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-6300 | Composition of Mapping with Inclusion is Restriction | Let $S$ and $T$ be sets.
Let $f: S \to T$ be a mapping.
Let $A \subseteq S$ be a subset of the domain of $S$.
Let $i_A: A \to S$ be the inclusion mapping from $A$ to $S$.
Then:
:$f \circ i_A = f \restriction_A$
where $f \restriction_A$ denotes the restriction of $f$ to $A$. | === Equality of Domains ===
{{begin-eqn}}
{{eqn | l = \Dom {f \circ i_A}
| r = \Dom {i_A}
| c = Domain of Composite Relation
}}
{{eqn | r = A
| c = Definition of $i_A$
}}
{{eqn | r = \Dom {f \restriction_A}
| c = {{Defof|Restriction of Mapping}}
}}
{{end-eqn}}
{{qed|lemma}} | Let $S$ and $T$ be [[Definition:Set|sets]].
Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
Let $A \subseteq S$ be a [[Definition:Subset|subset]] of the [[Definition:Domain of Mapping|domain]] of $S$.
Let $i_A: A \to S$ be the [[Definition:Inclusion Mapping|inclusion mapping]] from $A$ to $S$.
Then:
:$f \cir... | === Equality of Domains ===
{{begin-eqn}}
{{eqn | l = \Dom {f \circ i_A}
| r = \Dom {i_A}
| c = [[Domain of Composite Relation]]
}}
{{eqn | r = A
| c = Definition of $i_A$
}}
{{eqn | r = \Dom {f \restriction_A}
| c = {{Defof|Restriction of Mapping}}
}}
{{end-eqn}}
{{qed|lemma}} | Composition of Mapping with Inclusion is Restriction | https://proofwiki.org/wiki/Composition_of_Mapping_with_Inclusion_is_Restriction | https://proofwiki.org/wiki/Composition_of_Mapping_with_Inclusion_is_Restriction | [
"Composite Mappings",
"Inclusion Mappings"
] | [
"Definition:Set",
"Definition:Mapping",
"Definition:Subset",
"Definition:Domain (Set Theory)/Mapping",
"Definition:Inclusion Mapping",
"Definition:Restriction/Mapping"
] | [
"Domain of Composite Relation"
] |
proofwiki-6301 | Equivalence of Subobjects is Equivalence | Let $\mathbf C$ be a metacategory.
Let $C$ be an object of $\mathbf C$.
Let $\map {\mathbf {Sub}_{\mathbf C} } C$ be the category of subobjects of $C$.
The relation $\sim$ on $\map {\mathbf {Sub}_{\mathbf C} } C$ defined by:
:$m \sim m'$ {{iff}} $m$ and $m'$ are equivalent
is an equivalence.
{{improve|That's not a rela... | By Inclusion Relation on Subobjects is Preordering, $\subseteq$ is a preordering.
From Preorder Induces Equivalence, we see that $\sim$ is the equivalence induced by $\subseteq$.
In particular, $\sim$ is an equivalence.
{{qed}}
Category:Subobjects
Category:Examples of Equivalence Relations
9k9uru39wbxbdp6s08ugdiu78mwkp... | Let $\mathbf C$ be a [[Definition:Metacategory|metacategory]].
Let $C$ be an [[Definition:Object (Category Theory)|object]] of $\mathbf C$.
Let $\map {\mathbf {Sub}_{\mathbf C} } C$ be the [[Definition:Category of Subobjects|category of subobjects]] of $C$.
The [[Definition:Relation|relation]] $\sim$ on $\map {\mat... | By [[Inclusion Relation on Subobjects is Preordering]], $\subseteq$ is a [[Definition:Preordering|preordering]].
From [[Preorder Induces Equivalence]], we see that $\sim$ is the [[Definition:Equivalence Relation|equivalence]] induced by $\subseteq$.
In particular, $\sim$ is an [[Definition:Equivalence Relation|equiva... | Equivalence of Subobjects is Equivalence | https://proofwiki.org/wiki/Equivalence_of_Subobjects_is_Equivalence | https://proofwiki.org/wiki/Equivalence_of_Subobjects_is_Equivalence | [
"Examples of Equivalence Relations",
"Subobjects",
"Examples of Equivalence Relations"
] | [
"Definition:Metacategory",
"Definition:Object (Category Theory)",
"Definition:Category of Subobjects",
"Definition:Relation",
"Definition:Equivalent Subobjects",
"Definition:Equivalence Relation"
] | [
"Inclusion Relation on Subobjects is Preordering",
"Definition:Preordering",
"Preorder Induces Equivalence",
"Definition:Equivalence Relation",
"Definition:Equivalence Relation",
"Category:Subobjects",
"Category:Examples of Equivalence Relations"
] |
proofwiki-6302 | Equivalent Subobjects have Isomorphic Domains | Let $\mathbf C$ be a metacategory.
Let $C$ be an object of $\mathbf C$.
Let $\mathbf{Sub}_{\mathbf C} \left({C}\right)$ be the category of subobjects of $C$.
Let $m, m'$ be equivalent subobjects of $C$.
Then there exists an isomorphism $f: m \to m'$. | For brevity, let us write $\operatorname{dom} m = B$ and $\operatorname{dom} m' = B'$.
Since $m$ and $m'$ are equivalent subobjects of $C$, we have that:
:$m \subseteq m'$ and $m' \subseteq m$
where $\subseteq$ denotes the inclusion relation on subobjects.
That is, there exist morphisms $f: m \to m'$ and $f': m' \to m$... | Let $\mathbf C$ be a [[Definition:Metacategory|metacategory]].
Let $C$ be an [[Definition:Object (Category Theory)|object]] of $\mathbf C$.
Let $\mathbf{Sub}_{\mathbf C} \left({C}\right)$ be the [[Definition:Category of Subobjects|category of subobjects]] of $C$.
Let $m, m'$ be [[Definition:Equivalent Subobjects|eq... | For brevity, let us write $\operatorname{dom} m = B$ and $\operatorname{dom} m' = B'$.
Since $m$ and $m'$ are [[Definition:Equivalent Subobjects|equivalent subobjects]] of $C$, we have that:
:$m \subseteq m'$ and $m' \subseteq m$
where $\subseteq$ denotes the [[Definition:Inclusion Relation on Subobjects|inclusion r... | Equivalent Subobjects have Isomorphic Domains | https://proofwiki.org/wiki/Equivalent_Subobjects_have_Isomorphic_Domains | https://proofwiki.org/wiki/Equivalent_Subobjects_have_Isomorphic_Domains | [
"Categories of Subobjects",
"Subobjects"
] | [
"Definition:Metacategory",
"Definition:Object (Category Theory)",
"Definition:Category of Subobjects",
"Definition:Equivalent Subobjects",
"Definition:Isomorphism (Category Theory)"
] | [
"Definition:Equivalent Subobjects",
"Definition:Inclusion Relation on Subobjects",
"Definition:Morphism",
"Definition:Morphism",
"Category of Subobjects is Preorder Category",
"Definition:Unique",
"Definition:Morphism",
"Definition:Identity Morphism",
"Definition:Isomorphism (Category Theory)"
] |
proofwiki-6303 | Morphism Class Equivalence is Equivalence | Let $\mathbf C$ be a metacategory.
Let $C$ be an object of $\mathbf C$.
Let $\map {\mathbf {Sub}_{\mathbf C} } C$ be the category of subobjects of $C$.
Let $\sim$ denote morphism class equivalence on the morphisms of $\map {\mathbf {Sub}_{\mathbf C} } C$.
Then $\sim$ is an equivalence. | Let us check the three conditions for an equivalence in turn. | Let $\mathbf C$ be a [[Definition:Metacategory|metacategory]].
Let $C$ be an [[Definition:Object (Category Theory)|object]] of $\mathbf C$.
Let $\map {\mathbf {Sub}_{\mathbf C} } C$ be the [[Definition:Category of Subobjects|category of subobjects]] of $C$.
Let $\sim$ denote [[Definition:Morphism Class|morphism clas... | Let us check the three conditions for an [[Definition:Equivalence Relation|equivalence]] in turn. | Morphism Class Equivalence is Equivalence | https://proofwiki.org/wiki/Morphism_Class_Equivalence_is_Equivalence | https://proofwiki.org/wiki/Morphism_Class_Equivalence_is_Equivalence | [
"Subobjects"
] | [
"Definition:Metacategory",
"Definition:Object (Category Theory)",
"Definition:Category of Subobjects",
"Definition:Subobject Class/Morphism Class",
"Definition:Morphism",
"Definition:Equivalence Relation"
] | [
"Definition:Equivalence Relation",
"Definition:Equivalence Relation"
] |
proofwiki-6304 | Category of Subobject Classes is Category | Let $\mathbf C$ be a metacategory.
Let $C$ be an object of $\mathbf C$.
Let $\map {\overline {\mathbf {Sub} }_{\mathbf C} } C$ be the category of subobject classes of $C$.
:$g \circ f : X \to Z$
called the composition of $f$ and $g$.
Then $\map {\overline {\mathbf {Sub} }_{\mathbf C} } C$ is a metacategory. | Let us verify the axioms $(\text C 1)$ to $(\text C 3)$ for a metacategory.
{{finish|A feast of checking well-definedness}} | Let $\mathbf C$ be a [[Definition:Metacategory|metacategory]].
Let $C$ be an [[Definition:Object (Category Theory)|object]] of $\mathbf C$.
Let $\map {\overline {\mathbf {Sub} }_{\mathbf C} } C$ be the [[Definition:Category of Subobject Classes|category of subobject classes]] of $C$.
:$g \circ f : X \to Z$
called t... | Let us verify the axioms $(\text C 1)$ to $(\text C 3)$ for a [[Definition:Metacategory|metacategory]].
{{finish|A feast of checking well-definedness}} | Category of Subobject Classes is Category | https://proofwiki.org/wiki/Category_of_Subobject_Classes_is_Category | https://proofwiki.org/wiki/Category_of_Subobject_Classes_is_Category | [
"Categories of Subobjects"
] | [
"Definition:Metacategory",
"Definition:Object (Category Theory)",
"Definition:Category of Subobject Classes",
"Definition:Composition of Morphisms",
"Definition:Metacategory"
] | [
"Definition:Metacategory"
] |
proofwiki-6305 | Category of Subobject Classes is Order Category | Let $\mathbf C$ be a metacategory.
Let $C$ be an object of $\mathbf C$.
Let $\overline{\mathbf{Sub}}_{\mathbf C} \left({C}\right)$ be the category of subobject classes of $C$.
Then $\overline{\mathbf{Sub}}_{\mathbf C} \left({C}\right)$ is an order category. | {{ProofWanted|Preliminary (rather, partial) results missing}}
Category:Categories of Subobjects
Category:Order Categories
2l7bbazegfq66yg4yxfb335atdvs5h8 | Let $\mathbf C$ be a [[Definition:Metacategory|metacategory]].
Let $C$ be an [[Definition:Object (Category Theory)|object]] of $\mathbf C$.
Let $\overline{\mathbf{Sub}}_{\mathbf C} \left({C}\right)$ be the [[Definition:Category of Subobject Classes|category of subobject classes]] of $C$.
Then $\overline{\mathbf{Sub... | {{ProofWanted|Preliminary (rather, partial) results missing}}
[[Category:Categories of Subobjects]]
[[Category:Order Categories]]
2l7bbazegfq66yg4yxfb335atdvs5h8 | Category of Subobject Classes is Order Category | https://proofwiki.org/wiki/Category_of_Subobject_Classes_is_Order_Category | https://proofwiki.org/wiki/Category_of_Subobject_Classes_is_Order_Category | [
"Categories of Subobjects",
"Order Categories"
] | [
"Definition:Metacategory",
"Definition:Object (Category Theory)",
"Definition:Category of Subobject Classes",
"Definition:Order Category"
] | [
"Category:Categories of Subobjects",
"Category:Order Categories"
] |
proofwiki-6306 | Reverse Triangle Inequality/Real and Complex Fields | Let $x$ and $y$ be elements of either the real numbers $\R$ or the complex numbers $\C$.
