
"Definition:Hilbert Proof System/Predicate Logic/Instance 1",
"Reductio ad Absurdum/Proof Rule",
"Definition:Derived Rule"
] | [
"Contradictory Antecedent/Proof by Truth Table",
"Definition:Tautology/Formal Semantics/Boolean Interpretations",
"Definition:Axiom/Formal Systems",
"Modus Ponendo Ponens",
"Deduction Theorem for Hilbert Proof System for Predicate Logic",
"Definition:Tautology/Formal Semantics/Boolean Interpretations",
... |
proofwiki-19898 | Proof by Contradiction for Hilbert Proof System Instance 1 for Predicate Logic | Let $\LL$ be the language of predicate logic.
Let $\mathscr H$ be instance 1 of a Hilbert proof system for predicate logic.
Then Proof by Contradiction is a derived rule of $\mathscr H$:
{{:Proof by Contradiction/Proof Rule}} | {{MissingLinks|To the propositional tautologies}}
Suppose that $\Sigma, \phi \vdash_{\mathscr H} \bot$.
By Contradictory Antecedent, $\bot \implies \neg\phi$ is a tautology.
Therefore, $\bot \implies \neg\phi$ is an axiom of $\mathscr H$, so that by Modus Ponendo Ponens:
:$\Sigma, \phi \vdash_{\mathscr H} \neg \phi$
By... | Let $\LL$ be the [[Definition:Language of Predicate Logic|language of predicate logic]].
Let $\mathscr H$ be [[Definition:Hilbert Proof System Instance 1 for Predicate Logic|instance 1 of a Hilbert proof system for predicate logic]].
Then [[Proof by Contradiction/Proof Rule|Proof by Contradiction]] is a [[Definition... | {{MissingLinks|To the propositional tautologies}}
Suppose that $\Sigma, \phi \vdash_{\mathscr H} \bot$.
By [[Contradictory Antecedent/Proof by Truth Table|Contradictory Antecedent]], $\bot \implies \neg\phi$ is a [[Definition:Propositional Tautology|tautology]].
Therefore, $\bot \implies \neg\phi$ is an [[Definition:... | Proof by Contradiction for Hilbert Proof System Instance 1 for Predicate Logic | https://proofwiki.org/wiki/Proof_by_Contradiction_for_Hilbert_Proof_System_Instance_1_for_Predicate_Logic | https://proofwiki.org/wiki/Proof_by_Contradiction_for_Hilbert_Proof_System_Instance_1_for_Predicate_Logic | [
"Proof by Contradiction",
"Hilbert Proof System Instance 1 for Predicate Logic"
] | [
"Definition:Language of Predicate Logic",
"Definition:Hilbert Proof System/Predicate Logic/Instance 1",
"Proof by Contradiction/Proof Rule",
"Definition:Derived Rule"
] | [
"Contradictory Antecedent/Proof by Truth Table",
"Definition:Tautology/Formal Semantics/Boolean Interpretations",
"Definition:Axiom/Formal Systems",
"Modus Ponendo Ponens",
"Deduction Theorem for Hilbert Proof System for Predicate Logic",
"Definition:Tautology/Formal Semantics/Boolean Interpretations",
... |
proofwiki-19899 | 3 is Divisor of one of n, n+2, n+4 | Let $n \in \Z$ be an integer such that $n > 3$.
Then at least one of $n$, $n + 2$, $n + 4$ is divisible by $3$. | Let $n \in \Z_{>0}$.
Then $n$ can be represented as either:
{{begin-eqn}}
{{eqn | l = n
| r = 3 k
}}
{{eqn | l = n
| r = 3 k + 1
}}
{{eqn | l = n
| r = 3 k + 2
}}
{{end-eqn}}
Let $n = 3 k$.
Then $n$ is divisible by $3$.
Let $n = 3 k + 1$.
Then:
:$n + 2 = 3 k + 3 = 3 \paren {k + 1}$
which is divisible ... | Let $n \in \Z$ be an [[Definition:Integer|integer]] such that $n > 3$.
Then at least one of $n$, $n + 2$, $n + 4$ is [[Definition:Divisor of Integer|divisible]] by $3$. | Let $n \in \Z_{>0}$.
Then $n$ can be represented as either:
{{begin-eqn}}
{{eqn | l = n
| r = 3 k
}}
{{eqn | l = n
| r = 3 k + 1
}}
{{eqn | l = n
| r = 3 k + 2
}}
{{end-eqn}}
Let $n = 3 k$.
Then $n$ is [[Definition:Divisor of Integer|divisible]] by $3$.
Let $n = 3 k + 1$.
Then:
:$n + 2 = 3 k + 3... | 3 is Divisor of one of n, n+2, n+4 | https://proofwiki.org/wiki/3_is_Divisor_of_one_of_n,_n+2,_n+4 | https://proofwiki.org/wiki/3_is_Divisor_of_one_of_n,_n+2,_n+4 | [
"Examples of Divisors of Integers"
] | [
"Definition:Integer",
"Definition:Divisor (Algebra)/Integer"
] | [
"Definition:Divisor (Algebra)/Integer",
"Definition:Divisor (Algebra)/Integer",
"Definition:Divisor (Algebra)/Integer"
] |
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