Then:
:$\cmod {x - y} \ge \size {\cmod x - \cmod y}$ | From the Reverse Triangle Inequality:
:$\cmod {x - y} \ge \cmod {\cmod x - \cmod y}$
By the definition of both absolute value and complex modulus:
:$\cmod {\cmod x - \cmod y} \ge 0$
As:
:$\cmod x - \cmod y = \pm \cmod {\cmod x - \cmod y}$
it follows that:
:$\cmod {\cmod x - \cmod y} \ge \cmod x - \cmod y$
Hence the res... | Let $x$ and $y$ be elements of either the [[Definition:Real Number|real numbers]] $\R$ or the [[Definition:Complex Number|complex numbers]] $\C$.
Then:
:$\cmod {x - y} \ge \size {\cmod x - \cmod y}$ | From the [[Reverse Triangle Inequality/Real and Complex Fields/Proof 1|Reverse Triangle Inequality]]:
:$\cmod {x - y} \ge \cmod {\cmod x - \cmod y}$
By the definition of both [[Definition:Absolute Value|absolute value]] and [[Definition:Complex Modulus|complex modulus]]:
:$\cmod {\cmod x - \cmod y} \ge 0$
As:
:$\cmod... | Reverse Triangle Inequality/Real and Complex Fields/Corollary 1/Proof 1 | https://proofwiki.org/wiki/Reverse_Triangle_Inequality/Real_and_Complex_Fields | https://proofwiki.org/wiki/Reverse_Triangle_Inequality/Real_and_Complex_Fields/Corollary_1/Proof_1 | [
"Triangle Inequality"
] | [
"Definition:Real Number",
"Definition:Complex Number"
] | [
"Reverse Triangle Inequality/Real and Complex Fields/Proof 1",
"Definition:Absolute Value",
"Definition:Complex Modulus"
] |
proofwiki-6307 | Reverse Triangle Inequality/Real and Complex Fields | Let $x$ and $y$ be elements of either the real numbers $\R$ or the complex numbers $\C$.
Then:
:$\cmod {x - y} \ge \size {\cmod x - \cmod y}$ | By the Triangle Inequality:
:$\cmod {x + y} - \cmod y \le \cmod x$
Let $z = x + y$.
Then $x = z - y$ and so:
:$\cmod z - \cmod y \le \cmod {z - y}$
Renaming variables as appropriate gives:
:$\cmod {x - y} \ge \cmod x - \cmod y$
{{qed}} | Let $x$ and $y$ be elements of either the [[Definition:Real Number|real numbers]] $\R$ or the [[Definition:Complex Number|complex numbers]] $\C$.
Then:
:$\cmod {x - y} \ge \size {\cmod x - \cmod y}$ | By the [[Triangle Inequality]]:
:$\cmod {x + y} - \cmod y \le \cmod x$
Let $z = x + y$.
Then $x = z - y$ and so:
:$\cmod z - \cmod y \le \cmod {z - y}$
Renaming variables as appropriate gives:
:$\cmod {x - y} \ge \cmod x - \cmod y$
{{qed}} | Reverse Triangle Inequality/Real and Complex Fields/Corollary 1/Proof 2 | https://proofwiki.org/wiki/Reverse_Triangle_Inequality/Real_and_Complex_Fields | https://proofwiki.org/wiki/Reverse_Triangle_Inequality/Real_and_Complex_Fields/Corollary_1/Proof_2 | [
"Triangle Inequality"
] | [
"Definition:Real Number",
"Definition:Complex Number"
] | [
"Triangle Inequality"
] |
proofwiki-6308 | Reverse Triangle Inequality/Real and Complex Fields | Let $x$ and $y$ be elements of either the real numbers $\R$ or the complex numbers $\C$.
Then:
:$\cmod {x - y} \ge \size {\cmod x - \cmod y}$ | Let $z_1$ and $z_2$ be represented by the points $A$ and $B$ respectively in the complex plane.
From Geometrical Interpretation of Complex Subtraction, we can construct the parallelogram $OACB$ where:
:$OA$ and $OB$ represent $z_1$ and $z_2$ respectively
:$BA$ represents $z_1 - z_2$.
:400px
But $OA$, $OB$ and $BA$ form... | Let $x$ and $y$ be elements of either the [[Definition:Real Number|real numbers]] $\R$ or the [[Definition:Complex Number|complex numbers]] $\C$.
Then:
:$\cmod {x - y} \ge \size {\cmod x - \cmod y}$ | Let $z_1$ and $z_2$ be represented by the [[Definition:Point|points]] $A$ and $B$ respectively in the [[Definition:Complex Plane|complex plane]].
From [[Geometrical Interpretation of Complex Subtraction]], we can construct the [[Definition:Parallelogram|parallelogram]] $OACB$ where:
:$OA$ and $OB$ represent $z_1$ and ... | Reverse Triangle Inequality/Real and Complex Fields/Corollary 1/Proof 3 | https://proofwiki.org/wiki/Reverse_Triangle_Inequality/Real_and_Complex_Fields | https://proofwiki.org/wiki/Reverse_Triangle_Inequality/Real_and_Complex_Fields/Corollary_1/Proof_3 | [
"Triangle Inequality"
] | [
"Definition:Real Number",
"Definition:Complex Number"
] | [
"Definition:Point",
"Definition:Complex Number/Complex Plane",
"Geometrical Interpretation of Complex Subtraction",
"Definition:Quadrilateral/Parallelogram",
"File:Complex-Reverse-Triangle-Inequality-Corollary.png",
"Definition:Polygon/Side",
"Definition:Triangle (Geometry)",
"Sum of Two Sides of Tria... |
proofwiki-6309 | Reverse Triangle Inequality/Real and Complex Fields | Let $x$ and $y$ be elements of either the real numbers $\R$ or the complex numbers $\C$.
Then:
:$\cmod {x - y} \ge \size {\cmod x - \cmod y}$ | Let $X$ denote either $\R$ or $\C$ as appropriate.
From Real Number Line is Metric Space and Complex Plane is Metric Space the distance function $d: X \times X \to \R$ can be defined as:
:$\map d {x, y} = \size {x - y}$
From the Reverse Triangle Inequality as applied to metric spaces:
:$(1): \quad \forall x, y, z \in X... | Let $x$ and $y$ be elements of either the [[Definition:Real Number|real numbers]] $\R$ or the [[Definition:Complex Number|complex numbers]] $\C$.
Then:
:$\cmod {x - y} \ge \size {\cmod x - \cmod y}$ | Let $X$ denote either $\R$ or $\C$ as appropriate.
From [[Real Number Line is Metric Space]] and [[Complex Plane is Metric Space]] the [[Definition:Distance Function|distance function]] $d: X \times X \to \R$ can be defined as:
:$\map d {x, y} = \size {x - y}$
From the [[Reverse Triangle Inequality]] as applied to [... | Reverse Triangle Inequality/Real and Complex Fields/Proof 1 | https://proofwiki.org/wiki/Reverse_Triangle_Inequality/Real_and_Complex_Fields | https://proofwiki.org/wiki/Reverse_Triangle_Inequality/Real_and_Complex_Fields/Proof_1 | [
"Triangle Inequality"
] | [
"Definition:Real Number",
"Definition:Complex Number"
] | [
"Real Number Line is Metric Space",
"Complex Plane is Metric Space",
"Definition:Distance Function",
"Reverse Triangle Inequality",
"Definition:Metric Space"
] |
proofwiki-6310 | Reverse Triangle Inequality/Real and Complex Fields | Let $x$ and $y$ be elements of either the real numbers $\R$ or the complex numbers $\C$.
Then:
:$\cmod {x - y} \ge \size {\cmod x - \cmod y}$ | From proof $2$ of corollary $1$ to this result, which is derived independently:
:$\size {x - y} \ge \size x - \size y$
There are two cases:
$(1): \quad \size x \ge \size y$
We have :
:$\size {\size x - \size y} = \size x - \size y$
and the proof is finished.
{{qed|lemma}}
$(2): \quad \size y \ge \size x$
We have:
:$\si... | Let $x$ and $y$ be elements of either the [[Definition:Real Number|real numbers]] $\R$ or the [[Definition:Complex Number|complex numbers]] $\C$.
Then:
:$\cmod {x - y} \ge \size {\cmod x - \cmod y}$ | From [[Reverse Triangle Inequality/Real and Complex Fields/Corollary 1/Proof 2|proof $2$ of corollary $1$ to this result]], which is derived independently:
:$\size {x - y} \ge \size x - \size y$
There are two cases:
$(1): \quad \size x \ge \size y$
We have :
:$\size {\size x - \size y} = \size x - \size y$
and the ... | Reverse Triangle Inequality/Real and Complex Fields/Proof 2 | https://proofwiki.org/wiki/Reverse_Triangle_Inequality/Real_and_Complex_Fields | https://proofwiki.org/wiki/Reverse_Triangle_Inequality/Real_and_Complex_Fields/Proof_2 | [
"Triangle Inequality"
] | [
"Definition:Real Number",
"Definition:Complex Number"
] | [
"Reverse Triangle Inequality/Real and Complex Fields/Corollary 1/Proof 2",
"Negative of Absolute Value"
] |
proofwiki-6311 | Triangle Inequality/Real Numbers | Let $x, y \in \R$ be real numbers.
Let $\size x$ denote the absolute value of $x$.
Then:
:$\size {x + y} \le \size x + \size y$ | {{begin-eqn}}
{{eqn | l = \size {x + y}^2
| r = \paren {x + y}^2
| c = Square of Real Number is Non-Negative
}}
{{eqn | r = x^2 + 2 x y + y^2
| c = Square of Sum
}}
{{eqn | r = \size x^2 + 2 x y + \size y^2
| c = Square of Real Number is Non-Negative
}}
{{eqn | o = \le
| r = \size x^2 + 2 ... | Let $x, y \in \R$ be [[Definition:Real Number|real numbers]].
Let $\size x$ denote the [[Definition:Absolute Value|absolute value]] of $x$.
Then:
:$\size {x + y} \le \size x + \size y$ | {{begin-eqn}}
{{eqn | l = \size {x + y}^2
| r = \paren {x + y}^2
| c = [[Square of Real Number is Non-Negative]]
}}
{{eqn | r = x^2 + 2 x y + y^2
| c = [[Square of Sum]]
}}
{{eqn | r = \size x^2 + 2 x y + \size y^2
| c = [[Square of Real Number is Non-Negative]]
}}
{{eqn | o = \le
| r = \s... | Triangle Inequality/Real Numbers/Proof 1 | https://proofwiki.org/wiki/Triangle_Inequality/Real_Numbers | https://proofwiki.org/wiki/Triangle_Inequality/Real_Numbers/Proof_1 | [
"Triangle Inequality"
] | [
"Definition:Real Number",
"Definition:Absolute Value"
] | [
"Square of Real Number is Non-Negative",
"Square of Sum",
"Square of Real Number is Non-Negative",
"Negative of Absolute Value",
"Absolute Value Function is Completely Multiplicative",
"Square of Sum",
"Order is Preserved on Positive Reals by Squaring"
] |
proofwiki-6312 | Triangle Inequality/Real Numbers | Let $x, y \in \R$ be real numbers.
Let $\size x$ denote the absolute value of $x$.
Then:
:$\size {x + y} \le \size x + \size y$ | This can be seen to be a special case of Minkowski's Inequality for Sums, with $n = 1$.
{{qed}} | Let $x, y \in \R$ be [[Definition:Real Number|real numbers]].
Let $\size x$ denote the [[Definition:Absolute Value|absolute value]] of $x$.
Then:
:$\size {x + y} \le \size x + \size y$ | This can be seen to be a special case of [[Minkowski's Inequality for Sums]], with $n = 1$.
{{qed}} | Triangle Inequality/Real Numbers/Proof 2 | https://proofwiki.org/wiki/Triangle_Inequality/Real_Numbers | https://proofwiki.org/wiki/Triangle_Inequality/Real_Numbers/Proof_2 | [
"Triangle Inequality"
] | [
"Definition:Real Number",
"Definition:Absolute Value"
] | [
"Minkowski's Inequality for Sums"
] |
proofwiki-6313 | Triangle Inequality/Real Numbers | Let $x, y \in \R$ be real numbers.
Let $\size x$ denote the absolute value of $x$.
Then:
:$\size {x + y} \le \size x + \size y$ | We have that Real Numbers form Ordered Integral Domain.
Therefore Sum of Absolute Values on Ordered Integral Domain applies directly.
{{qed}} | Let $x, y \in \R$ be [[Definition:Real Number|real numbers]].
Let $\size x$ denote the [[Definition:Absolute Value|absolute value]] of $x$.
Then:
:$\size {x + y} \le \size x + \size y$ | We have that [[Real Numbers form Ordered Integral Domain]].
Therefore [[Sum of Absolute Values on Ordered Integral Domain]] applies directly.
{{qed}} | Triangle Inequality/Real Numbers/Proof 3 | https://proofwiki.org/wiki/Triangle_Inequality/Real_Numbers | https://proofwiki.org/wiki/Triangle_Inequality/Real_Numbers/Proof_3 | [
"Triangle Inequality"
] | [
"Definition:Real Number",
"Definition:Absolute Value"
] | [
"Real Numbers form Ordered Integral Domain",
"Sum of Absolute Values on Ordered Integral Domain"
] |
proofwiki-6314 | Triangle Inequality/Real Numbers | Let $x, y \in \R$ be real numbers.
Let $\size x$ denote the absolute value of $x$.
Then:
:$\size {x + y} \le \size x + \size y$ | We do a case analysis.
=== $(1): \quad x \ge 0, y \ge 0$ ===
{{begin-eqn}}
{{eqn | l = x
| o = \ge
| r = 0
| c =
}}
{{eqn | l = y
| o = \ge
| r = 0
| c =
}}
{{eqn | ll= \leadsto
| l = \size {x + y}
| r = x + y
| c =
}}
{{eqn | r = \size x + \size y
| c =
}... | Let $x, y \in \R$ be [[Definition:Real Number|real numbers]].
Let $\size x$ denote the [[Definition:Absolute Value|absolute value]] of $x$.
Then:
:$\size {x + y} \le \size x + \size y$ | We do a case analysis.
=== $(1): \quad x \ge 0, y \ge 0$ ===
{{begin-eqn}}
{{eqn | l = x
| o = \ge
| r = 0
| c =
}}
{{eqn | l = y
| o = \ge
| r = 0
| c =
}}
{{eqn | ll= \leadsto
| l = \size {x + y}
| r = x + y
| c =
}}
{{eqn | r = \size x + \size y
| c =... | Triangle Inequality/Real Numbers/Proof 4 | https://proofwiki.org/wiki/Triangle_Inequality/Real_Numbers | https://proofwiki.org/wiki/Triangle_Inequality/Real_Numbers/Proof_4 | [
"Triangle Inequality"
] | [
"Definition:Real Number",
"Definition:Absolute Value"
] | [
"Definition:Positive/Real Number"
] |
proofwiki-6315 | Triangle Inequality/Real Numbers | Let $x, y \in \R$ be real numbers.
Let $\size x$ denote the absolute value of $x$.
Then:
:$\size {x + y} \le \size x + \size y$ | From Negative of Absolute Value, it is sufficient to prove that:
:$\size x + \size y \ge x + y$
and:
:$\size x + \size y \ge -\paren {x + y}$
By definition of absolute value:
:$x \le \size x$
and:
:$y \le \size y$
Then:
:$x + y \le \size x + \size y$
We also have that:
{{begin-eqn}}
{{eqn | l = -x
| o = \le
... | Let $x, y \in \R$ be [[Definition:Real Number|real numbers]].
Let $\size x$ denote the [[Definition:Absolute Value|absolute value]] of $x$.
Then:
:$\size {x + y} \le \size x + \size y$ | From [[Negative of Absolute Value]], it is sufficient to prove that:
:$\size x + \size y \ge x + y$
and:
:$\size x + \size y \ge -\paren {x + y}$
By definition of [[Definition:Absolute Value|absolute value]]:
:$x \le \size x$
and:
:$y \le \size y$
Then:
:$x + y \le \size x + \size y$
We also have that:
{{begin-e... | Triangle Inequality/Real Numbers/Proof 5 | https://proofwiki.org/wiki/Triangle_Inequality/Real_Numbers | https://proofwiki.org/wiki/Triangle_Inequality/Real_Numbers/Proof_5 | [
"Triangle Inequality"
] | [
"Definition:Real Number",
"Definition:Absolute Value"
] | [
"Negative of Absolute Value",
"Definition:Absolute Value",
"Absolute Value of Negative",
"Absolute Value of Negative"
] |
proofwiki-6316 | Triangle Inequality/Complex Numbers | Let $z_1, z_2 \in \C$ be complex numbers.
Let $\cmod z$ denote the modulus of $z$.
Then:
:$\cmod {z_1 + z_2} \le \cmod {z_1} + \cmod {z_2}$ | This is an instance of the General Triangle Inequality for Complex Numbers:
:$\cmod {z_1 + z_2 + \dotsb + z_n} \le \cmod {z_1} + \cmod {z_2} + \dotsb + \cmod {z_n}$
setting $n = 3$.
{{qed}} | Let $z_1, z_2 \in \C$ be [[Definition:Complex Number|complex numbers]].
Let $\cmod z$ denote the [[Definition:Modulus of Complex Number|modulus]] of $z$.
Then:
:$\cmod {z_1 + z_2} \le \cmod {z_1} + \cmod {z_2}$ | This is an instance of the [[General Triangle Inequality for Complex Numbers]]:
:$\cmod {z_1 + z_2 + \dotsb + z_n} \le \cmod {z_1} + \cmod {z_2} + \dotsb + \cmod {z_n}$
setting $n = 3$.
{{qed}} | Triangle Inequality/Complex Numbers/Examples/3 Arguments/Proof 1 | https://proofwiki.org/wiki/Triangle_Inequality/Complex_Numbers | https://proofwiki.org/wiki/Triangle_Inequality/Complex_Numbers/Examples/3_Arguments/Proof_1 | [
"Triangle Inequality"
] | [
"Definition:Complex Number",
"Definition:Complex Modulus"
] | [
"Triangle Inequality/Complex Numbers/General Result"
] |
proofwiki-6317 | Triangle Inequality/Complex Numbers | Let $z_1, z_2 \in \C$ be complex numbers.
Let $\cmod z$ denote the modulus of $z$.
Then:
:$\cmod {z_1 + z_2} \le \cmod {z_1} + \cmod {z_2}$ | {{begin-eqn}}
{{eqn | l = \cmod {z_1 + z_2 + z_3}
| r = \cmod {z_1 + \paren {z_2 + z_3} }
| c =
}}
{{eqn | o = \le
| r = \cmod {z_1} + \cmod {z_2 + z_3}
| c = Triangle Inequality for Complex Numbers
}}
{{eqn | o = \le
| r = \cmod {z_1} + \cmod {z_2} + \cmod {z_3}
| c = Triangle Ineq... | Let $z_1, z_2 \in \C$ be [[Definition:Complex Number|complex numbers]].
Let $\cmod z$ denote the [[Definition:Modulus of Complex Number|modulus]] of $z$.
Then:
:$\cmod {z_1 + z_2} \le \cmod {z_1} + \cmod {z_2}$ | {{begin-eqn}}
{{eqn | l = \cmod {z_1 + z_2 + z_3}
| r = \cmod {z_1 + \paren {z_2 + z_3} }
| c =
}}
{{eqn | o = \le
| r = \cmod {z_1} + \cmod {z_2 + z_3}
| c = [[Triangle Inequality for Complex Numbers]]
}}
{{eqn | o = \le
| r = \cmod {z_1} + \cmod {z_2} + \cmod {z_3}
| c = [[Triangl... | Triangle Inequality/Complex Numbers/Examples/3 Arguments/Proof 2 | https://proofwiki.org/wiki/Triangle_Inequality/Complex_Numbers | https://proofwiki.org/wiki/Triangle_Inequality/Complex_Numbers/Examples/3_Arguments/Proof_2 | [
"Triangle Inequality"
] | [
"Definition:Complex Number",
"Definition:Complex Modulus"
] | [
"Triangle Inequality/Complex Numbers",
"Triangle Inequality/Complex Numbers"
] |
proofwiki-6318 | Triangle Inequality/Complex Numbers | Let $z_1, z_2 \in \C$ be complex numbers.
Let $\cmod z$ denote the modulus of $z$.
Then:
:$\cmod {z_1 + z_2} \le \cmod {z_1} + \cmod {z_2}$ | Let $z_1$, $z_2$ and $z_3$ be represented by the points $A$, $B$ and $C$ respectively in the complex plane.
From Geometrical Interpretation of Complex Addition, we can construct the parallelogram $OADB$ where:
:$OA$ and $OB$ represent $z_1$ and $z_2$ respectively
:$OD$ represents $z_1 + z_2$.
Also from Geometrical Inte... | Let $z_1, z_2 \in \C$ be [[Definition:Complex Number|complex numbers]].
Let $\cmod z$ denote the [[Definition:Modulus of Complex Number|modulus]] of $z$.
Then:
:$\cmod {z_1 + z_2} \le \cmod {z_1} + \cmod {z_2}$ | Let $z_1$, $z_2$ and $z_3$ be represented by the [[Definition:Point|points]] $A$, $B$ and $C$ respectively in the [[Definition:Complex Plane|complex plane]].
From [[Geometrical Interpretation of Complex Addition]], we can construct the [[Definition:Parallelogram|parallelogram]] $OADB$ where:
:$OA$ and $OB$ represent $... | Triangle Inequality/Complex Numbers/Examples/3 Arguments/Proof 3 | https://proofwiki.org/wiki/Triangle_Inequality/Complex_Numbers | https://proofwiki.org/wiki/Triangle_Inequality/Complex_Numbers/Examples/3_Arguments/Proof_3 | [
"Triangle Inequality"
] | [
"Definition:Complex Number",
"Definition:Complex Modulus"
] | [
"Definition:Point",
"Definition:Complex Number/Complex Plane",
"Geometrical Interpretation of Complex Addition",
"Definition:Quadrilateral/Parallelogram",
"Geometrical Interpretation of Complex Addition",
"Definition:Quadrilateral/Parallelogram",
"File:Triangle-Inequality-Complex-3-Arguments.png",
"De... |
proofwiki-6319 | Triangle Inequality/Complex Numbers | Let $z_1, z_2 \in \C$ be complex numbers.
Let $\cmod z$ denote the modulus of $z$.
Then:
:$\cmod {z_1 + z_2} \le \cmod {z_1} + \cmod {z_2}$ | Let $z_1 = a_1 + i a_2, z_2 = b_1 + i b_2$.
Then from the definition of the modulus, the above equation translates into:
:$\paren {\paren {a_1 + b_1}^2 + \paren {a_2 + b_2}^2}^{\frac 1 2} \le \paren { {a_1}^2 + {a_2}^2}^{\frac 1 2} + \paren { {b_1}^2 + {b_2}^2}^{\frac 1 2}$
This is a special case of Minkowski's Inequal... | Let $z_1, z_2 \in \C$ be [[Definition:Complex Number|complex numbers]].
Let $\cmod z$ denote the [[Definition:Modulus of Complex Number|modulus]] of $z$.
Then:
:$\cmod {z_1 + z_2} \le \cmod {z_1} + \cmod {z_2}$ | Let $z_1 = a_1 + i a_2, z_2 = b_1 + i b_2$.
Then from the definition of the [[Definition:Modulus of Complex Number|modulus]], the above equation translates into:
:$\paren {\paren {a_1 + b_1}^2 + \paren {a_2 + b_2}^2}^{\frac 1 2} \le \paren { {a_1}^2 + {a_2}^2}^{\frac 1 2} + \paren { {b_1}^2 + {b_2}^2}^{\frac 1 2}$
Th... | Triangle Inequality/Complex Numbers/Proof 1 | https://proofwiki.org/wiki/Triangle_Inequality/Complex_Numbers | https://proofwiki.org/wiki/Triangle_Inequality/Complex_Numbers/Proof_1 | [
"Triangle Inequality"
] | [
"Definition:Complex Number",
"Definition:Complex Modulus"
] | [
"Definition:Complex Modulus",
"Minkowski's Inequality for Sums"
] |
proofwiki-6320 | Triangle Inequality/Complex Numbers | Let $z_1, z_2 \in \C$ be complex numbers.
Let $\cmod z$ denote the modulus of $z$.
Then:
:$\cmod {z_1 + z_2} \le \cmod {z_1} + \cmod {z_2}$ | Let $z_1 = a_1 + i a_2, z_2 = b_1 + i b_2$.
{{begin-eqn}}
{{eqn | l = \cmod {z_1 + z_2}
| o = \le
| r = \cmod {z_1} + \cmod {z_2}
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = \paren {\paren {a_1 + b_1}^2 + \paren {a_2 + b_2}^2}^{\frac 1 2}
| o = \le
| r = \paren { {a_1}^2 + {a_2}^2}^{... | Let $z_1, z_2 \in \C$ be [[Definition:Complex Number|complex numbers]].
Let $\cmod z$ denote the [[Definition:Modulus of Complex Number|modulus]] of $z$.
Then:
:$\cmod {z_1 + z_2} \le \cmod {z_1} + \cmod {z_2}$ | Let $z_1 = a_1 + i a_2, z_2 = b_1 + i b_2$.
{{begin-eqn}}
{{eqn | l = \cmod {z_1 + z_2}
| o = \le
| r = \cmod {z_1} + \cmod {z_2}
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = \paren {\paren {a_1 + b_1}^2 + \paren {a_2 + b_2}^2}^{\frac 1 2}
| o = \le
| r = \paren { {a_1}^2 + {a_2}^2}^... | Triangle Inequality/Complex Numbers/Proof 2 | https://proofwiki.org/wiki/Triangle_Inequality/Complex_Numbers | https://proofwiki.org/wiki/Triangle_Inequality/Complex_Numbers/Proof_2 | [
"Triangle Inequality"
] | [
"Definition:Complex Number",
"Definition:Complex Modulus"
] | [
"Cauchy's Inequality",
"Definition:Tautology"
] |
proofwiki-6321 | Triangle Inequality/Complex Numbers | Let $z_1, z_2 \in \C$ be complex numbers.
Let $\cmod z$ denote the modulus of $z$.
Then:
:$\cmod {z_1 + z_2} \le \cmod {z_1} + \cmod {z_2}$ | Let $z_1$ and $z_2$ be represented by the points $A$ and $B$ respectively in the complex plane.
From Geometrical Interpretation of Complex Addition, we can construct the parallelogram $OACB$ where:
:$OA$ and $OB$ represent $z_1$ and $z_2$ respectively
:$OC$ represents $z_1 + z_2$.
:400px
As $OACB$ is a parallelogram, w... | Let $z_1, z_2 \in \C$ be [[Definition:Complex Number|complex numbers]].
Let $\cmod z$ denote the [[Definition:Modulus of Complex Number|modulus]] of $z$.
Then:
:$\cmod {z_1 + z_2} \le \cmod {z_1} + \cmod {z_2}$ | Let $z_1$ and $z_2$ be represented by the [[Definition:Point|points]] $A$ and $B$ respectively in the [[Definition:Complex Plane|complex plane]].
From [[Geometrical Interpretation of Complex Addition]], we can construct the [[Definition:Parallelogram|parallelogram]] $OACB$ where:
:$OA$ and $OB$ represent $z_1$ and $z_... | Triangle Inequality/Complex Numbers/Proof 3 | https://proofwiki.org/wiki/Triangle_Inequality/Complex_Numbers | https://proofwiki.org/wiki/Triangle_Inequality/Complex_Numbers/Proof_3 | [
"Triangle Inequality"
] | [
"Definition:Complex Number",
"Definition:Complex Modulus"
] | [
"Definition:Point",
"Definition:Complex Number/Complex Plane",
"Geometrical Interpretation of Complex Addition",
"Definition:Quadrilateral/Parallelogram",
"File:Triangle-Inequality-Complex.png",
"Definition:Quadrilateral/Parallelogram",
"Definition:Linear Measure/Length",
"Definition:Polygon/Side",
... |
proofwiki-6322 | Triangle Inequality/Complex Numbers | Let $z_1, z_2 \in \C$ be complex numbers.
Let $\cmod z$ denote the modulus of $z$.
Then:
:$\cmod {z_1 + z_2} \le \cmod {z_1} + \cmod {z_2}$ | {{begin-eqn}}
{{eqn | l = \cmod {z + w}^2
| r = \paren {z + w} \paren {\overline z + \overline w}
| c = Product of Complex Number with Conjugate
}}
{{eqn | r = z \overline z + w \overline w + w \overline z + z \overline w
| c =
}}
{{eqn | r = \cmod z^2 + \cmod w^2 + w \overline z + z \overline w
... | Let $z_1, z_2 \in \C$ be [[Definition:Complex Number|complex numbers]].
Let $\cmod z$ denote the [[Definition:Modulus of Complex Number|modulus]] of $z$.
Then:
:$\cmod {z_1 + z_2} \le \cmod {z_1} + \cmod {z_2}$ | {{begin-eqn}}
{{eqn | l = \cmod {z + w}^2
| r = \paren {z + w} \paren {\overline z + \overline w}
| c = [[Product of Complex Number with Conjugate]]
}}
{{eqn | r = z \overline z + w \overline w + w \overline z + z \overline w
| c =
}}
{{eqn | r = \cmod z^2 + \cmod w^2 + w \overline z + z \overline w
... | Triangle Inequality/Complex Numbers/Proof 4 | https://proofwiki.org/wiki/Triangle_Inequality/Complex_Numbers | https://proofwiki.org/wiki/Triangle_Inequality/Complex_Numbers/Proof_4 | [
"Triangle Inequality"
] | [
"Definition:Complex Number",
"Definition:Complex Modulus"
] | [
"Product of Complex Number with Conjugate",
"Product of Complex Number with Conjugate",
"Complex Conjugation is Involution",
"Product of Complex Conjugates",
"Complex Multiplication is Commutative",
"Sum of Complex Number with Conjugate",
"Modulus Larger than Real Part",
"Complex Modulus of Product of... |
proofwiki-6323 | Triangle Inequality/Vectors in Euclidean Space | Let $\mathbf x, \mathbf y$ be vectors in the real Euclidean space $\R^n$.
Let $\norm {\, \cdot \,}$ denote vector length.
Then:
:$\norm {\mathbf x + \mathbf y} \le \norm {\mathbf x} + \norm {\mathbf y}$
If the two vectors are scalar multiples where said scalar is non-negative, an equality holds:
:$\exists \lambda \in \... | Let $\mathbf x, \mathbf y \in \R^n$.
We have:
{{begin-eqn}}
{{eqn | l = \norm {\mathbf x + \mathbf y}^2
| r = \paren {\mathbf x + \mathbf y} \cdot \paren {\mathbf x + \mathbf y}
| c = Dot Product of Vector with Itself
}}
{{eqn | r = \mathbf x \cdot \mathbf x + \mathbf x \cdot \mathbf y + \mathbf y \cdot \ma... | Let $\mathbf x, \mathbf y$ be [[Definition:Vector (Real Euclidean Space)|vectors]] in the [[Definition:Real Euclidean Space|real Euclidean space]] $\R^n$.
Let $\norm {\, \cdot \,}$ denote [[Definition:Vector Length|vector length]].
Then:
:$\norm {\mathbf x + \mathbf y} \le \norm {\mathbf x} + \norm {\mathbf y}$
If ... | Let $\mathbf x, \mathbf y \in \R^n$.
We have:
{{begin-eqn}}
{{eqn | l = \norm {\mathbf x + \mathbf y}^2
| r = \paren {\mathbf x + \mathbf y} \cdot \paren {\mathbf x + \mathbf y}
| c = [[Dot Product of Vector with Itself]]
}}
{{eqn | r = \mathbf x \cdot \mathbf x + \mathbf x \cdot \mathbf y + \mathbf y \cd... | Triangle Inequality/Vectors in Euclidean Space | https://proofwiki.org/wiki/Triangle_Inequality/Vectors_in_Euclidean_Space | https://proofwiki.org/wiki/Triangle_Inequality/Vectors_in_Euclidean_Space | [
"Linear Algebra",
"Triangle Inequality"
] | [
"Definition:Vector/Real Euclidean Space",
"Definition:Euclidean Space/Real",
"Definition:Vector Length",
"Definition:Scalar Multiplication/Vector Space"
] | [
"Dot Product of Vector with Itself",
"Dot Product Distributes over Addition",
"Dot Product Operator is Commutative",
"Dot Product of Vector with Itself",
"Cauchy-Bunyakovsky-Schwarz Inequality",
"Negative of Absolute Value",
"Definition:Square Root",
"Definition:Vector/Real Euclidean Space",
"Defini... |
proofwiki-6324 | Negative of Complex Modulus | Let $z \in \C$ be a complex number.
Then:
:$-\cmod z \le \cmod z$
where $\cmod z$ denotes the complex modulus of $z$.
The equality holds {{iff}} $z = 0$. | Let $z = x + i y$.
By definition of complex modulus:
:$\cmod z = \sqrt {x^2 + y^2}$
and so:
:$\cmod z \ge 0$
Thus:
:$-\cmod z\le 0$
Hence the result.
{{qed}} | Let $z \in \C$ be a [[Definition:Complex Number|complex number]].
Then:
:$-\cmod z \le \cmod z$
where $\cmod z$ denotes the [[Definition:Complex Modulus|complex modulus]] of $z$.
The equality holds {{iff}} $z = 0$. | Let $z = x + i y$.
By definition of [[Definition:Complex Modulus|complex modulus]]:
:$\cmod z = \sqrt {x^2 + y^2}$
and so:
:$\cmod z \ge 0$
Thus:
:$-\cmod z\le 0$
Hence the result.
{{qed}} | Negative of Complex Modulus | https://proofwiki.org/wiki/Negative_of_Complex_Modulus | https://proofwiki.org/wiki/Negative_of_Complex_Modulus | [
"Complex Modulus"
] | [
"Definition:Complex Number",
"Definition:Complex Modulus"
] | [
"Definition:Complex Modulus"
] |
proofwiki-6325 | Limit of Function by Convergent Sequences/Corollary | Let $\openint a b$ be an open real interval.
Let $f: \openint a b \to \R$ be a real function.
Let $l \in \R$.
Then:
:$(1): \quad \ds \lim_{x \mathop \to a^+} \map f x = l \iff \forall \sequence {x_n} \subseteq \openint a b: \lim_{n \mathop \to \infty} x_n = a \implies \lim_{n \mathop \to \infty} \map f {x_n} = l$
:$(2)... | We have that $\struct {\R, d}$ is a metric space, where $d$ is the Euclidean metric on $\R$.
Thus the theorem follows immediately from Limit of Real Function by Convergent Sequences.
{{qed}}
{{ACC|Limit of Real Function by Convergent Sequences}} | Let $\openint a b$ be an [[Definition:Open Real Interval|open real interval]].
Let $f: \openint a b \to \R$ be a [[Definition:Real Function|real function]].
Let $l \in \R$.
Then:
:$(1): \quad \ds \lim_{x \mathop \to a^+} \map f x = l \iff \forall \sequence {x_n} \subseteq \openint a b: \lim_{n \mathop \to \infty} x... | We have that [[Distance on Real Numbers is Metric|$\struct {\R, d}$ is a metric space]], where $d$ is the [[Definition:Euclidean Metric on Real Number Line|Euclidean metric on $\R$]].
Thus the theorem follows immediately from [[Limit of Real Function by Convergent Sequences]].
{{qed}}
{{ACC|Limit of Real Function by ... | Limit of Function by Convergent Sequences/Corollary | https://proofwiki.org/wiki/Limit_of_Function_by_Convergent_Sequences/Corollary | https://proofwiki.org/wiki/Limit_of_Function_by_Convergent_Sequences/Corollary | [
"Real Analysis",
"Limits of Sequences"
] | [
"Definition:Real Interval/Open",
"Definition:Real Function",
"Definition:Limit of Real Function/Right",
"Definition:Limit of Real Function/Left"
] | [
"Distance on Real Numbers is Metric",
"Definition:Euclidean Metric/Real Number Line",
"Limit of Function by Convergent Sequences/Real Number Line"
] |
proofwiki-6326 | Infinite Set in Compact Space has Omega-Accumulation Point | Let $\struct {X, \tau}$ be a compact topological space.
Let $A \subseteq X$ be infinite.
Then $A$ has an $\omega$-accumulation point in $X$. | {{AimForCont}} $A$ has no $\omega$-accumulation points in $X$.
Then for any $x \in X$, there exists some open set $U_x$ such that $U_x$ only contains a finite number of points in $A$.
The collection of all such open sets $\CC = \set {U_x: x \in X}$ is an open cover for $X$.
Since $X$ is compact, $\CC$ has a finite subc... | Let $\struct {X, \tau}$ be a [[Definition:Compact Topological Space|compact topological space]].
Let $A \subseteq X$ be [[Definition:Infinite Set|infinite]].
Then $A$ has an [[Definition:Omega-Accumulation Point|$\omega$-accumulation point]] in $X$. | {{AimForCont}} $A$ has no [[Definition:Omega-Accumulation Point|$\omega$-accumulation points]] in $X$.
Then for any $x \in X$, there exists some [[Definition:Open Set (Topology)|open set]] $U_x$ such that $U_x$ only contains a [[Definition:Finite|finite]] number of points in $A$.
The collection of all such [[Definiti... | Infinite Set in Compact Space has Omega-Accumulation Point | https://proofwiki.org/wiki/Infinite_Set_in_Compact_Space_has_Omega-Accumulation_Point | https://proofwiki.org/wiki/Infinite_Set_in_Compact_Space_has_Omega-Accumulation_Point | [
"Compact Topological Spaces",
"Omega-Accumulation Points"
] | [
"Definition:Compact Topological Space",
"Definition:Infinite Set",
"Definition:Omega-Accumulation Point"
] | [
"Definition:Omega-Accumulation Point",
"Definition:Open Set/Topology",
"Definition:Finite",
"Definition:Open Set/Topology",
"Definition:Open Cover",
"Definition:Compact Topological Space",
"Definition:Subcover/Finite",
"Definition:Finite",
"Definition:Finite",
"Definition:Finite",
"Definition:Su... |
proofwiki-6327 | Half-Open Real Interval is not Open Set | Let $\R$ be the real number line considered as an Euclidean space.
Let $\hointr a b \subset \R$ be a half-open interval of $\R$.
Then $\hointr a b$ is not an open set of $\R$.
Similarly, the half-open interval $\hointl a b \subset \R$ is not an open set of $\R$. | Let $\epsilon \in \R_{>0}$.
Let $\map {B_\epsilon} a$ be the open $\epsilon$-ball of $a$.
We have that $a - \epsilon < a$ and so $\map {B_\epsilon} a = \openint {a - \epsilon} {a + \epsilon}$ does not lie entirely in $\hointr a b$.
Thus $\hointr a b$ is not a neighborhood $a$.
It follows that $\hointr a b$ is not an op... | Let $\R$ be the [[Definition:Real Number Line|real number line]] considered as an [[Definition:Euclidean Space|Euclidean space]].
Let $\hointr a b \subset \R$ be a [[Definition:Half-Open Real Interval|half-open interval]] of $\R$.
Then $\hointr a b$ is not an [[Definition:Open Set (Metric Space)|open set]] of $\R$.
... | Let $\epsilon \in \R_{>0}$.
Let $\map {B_\epsilon} a$ be the [[Definition:Open Ball|open $\epsilon$-ball]] of $a$.
We have that $a - \epsilon < a$ and so $\map {B_\epsilon} a = \openint {a - \epsilon} {a + \epsilon}$ does not lie entirely in $\hointr a b$.
Thus $\hointr a b$ is not a [[Definition:Neighborhood (Metri... | Half-Open Real Interval is not Open Set | https://proofwiki.org/wiki/Half-Open_Real_Interval_is_not_Open_Set | https://proofwiki.org/wiki/Half-Open_Real_Interval_is_not_Open_Set | [
"Real Intervals",
"Open Sets"
] | [
"Definition:Real Number/Real Number Line",
"Definition:Euclidean Space",
"Definition:Real Interval/Half-Open",
"Definition:Open Set/Metric Space",
"Definition:Real Interval/Half-Open",
"Definition:Open Set/Metric Space"
] | [
"Definition:Open Ball",
"Definition:Neighborhood (Metric Space)",
"Definition:Open Set/Metric Space",
"Definition:Open Set/Metric Space"
] |
proofwiki-6328 | Empty Set is Open in Metric Space | Let $M = \struct {A, d}$ be a metric space.
Then the empty set $\O$ is an open set of $M$. | By definition, an open set $S \subseteq A$ is one where every point inside it is an element of an open ball contained entirely within that set.
That is, there are no points in $S$ which have an open ball some of whose elements are not in $S$.
As there are no elements in $\O$, the result follows vacuously.
{{qed}} | Let $M = \struct {A, d}$ be a [[Definition:Metric Space|metric space]].
Then the [[Definition:Empty Set|empty set]] $\O$ is an [[Definition:Open Set (Metric Space)|open set]] of $M$. | By definition, an [[Definition:Open Set (Metric Space)|open set]] $S \subseteq A$ is one where every point inside it is an element of an [[Definition:Open Ball|open ball]] contained entirely within that set.
That is, there are no points in $S$ which have an [[Definition:Open Ball|open ball]] some of whose elements are... | Empty Set is Open in Metric Space | https://proofwiki.org/wiki/Empty_Set_is_Open_in_Metric_Space | https://proofwiki.org/wiki/Empty_Set_is_Open_in_Metric_Space | [
"Empty Set",
"Open Sets (Metric Spaces)",
"Metric Spaces"
] | [
"Definition:Metric Space",
"Definition:Empty Set",
"Definition:Open Set/Metric Space"
] | [
"Definition:Open Set/Metric Space",
"Definition:Open Ball",
"Definition:Open Ball",
"Definition:Vacuous Truth"
] |
proofwiki-6329 | Metric Space is Open in Itself | Let $M = \struct {A, d}$ be a metric space.
Then the set $A$ is an open set of $M$. | By definition, an open set $S \subseteq A$ is one where every point inside it is an element of an open ball contained entirely within that set.
Let $x \in A$.
An open ball of $x$ in $M$ is by definition a subset of $A$.
Hence the result.
{{qed}} | Let $M = \struct {A, d}$ be a [[Definition:Metric Space|metric space]].
Then the [[Definition:Set|set]] $A$ is an [[Definition:Open Set (Metric Space)|open set]] of $M$. | By definition, an [[Definition:Open Set (Metric Space)|open set]] $S \subseteq A$ is one where every point inside it is an element of an [[Definition:Open Ball of Metric Space|open ball]] contained entirely within that set.
Let $x \in A$.
An [[Definition:Open Ball of Metric Space|open ball]] of $x$ in $M$ is by defin... | Metric Space is Open in Itself | https://proofwiki.org/wiki/Metric_Space_is_Open_in_Itself | https://proofwiki.org/wiki/Metric_Space_is_Open_in_Itself | [
"Metric Space is Open in Itself",
"Metric Spaces",
"Open Sets (Metric Spaces)"
] | [
"Definition:Metric Space",
"Definition:Set",
"Definition:Open Set/Metric Space"
] | [
"Definition:Open Set/Metric Space",
"Definition:Open Ball",
"Definition:Open Ball",
"Definition:Subset"
] |
proofwiki-6330 | Subset of Standard Discrete Metric Space is Open | Let $M = \struct {A, d}$ be a standard discrete metric space.
Let $S \subseteq A$ be a subset of $A$.
Then $S$ is an open set of $M$. | From the definition of standard discrete metric:
:$\forall x, y \in A: \map d {x, y} = \begin {cases} 0 & : x = y \\ 1 & : x \ne y \end {cases}$
Let $\epsilon \in \R_{>0}$ be such that $0 < \epsilon \le 1$.
Let $x \in S$.
Let $\map {B_\epsilon} x$ be the open $\epsilon$-ball of $x$.
Then by definition of $\epsilon$ and... | Let $M = \struct {A, d}$ be a [[Definition:Standard Discrete Metric|standard discrete metric space]].
Let $S \subseteq A$ be a [[Definition:Subset|subset]] of $A$.
Then $S$ is an [[Definition:Open Set (Metric Space)|open set]] of $M$. | From the definition of [[Definition:Standard Discrete Metric|standard discrete metric]]:
:$\forall x, y \in A: \map d {x, y} = \begin {cases} 0 & : x = y \\ 1 & : x \ne y \end {cases}$
Let $\epsilon \in \R_{>0}$ be such that $0 < \epsilon \le 1$.
Let $x \in S$.
Let $\map {B_\epsilon} x$ be the [[Definition:Open Bal... | Subset of Standard Discrete Metric Space is Open | https://proofwiki.org/wiki/Subset_of_Standard_Discrete_Metric_Space_is_Open | https://proofwiki.org/wiki/Subset_of_Standard_Discrete_Metric_Space_is_Open | [
"Standard Discrete Metric"
] | [
"Definition:Standard Discrete Metric",
"Definition:Subset",
"Definition:Open Set/Metric Space"
] | [
"Definition:Standard Discrete Metric",
"Definition:Open Ball",
"Definition:Open Set/Metric Space"
] |
proofwiki-6331 | Image of Open Set under Continuous Mapping in Metric Space may not be Open | Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces.
Let $f: A_1 \to A_2$ be a $\tuple {d_1, d_2}$-continuous mapping from $A_1$ to $A_2$.
Let $U \subseteq A_1$ be an open set of $M_1$.
Then it is not necessarily the case that $f \sqbrk U$ is an open set of $M_2$. | Consider the constant mapping $f_0: \R^2 \to \R$ defined as:
:$\forall \tuple {x, y} \in \R^2: \map f {x, y} = 0$
Then by Constant Mapping is Continuous $f$ is a continuous mapping.
But consider any open set $U \subseteq A_1$ of $M_1$ such that $U \ne \O$.
Then $f \sqbrk U = \set 0 = \closedint 0 0$ which is a closed i... | Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be [[Definition:Metric Space|metric spaces]].
Let $f: A_1 \to A_2$ be a [[Definition:Continuous Mapping (Metric Spaces)|$\tuple {d_1, d_2}$-continuous mapping]] from $A_1$ to $A_2$.
Let $U \subseteq A_1$ be an [[Definition:Open Set (Metric Space)|open set]... | Consider the [[Definition:Constant Mapping|constant mapping]] $f_0: \R^2 \to \R$ defined as:
:$\forall \tuple {x, y} \in \R^2: \map f {x, y} = 0$
Then by [[Constant Mapping is Continuous]] $f$ is a [[Definition:Continuous Mapping (Metric Spaces)|continuous mapping]].
But consider any [[Definition:Open Set (Metric Spa... | Image of Open Set under Continuous Mapping in Metric Space may not be Open | https://proofwiki.org/wiki/Image_of_Open_Set_under_Continuous_Mapping_in_Metric_Space_may_not_be_Open | https://proofwiki.org/wiki/Image_of_Open_Set_under_Continuous_Mapping_in_Metric_Space_may_not_be_Open | [
"Open Sets (Metric Spaces)",
"Continuity"
] | [
"Definition:Metric Space",
"Definition:Continuous Mapping (Metric Space)",
"Definition:Open Set/Metric Space",
"Definition:Open Set/Metric Space"
] | [
"Definition:Constant Mapping",
"Constant Mapping is Continuous",
"Definition:Continuous Mapping (Metric Space)",
"Definition:Open Set/Metric Space",
"Definition:Real Interval/Closed",
"Closed Real Interval is not Open Set"
] |
proofwiki-6332 | Infinite Intersection of Open Sets of Metric Space may not be Open | Let $M = \struct {A, d}$ be a metric space.
Let $\sequence {U_n}_{n \mathop \in \N}$ be an infinite sequence of open sets of $M$.
Then it is not necessarily the case that $\ds \bigcap_{n \mathop \in \N} U_n$ is itself an open set of $M$. | Consider the open real interval $\openint {-\dfrac 1 n} {\dfrac 1 n} \subseteq \R$.
From Open Real Interval is Open Set, $\openint {-\dfrac 1 n} {\dfrac 1 n}$ is open in $\R$ for all $n \in \N_{> 0}$.
But:
:$\ds \bigcap_{n \mathop \in \N_{> 0}} \openint {-\dfrac 1 n} {\dfrac 1 n} = \set 0 = \closedint 0 0$
which is a ... | Let $M = \struct {A, d}$ be a [[Definition:Metric Space|metric space]].
Let $\sequence {U_n}_{n \mathop \in \N}$ be an [[Definition:Infinite Sequence|infinite sequence]] of [[Definition:Open Set (Metric Space)|open sets]] of $M$.
Then it is not necessarily the case that $\ds \bigcap_{n \mathop \in \N} U_n$ is itself... | Consider the [[Definition:Open Real Interval|open real interval]] $\openint {-\dfrac 1 n} {\dfrac 1 n} \subseteq \R$.
From [[Open Real Interval is Open Set]], $\openint {-\dfrac 1 n} {\dfrac 1 n}$ is [[Definition:Open Set (Metric Space)|open]] in $\R$ for all $n \in \N_{> 0}$.
But:
:$\ds \bigcap_{n \mathop \in \N_{>... | Infinite Intersection of Open Sets of Metric Space may not be Open | https://proofwiki.org/wiki/Infinite_Intersection_of_Open_Sets_of_Metric_Space_may_not_be_Open | https://proofwiki.org/wiki/Infinite_Intersection_of_Open_Sets_of_Metric_Space_may_not_be_Open | [
"Open Sets (Metric Spaces)",
"Set Intersection"
] | [
"Definition:Metric Space",
"Definition:Sequence/Infinite Sequence",
"Definition:Open Set/Metric Space",
"Definition:Open Set/Metric Space"
] | [
"Definition:Real Interval/Open",
"Open Real Interval is Open Set",
"Definition:Open Set/Metric Space",
"Definition:Real Interval/Closed",
"Closed Real Interval is not Open Set"
] |
proofwiki-6333 | Equivalence of Definitions of Adherent Point of Set | Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq S$.
{{TFAE|def = Adherent Point of Set|view = adherent point of $H$}}
=== Definition by Open Neighborhood ===
{{:Definition:Adherent Point of Set/Definition by Open Neighborhood}}
=== Definition by Closure ===
{{:Definition:Adherent Point of Set/Defin... | === Definition by Open Neighborhood is equivalent to Definition by Closure ===
This is shown in Condition for Point being in Closure.
{{qed|lemma}} | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $H \subseteq S$.
{{TFAE|def = Adherent Point of Set|view = adherent point of $H$}}
=== [[Definition:Adherent Point of Set/Definition by Open Neighborhood|Definition by Open Neighborhood]] ===
{{:Definition:Adherent Point of Set... | === Definition by Open Neighborhood is equivalent to Definition by Closure ===
This is shown in [[Condition for Point being in Closure]].
{{qed|lemma}} | Equivalence of Definitions of Adherent Point of Set | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Adherent_Point_of_Set | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Adherent_Point_of_Set | [
"Equivalence of Definitions of Adherent Point of Set",
"Adherent Points of Sets"
] | [
"Definition:Topological Space",
"Definition:Adherent Point of Set/Definition by Open Neighborhood",
"Definition:Adherent Point of Set/Definition by Closure",
"Definition:Adherent Point of Set/Definition by Neighborhood"
] | [
"Condition for Point being in Closure"
] |
proofwiki-6334 | Metric Space is Compact iff Countably Compact | A metric space is compact {{iff}} it is countably compact. | Follows directly from:
:Compact Space is Countably Compact
:Countably Compact Metric Space is Compact
{{qed}}
{{ACC|Countably Compact Metric Space is Compact}} | A [[Definition:Metric Space|metric space]] is [[Definition:Compact Metric Space|compact]] {{iff}} it is [[Definition:Countably Compact Space|countably compact]]. | Follows directly from:
:[[Compact Space is Countably Compact]]
:[[Countably Compact Metric Space is Compact]]
{{qed}}
{{ACC|Countably Compact Metric Space is Compact}} | Metric Space is Compact iff Countably Compact | https://proofwiki.org/wiki/Metric_Space_is_Compact_iff_Countably_Compact | https://proofwiki.org/wiki/Metric_Space_is_Compact_iff_Countably_Compact | [
"Metric Spaces",
"Compact Metric Spaces",
"Countably Compact Spaces",
"Sequence of Implications of Metric Space Compactness Properties"
] | [
"Definition:Metric Space",
"Definition:Compact Space/Metric Space",
"Definition:Countably Compact Space"
] | [
"Compact Space is Countably Compact",
"Countably Compact Metric Space is Compact"
] |
proofwiki-6335 | Countably Compact Metric Space is Sequentially Compact | Let $M$ be a countably compact metric space.
Then $M$ is sequentially compact. | This follows directly from the results:
:Metric Space is First-Countable
:Countably Compact First-Countable Space is Sequentially Compact
{{qed}} | Let $M$ be a [[Definition:Countably Compact Space|countably compact]] [[Definition:Metric Space|metric space]].
Then $M$ is [[Definition:Sequentially Compact Space|sequentially compact]]. | This follows directly from the results:
:[[Metric Space is First-Countable]]
:[[Countably Compact First-Countable Space is Sequentially Compact]]
{{qed}} | Countably Compact Metric Space is Sequentially Compact | https://proofwiki.org/wiki/Countably_Compact_Metric_Space_is_Sequentially_Compact | https://proofwiki.org/wiki/Countably_Compact_Metric_Space_is_Sequentially_Compact | [
"Metric Spaces",
"Countably Compact Spaces",
"Sequentially Compact Spaces"
] | [
"Definition:Countably Compact Space",
"Definition:Metric Space",
"Definition:Sequentially Compact Space"
] | [
"Metric Space is First-Countable",
"Countably Compact First-Countable Space is Sequentially Compact"
] |
proofwiki-6336 | Countably Infinite Set in Countably Compact Space has Omega-Accumulation Point | Let $\struct {X, \tau}$ be a countably compact topological space.
Let $A \subseteq X$ be countably infinite.
Then $A$ has an $\omega$-accumulation point in $X$. | ;Proof by Contradiction
{{AimForCont}} that $A$ does not have an $\omega$-accumulation point in $X$.
Let $\SS \subseteq \powerset A$ be the set of all finite subsets of $A$.
By Set of Finite Subsets of Countable Set is Countable, we have that $\SS$ is countable.
For all (finite) $F \in \SS$, define:
:$U_F = \paren {F \... | Let $\struct {X, \tau}$ be a [[Definition:Countably Compact Space|countably compact]] [[Definition:Topological Space|topological space]].
Let $A \subseteq X$ be [[Definition:Countable Set|countably infinite]].
Then $A$ has an [[Definition:Omega-Accumulation Point|$\omega$-accumulation point]] in $X$. | ;[[Proof by Contradiction]]
{{AimForCont}} that $A$ does not have an [[Definition:Omega-Accumulation Point|$\omega$-accumulation point]] in $X$.
Let $\SS \subseteq \powerset A$ be the [[Definition:Set|set]] of all [[Definition:Finite Set|finite]] [[Definition:Subset|subsets]] of $A$.
By [[Set of Finite Subsets of C... | Countably Infinite Set in Countably Compact Space has Omega-Accumulation Point | https://proofwiki.org/wiki/Countably_Infinite_Set_in_Countably_Compact_Space_has_Omega-Accumulation_Point | https://proofwiki.org/wiki/Countably_Infinite_Set_in_Countably_Compact_Space_has_Omega-Accumulation_Point | [
"Countably Compact Spaces",
"Infinite Sets",
"Countable Sets",
"Omega-Accumulation Points"
] | [
"Definition:Countably Compact Space",
"Definition:Topological Space",
"Definition:Countable Set",
"Definition:Omega-Accumulation Point"
] | [
"Proof by Contradiction",
"Definition:Omega-Accumulation Point",
"Definition:Set",
"Definition:Finite Set",
"Definition:Subset",
"Set of Finite Subsets of Countable Set is Countable",
"Definition:Countable Set",
"Definition:Finite Set",
"Definition:Interior (Topology)",
"Image of Countable Set und... |
proofwiki-6337 | First-Countable Space is Sequentially Compact iff Countably Compact | A first-countable topological space is sequentially compact {{iff}} it is countably compact. | Follows directly from:
:Sequentially Compact Space is Countably Compact
:Countably Compact First-Countable Space is Sequentially Compact
{{qed}}
{{ACC|Sequentially Compact Space is Countably Compact}} | A [[Definition:First-Countable Space|first-countable]] [[Definition:Topological Space|topological space]] is [[Definition:Sequentially Compact Space|sequentially compact]] {{iff}} it is [[Definition:Countably Compact Space|countably compact]]. | Follows directly from:
:[[Sequentially Compact Space is Countably Compact]]
:[[Countably Compact First-Countable Space is Sequentially Compact]]
{{qed}}
{{ACC|Sequentially Compact Space is Countably Compact}} | First-Countable Space is Sequentially Compact iff Countably Compact | https://proofwiki.org/wiki/First-Countable_Space_is_Sequentially_Compact_iff_Countably_Compact | https://proofwiki.org/wiki/First-Countable_Space_is_Sequentially_Compact_iff_Countably_Compact | [
"Sequentially Compact Spaces",
"Countably Compact Spaces",
"First-Countable Spaces"
] | [
"Definition:First-Countable Space",
"Definition:Topological Space",
"Definition:Sequentially Compact Space",
"Definition:Countably Compact Space"
] | [
"Sequentially Compact Space is Countably Compact",
"Countably Compact First-Countable Space is Sequentially Compact"
] |
proofwiki-6338 | Accumulation Point of Infinite Sequence in First-Countable Space is Subsequential Limit | Let $\struct {X, \tau}$ be a first-countable topological space.
Let $\sequence {x_n}_{n \mathop \in \N}$ be an infinite sequence in $X$.
Let $x$ be an accumulation point of $\sequence {x_n}$.
Then $x$ is a subsequential limit of $\sequence {x_n}$. | By the definition of a first-countable space, there exists a countable local basis $\BB$ at $x$.
By Surjection from Natural Numbers iff Countable, there exists a surjection $\phi: \N \to \BB$.
For all $n \in \N$, define the set:
:$\ds U_n = \bigcap_{k \mathop = 0}^n \map \phi k$
By General Intersection Property of Topo... | Let $\struct {X, \tau}$ be a [[Definition:First-Countable Space|first-countable]] [[Definition:Topological Space|topological space]].
Let $\sequence {x_n}_{n \mathop \in \N}$ be an [[Definition:Infinite Sequence|infinite sequence]] in $X$.
Let $x$ be an [[Definition:Accumulation Point of Sequence|accumulation point]]... | By the definition of a [[Definition:First-Countable Space|first-countable space]], there exists a [[Definition:Countable Set|countable]] [[Definition:Local Basis|local basis]] $\BB$ at $x$.
By [[Surjection from Natural Numbers iff Countable]], there exists a [[Definition:Surjection|surjection]] $\phi: \N \to \BB$.
F... | Accumulation Point of Infinite Sequence in First-Countable Space is Subsequential Limit | https://proofwiki.org/wiki/Accumulation_Point_of_Infinite_Sequence_in_First-Countable_Space_is_Subsequential_Limit | https://proofwiki.org/wiki/Accumulation_Point_of_Infinite_Sequence_in_First-Countable_Space_is_Subsequential_Limit | [
"First-Countable Spaces",
"Sequences"
] | [
"Definition:First-Countable Space",
"Definition:Topological Space",
"Definition:Sequence/Infinite Sequence",
"Definition:Accumulation Point/Sequence",
"Definition:Subsequential Limit"
] | [
"Definition:First-Countable Space",
"Definition:Countable Set",
"Definition:Local Basis",
"Surjection from Natural Numbers iff Countable",
"Definition:Surjection",
"Definition:Set",
"General Intersection Property of Topological Space",
"Definition:Open Neighborhood/Point",
"Principle of Recursive De... |
proofwiki-6339 | Sequentially Compact Metric Space is Totally Bounded | Let $M = \struct {A, d}$ be a metric space.
Let $M$ be sequentially compact.
Then $M$ is totally bounded. | Let $\epsilon \in \R_{>0}$ be a strictly positive real number.
By definition, $M$ is totally bounded only if there exists a finite $\epsilon$-net for $M$.
{{AimForCont}} there exists no finite $\epsilon$-net for $M$.
The aim is to construct an infinite sequence $\sequence {x_n}_{n \ge 1}$ in $A$ that has no convergent ... | Let $M = \struct {A, d}$ be a [[Definition:Metric Space|metric space]].
Let $M$ be [[Definition:Sequentially Compact Space|sequentially compact]].
Then $M$ is [[Definition:Totally Bounded Metric Space|totally bounded]]. | Let $\epsilon \in \R_{>0}$ be a [[Definition:Strictly Positive Real Number|strictly positive real number]].
By definition, $M$ is [[Definition:Totally Bounded Metric Space|totally bounded]] only if there exists a [[Definition:Finite Epsilon-Net|finite $\epsilon$-net]] for $M$.
{{AimForCont}} there exists no [[Defini... | Sequentially Compact Metric Space is Totally Bounded/Proof 1 | https://proofwiki.org/wiki/Sequentially_Compact_Metric_Space_is_Totally_Bounded | https://proofwiki.org/wiki/Sequentially_Compact_Metric_Space_is_Totally_Bounded/Proof_1 | [
"Metric Spaces",
"Sequentially Compact Spaces",
"Totally Bounded Metric Spaces",
"Sequentially Compact Metric Space is Totally Bounded"
] | [
"Definition:Metric Space",
"Definition:Sequentially Compact Space",
"Definition:Totally Bounded Metric Space"
] | [
"Definition:Strictly Positive/Real Number",
"Definition:Totally Bounded Metric Space",
"Definition:Epsilon-Net/Finite Net",
"Definition:Epsilon-Net/Finite Net",
"Definition:Sequence/Infinite Sequence",
"Definition:Convergent Sequence/Metric Space",
"Definition:Subsequence",
"Definition:Natural Numbers... |
proofwiki-6340 | Sequentially Compact Metric Space is Totally Bounded | Let $M = \struct {A, d}$ be a metric space.
Let $M$ be sequentially compact.
Then $M$ is totally bounded. | We have {{hypothesis}} that $M$ is a sequentially compact space.
So {{afortiori}} every infinite sequence in $M$ has a subsequence which converges to a point in $A$.
Let $\epsilon \in \R_{>0}$ be a strictly positive real number.
By definition, $M$ is totally bounded {{iff}} there exists a finite $\epsilon$-net for $M$.... | Let $M = \struct {A, d}$ be a [[Definition:Metric Space|metric space]].
Let $M$ be [[Definition:Sequentially Compact Space|sequentially compact]].
Then $M$ is [[Definition:Totally Bounded Metric Space|totally bounded]]. | We have {{hypothesis}} that $M$ is a [[Definition:Sequentially Compact Space|sequentially compact space]].
So {{afortiori}} every [[Definition:Infinite Sequence|infinite sequence]] in $M$ has a [[Definition:Subsequence|subsequence]] which [[Definition:Convergent Sequence (Metric Space)|converges]] to a point in $A$.
... | Sequentially Compact Metric Space is Totally Bounded/Proof 2 | https://proofwiki.org/wiki/Sequentially_Compact_Metric_Space_is_Totally_Bounded | https://proofwiki.org/wiki/Sequentially_Compact_Metric_Space_is_Totally_Bounded/Proof_2 | [
"Metric Spaces",
"Sequentially Compact Spaces",
"Totally Bounded Metric Spaces",
"Sequentially Compact Metric Space is Totally Bounded"
] | [
"Definition:Metric Space",
"Definition:Sequentially Compact Space",
"Definition:Totally Bounded Metric Space"
] | [
"Definition:Sequentially Compact Space",
"Definition:Sequence/Infinite Sequence",
"Definition:Subsequence",
"Definition:Convergent Sequence/Metric Space",
"Definition:Strictly Positive/Real Number",
"Definition:Totally Bounded Metric Space",
"Definition:Epsilon-Net/Finite Net",
"Definition:Epsilon-Net... |
proofwiki-6341 | Separable Metric Space is Second-Countable | Let $M = \struct {A, d}$ be a metric space.
Let $M$ be separable.
Then $M$ is second-countable. | {{Recall|Second-Countable Space|second-countable space}}
{{:Definition:Second-Countable Space}}
By the definition of separability, we can choose a subset $S \subseteq A$ that is countable and everywhere dense.
Define:
:$\BB = \set {\map {B_{1/n} } x: x \in S, \, n \in \N_{>0} }$
where $\map {B_\epsilon } x$ denotes the... | Let $M = \struct {A, d}$ be a [[Definition:Metric Space|metric space]].
Let $M$ be [[Definition:Separable Space|separable]].
Then $M$ is [[Definition:Second-Countable Space|second-countable]]. | {{Recall|Second-Countable Space|second-countable space}}
{{:Definition:Second-Countable Space}}
By the definition of [[Definition:Separable Space|separability]], we can choose a [[Definition:Subset|subset]] $S \subseteq A$ that is [[Definition:Countable Set|countable]] and [[Definition:Everywhere Dense|everywhere dens... | Separable Metric Space is Second-Countable | https://proofwiki.org/wiki/Separable_Metric_Space_is_Second-Countable | https://proofwiki.org/wiki/Separable_Metric_Space_is_Second-Countable | [
"Metric Spaces",
"Second-Countable Spaces",
"Separable Spaces"
] | [
"Definition:Metric Space",
"Definition:Separable Space",
"Definition:Second-Countable Space"
] | [
"Definition:Separable Space",
"Definition:Subset",
"Definition:Countable Set",
"Definition:Everywhere Dense",
"Definition:Open Ball",
"Cartesian Product of Countable Sets is Countable",
"Image of Countable Set under Mapping is Countable",
"Definition:Countable Set",
"Definition:Topology Induced by M... |
proofwiki-6342 | Lipschitz Equivalence is Equivalence Relation | Let $A$ be a set.
Let $\DD$ be the set of all metrics on $A$.
Let $\sim$ be the relation on $\DD$ defined as:
:$\forall d_1, d_2 \in \DD: d_1 \sim d_2 \iff d_1$ is Lipschitz equivalent to $d_2$
Then $\sim$ is an equivalence relation. | Let $A$ be a set and let $\DD$ be the set of all metrics on $A$.
In the following, let $d_1, d_2, d_3 \subseteq \DD$ be arbitrary.
Checking in turn each of the criteria for equivalence: | Let $A$ be a [[Definition:Set|set]].
Let $\DD$ be the [[Definition:Set|set]] of all [[Definition:Metric|metrics]] on $A$.
Let $\sim$ be the [[Definition:Relation|relation]] on $\DD$ defined as:
:$\forall d_1, d_2 \in \DD: d_1 \sim d_2 \iff d_1$ is [[Definition:Lipschitz Equivalent Metrics|Lipschitz equivalent]] to $d... | Let $A$ be a [[Definition:Set|set]] and let $\DD$ be the [[Definition:Set|set]] of all [[Definition:Metric|metrics]] on $A$.
In the following, let $d_1, d_2, d_3 \subseteq \DD$ be arbitrary.
Checking in turn each of the criteria for [[Definition:Equivalence Relation|equivalence]]: | Lipschitz Equivalence is Equivalence Relation | https://proofwiki.org/wiki/Lipschitz_Equivalence_is_Equivalence_Relation | https://proofwiki.org/wiki/Lipschitz_Equivalence_is_Equivalence_Relation | [
"Lipschitz Equivalence",
"Examples of Equivalence Relations"
] | [
"Definition:Set",
"Definition:Set",
"Definition:Metric Space/Metric",
"Definition:Relation",
"Definition:Lipschitz Equivalence/Metrics",
"Definition:Equivalence Relation"
] | [
"Definition:Set",
"Definition:Set",
"Definition:Metric Space/Metric",
"Definition:Equivalence Relation",
"Definition:Metric Space/Metric",
"Definition:Equivalence Relation"
] |
proofwiki-6343 | Local Membership of Equalizer | Let $\mathbf C$ be a metacategory.
Let $e: E \to C$ be the equalizer of $f, g : C \to D$.
Then a variable element $z: Z \to C$ is a local member of $e$ {{iff}} $f \circ z = g \circ z$:
:$z \in_C e \iff f \circ z = g \circ z$ | Firstly, note that by Equalizer is Monomorphism, local membership of $e$ is defined. | Let $\mathbf C$ be a [[Definition:Metacategory|metacategory]].
Let $e: E \to C$ be the [[Definition:Equalizer|equalizer]] of $f, g : C \to D$.
Then a [[Definition:Variable Element|variable element]] $z: Z \to C$ is a [[Definition:Local Membership Relation|local member]] of $e$ {{iff}} $f \circ z = g \circ z$:
:$z \... | Firstly, note that by [[Equalizer is Monomorphism]], [[Definition:Local Membership Relation|local membership]] of $e$ is defined. | Local Membership of Equalizer | https://proofwiki.org/wiki/Local_Membership_of_Equalizer | https://proofwiki.org/wiki/Local_Membership_of_Equalizer | [
"Subobjects",
"Category Theory"
] | [
"Definition:Metacategory",
"Definition:Equalizer",
"Definition:Variable Element",
"Definition:Local Membership Relation"
] | [
"Equalizer is Monomorphism",
"Definition:Local Membership Relation",
"Definition:Local Membership Relation",
"Definition:Local Membership Relation"
] |
proofwiki-6344 | Sequentially Compact Metric Space is Separable | A sequentially compact metric space is separable. | This follows directly from:
* Sequentially Compact Metric Space is Totally Bounded
* Totally Bounded Metric Space is Separable.
{{qed}}
{{ACC|Totally Bounded Metric Space is Separable}}
Category:Separable Spaces
Category:Metric Spaces
Category:Sequentially Compact Spaces
5ngsevn0ih6kfj87bntzl2pno8t3wip | A [[Definition:Sequentially Compact Space|sequentially compact]] [[Definition:Metric Space|metric space]] is [[Definition:Separable Space|separable]]. | This follows directly from:
* [[Sequentially Compact Metric Space is Totally Bounded]]
* [[Totally Bounded Metric Space is Separable]].
{{qed}}
{{ACC|Totally Bounded Metric Space is Separable}}
[[Category:Separable Spaces]]
[[Category:Metric Spaces]]
[[Category:Sequentially Compact Spaces]]
5ngsevn0ih6kfj87bntzl2pno8... | Sequentially Compact Metric Space is Separable | https://proofwiki.org/wiki/Sequentially_Compact_Metric_Space_is_Separable | https://proofwiki.org/wiki/Sequentially_Compact_Metric_Space_is_Separable | [
"Separable Spaces",
"Metric Spaces",
"Sequentially Compact Spaces"
] | [
"Definition:Sequentially Compact Space",
"Definition:Metric Space",
"Definition:Separable Space"
] | [
"Sequentially Compact Metric Space is Totally Bounded",
"Totally Bounded Metric Space is Separable",
"Category:Separable Spaces",
"Category:Metric Spaces",
"Category:Sequentially Compact Spaces"
] |
proofwiki-6345 | Sequentially Compact Metric Space is Second-Countable | A sequentially compact metric space is second-countable. | This follows directly from:
: Sequentially Compact Metric Space is Separable
: Separable Metric Space is Second-Countable.
{{qed}}
{{ACC|Sequentially Compact Metric Space is Separable}}
Category:Second-Countable Spaces
Category:Metric Spaces
Category:Sequentially Compact Spaces
au7mekk5yhl715pl6nvwpjyl5tw6cpe | A [[Definition:Sequentially Compact Space|sequentially compact]] [[Definition:Metric Space|metric space]] is [[Definition:Second-Countable Space|second-countable]]. | This follows directly from:
: [[Sequentially Compact Metric Space is Separable]]
: [[Separable Metric Space is Second-Countable]].
{{qed}}
{{ACC|Sequentially Compact Metric Space is Separable}}
[[Category:Second-Countable Spaces]]
[[Category:Metric Spaces]]
[[Category:Sequentially Compact Spaces]]
au7mekk5yhl715pl6nv... | Sequentially Compact Metric Space is Second-Countable | https://proofwiki.org/wiki/Sequentially_Compact_Metric_Space_is_Second-Countable | https://proofwiki.org/wiki/Sequentially_Compact_Metric_Space_is_Second-Countable | [
"Second-Countable Spaces",
"Metric Spaces",
"Sequentially Compact Spaces"
] | [
"Definition:Sequentially Compact Space",
"Definition:Metric Space",
"Definition:Second-Countable Space"
] | [
"Sequentially Compact Metric Space is Separable",
"Separable Metric Space is Second-Countable",
"Category:Second-Countable Spaces",
"Category:Metric Spaces",
"Category:Sequentially Compact Spaces"
] |
proofwiki-6346 | Topological Equivalence is Equivalence Relation | Let $A$ be a set.
Let $\DD$ be the set of all metrics on $A$.
Let $\sim$ be the relation on $\DD$ defined as:
:$\forall d_1, d_2 \in \DD: d_1 \sim d_2 \iff d_1$ is topologically equivalent to $d_2$
Then $\sim$ is an equivalence relation. | Let $A$ be a set and let $\DD$ be the set of all metrics on $A$.
In the following, let $d_1, d_2, d_3 \subseteq \DD$ be arbitrary.
Checking in turn each of the criteria for equivalence: | Let $A$ be a [[Definition:Set|set]].
Let $\DD$ be the [[Definition:Set|set]] of all [[Definition:Metric|metrics]] on $A$.
Let $\sim$ be the [[Definition:Relation|relation]] on $\DD$ defined as:
:$\forall d_1, d_2 \in \DD: d_1 \sim d_2 \iff d_1$ is [[Definition:Topologically Equivalent Metrics|topologically equivalent... | Let $A$ be a [[Definition:Set|set]] and let $\DD$ be the [[Definition:Set|set]] of all [[Definition:Metric|metrics]] on $A$.
In the following, let $d_1, d_2, d_3 \subseteq \DD$ be arbitrary.
Checking in turn each of the criteria for [[Definition:Equivalence Relation|equivalence]]: | Topological Equivalence is Equivalence Relation | https://proofwiki.org/wiki/Topological_Equivalence_is_Equivalence_Relation | https://proofwiki.org/wiki/Topological_Equivalence_is_Equivalence_Relation | [
"Topologically Equivalent Metrics",
"Examples of Equivalence Relations"
] | [
"Definition:Set",
"Definition:Set",
"Definition:Metric Space/Metric",
"Definition:Relation",
"Definition:Topologically Equivalent Metrics",
"Definition:Equivalence Relation"
] | [
"Definition:Set",
"Definition:Set",
"Definition:Metric Space/Metric",
"Definition:Equivalence Relation",
"Definition:Metric Space/Metric",
"Definition:Equivalence Relation"
] |
proofwiki-6347 | Lipschitz Equivalent Metrics are Topologically Equivalent | Let $M_1 = \struct {A, d_1}$ and $M_2 = \struct {A, d_2}$ be metric spaces on the same underlying set $A$.
Let $d_1$ and $d_2$ be Lipschitz equivalent.
Then $d_1$ and $d_2$ are topologically equivalent. | Consider the identity mapping:
: $f: A \to A: \forall x \in A: f \left({x}\right) = x$
Then $f: \left({A, d_1}\right) \to \left({A, d_2}\right)$ can be considered as a Lipschitz equivalence.
The result then follows from Lipschitz Equivalent Metric Spaces are Homeomorphic.
{{qed}} | Let $M_1 = \struct {A, d_1}$ and $M_2 = \struct {A, d_2}$ be [[Definition:Metric Space|metric spaces]] on the same [[Definition:Underlying Set of Metric Space|underlying set]] $A$.
Let $d_1$ and $d_2$ be [[Definition:Lipschitz Equivalent Metrics|Lipschitz equivalent]].
Then $d_1$ and $d_2$ are [[Definition:Topologic... | Consider the [[Definition:Identity Mapping|identity mapping]]:
: $f: A \to A: \forall x \in A: f \left({x}\right) = x$
Then $f: \left({A, d_1}\right) \to \left({A, d_2}\right)$ can be considered as a [[Definition:Lipschitz Equivalent Metric Spaces|Lipschitz equivalence]].
The result then follows from [[Lipschitz Equi... | Lipschitz Equivalent Metrics are Topologically Equivalent/Proof 1 | https://proofwiki.org/wiki/Lipschitz_Equivalent_Metrics_are_Topologically_Equivalent | https://proofwiki.org/wiki/Lipschitz_Equivalent_Metrics_are_Topologically_Equivalent/Proof_1 | [
"Lipschitz Equivalence",
"Topologically Equivalent Metrics",
"Lipschitz Equivalent Metrics are Topologically Equivalent"
] | [
"Definition:Metric Space",
"Definition:Underlying Set/Metric Space",
"Definition:Lipschitz Equivalence/Metrics",
"Definition:Topologically Equivalent Metrics"
] | [
"Definition:Identity Mapping",
"Definition:Lipschitz Equivalence/Metric Spaces",
"Lipschitz Equivalent Metric Spaces are Homeomorphic"
] |
proofwiki-6348 | Lipschitz Equivalent Metrics are Topologically Equivalent | Let $M_1 = \struct {A, d_1}$ and $M_2 = \struct {A, d_2}$ be metric spaces on the same underlying set $A$.
Let $d_1$ and $d_2$ be Lipschitz equivalent.
Then $d_1$ and $d_2$ are topologically equivalent. | By definition of Lipschitz equivalence:
:$\forall x, y \in A: h \map {d_2} {x, y} \le \map {d_1} {x, y} \le k \map {d_2} {x, y}$
for some $h, k \in \R_{>0}$.
Let $x \in A$.
Let $\epsilon \in \R_{>0}$.
Let $\map {B_{h \epsilon} } {x; d_1}$ denote the open $h \epsilon$-ball with respect to $d_1$ of $x \in A$.
Then:
{{beg... | Let $M_1 = \struct {A, d_1}$ and $M_2 = \struct {A, d_2}$ be [[Definition:Metric Space|metric spaces]] on the same [[Definition:Underlying Set of Metric Space|underlying set]] $A$.
Let $d_1$ and $d_2$ be [[Definition:Lipschitz Equivalent Metrics|Lipschitz equivalent]].
Then $d_1$ and $d_2$ are [[Definition:Topologic... | By definition of [[Definition:Lipschitz Equivalent Metrics|Lipschitz equivalence]]:
:$\forall x, y \in A: h \map {d_2} {x, y} \le \map {d_1} {x, y} \le k \map {d_2} {x, y}$
for some $h, k \in \R_{>0}$.
Let $x \in A$.
Let $\epsilon \in \R_{>0}$.
Let $\map {B_{h \epsilon} } {x; d_1}$ denote the [[Definition:Open Ball ... | Lipschitz Equivalent Metrics are Topologically Equivalent/Proof 2 | https://proofwiki.org/wiki/Lipschitz_Equivalent_Metrics_are_Topologically_Equivalent | https://proofwiki.org/wiki/Lipschitz_Equivalent_Metrics_are_Topologically_Equivalent/Proof_2 | [
"Lipschitz Equivalence",
"Topologically Equivalent Metrics",
"Lipschitz Equivalent Metrics are Topologically Equivalent"
] | [
"Definition:Metric Space",
"Definition:Underlying Set/Metric Space",
"Definition:Lipschitz Equivalence/Metrics",
"Definition:Topologically Equivalent Metrics"
] | [
"Definition:Lipschitz Equivalence/Metrics",
"Definition:Open Ball",
"Definition:Open Set/Metric Space",
"Definition:Open Set/Metric Space",
"Definition:Open Set/Metric Space",
"Definition:Open Set/Metric Space"
] |
proofwiki-6349 | Lipschitz Equivalent Metrics are Topologically Equivalent | Let $M_1 = \struct {A, d_1}$ and $M_2 = \struct {A, d_2}$ be metric spaces on the same underlying set $A$.
Let $d_1$ and $d_2$ be Lipschitz equivalent.
Then $d_1$ and $d_2$ are topologically equivalent. | By definition of Lipschitz equivalence:
$\exists K_1, K_2 \in \R_{>0}$ such that:
:$(1): \quad \forall x, y \in A: \map {d_2} {x, y} \le K_1 \map {d_1} {x, y}$
:$(2): \quad \forall x, y \in A: \map {d_1} {x, y} \le K_2 \map {d_2} {x, y}$
By Identity Mapping between Metrics separated by Scale Factor is Continuous:
:the ... | Let $M_1 = \struct {A, d_1}$ and $M_2 = \struct {A, d_2}$ be [[Definition:Metric Space|metric spaces]] on the same [[Definition:Underlying Set of Metric Space|underlying set]] $A$.
Let $d_1$ and $d_2$ be [[Definition:Lipschitz Equivalent Metrics|Lipschitz equivalent]].
Then $d_1$ and $d_2$ are [[Definition:Topologic... | By definition of [[Definition:Lipschitz Equivalent Metrics|Lipschitz equivalence]]:
$\exists K_1, K_2 \in \R_{>0}$ such that:
:$(1): \quad \forall x, y \in A: \map {d_2} {x, y} \le K_1 \map {d_1} {x, y}$
:$(2): \quad \forall x, y \in A: \map {d_1} {x, y} \le K_2 \map {d_2} {x, y}$
By [[Identity Mapping between Metric... | Lipschitz Equivalent Metrics are Topologically Equivalent/Proof 3 | https://proofwiki.org/wiki/Lipschitz_Equivalent_Metrics_are_Topologically_Equivalent | https://proofwiki.org/wiki/Lipschitz_Equivalent_Metrics_are_Topologically_Equivalent/Proof_3 | [
"Lipschitz Equivalence",
"Topologically Equivalent Metrics",
"Lipschitz Equivalent Metrics are Topologically Equivalent"
] | [
"Definition:Metric Space",
"Definition:Underlying Set/Metric Space",
"Definition:Lipschitz Equivalence/Metrics",
"Definition:Topologically Equivalent Metrics"
] | [
"Definition:Lipschitz Equivalence/Metrics",
"Identity Mapping between Metrics separated by Scale Factor is Continuous",
"Definition:Identity Mapping",
"Definition:Continuous Mapping (Metric Space)/Space",
"Definition:Identity Mapping",
"Definition:Continuous Mapping (Metric Space)/Space",
"Definition:To... |
proofwiki-6350 | Compact Metric Space is Complete | A compact metric space is complete. | Follows directly from:
:Compact Space is Countably Compact
:Countably Compact Metric Space is Sequentially Compact
:Sequentially Compact Metric Space is Complete
{{qed}} | A [[Definition:Compact Metric Space|compact metric space]] is [[Definition:Complete Metric Space|complete]]. | Follows directly from:
:[[Compact Space is Countably Compact]]
:[[Countably Compact Metric Space is Sequentially Compact]]
:[[Sequentially Compact Metric Space is Complete]]
{{qed}} | Compact Metric Space is Complete | https://proofwiki.org/wiki/Compact_Metric_Space_is_Complete | https://proofwiki.org/wiki/Compact_Metric_Space_is_Complete | [
"Complete Metric Spaces",
"Compact Metric Spaces"
] | [
"Definition:Compact Space/Metric Space",
"Definition:Complete Metric Space"
] | [
"Compact Space is Countably Compact",
"Countably Compact Metric Space is Sequentially Compact",
"Sequentially Compact Metric Space is Complete"
] |
proofwiki-6351 | Pullback as Equalizer | Let $\mathbf C$ be a metacategory with binary products.
Suppose that the following diagram commutes (possibly except for the square):
$\quad\quad \begin{xy}\xymatrix@+1em{
P
\ar@/^/[rrd]^*+{p_1}
\ar@/_/[ddr]_*+{p_2}
\ar[rd]^*+{e}
\\
&
A \times B
\ar[r]_*+{\pi_1}
\ar[d]^*+{\pi_2}
&
A
\ar[d]^*+{f}
\\
&
B
\ar[r]_*+{g}
&
C... | Compare the pullback UMP:
:::<nowiki>$\begin{xy}\xymatrix{
Q
\ar@/^/[rrd]^*+{q_1}
\ar@/_/[ddr]_*+{q_2}
\ar@{.>}[dr]^*+{u}
\\
&
P
\ar[r]_*+{p_1}
\ar[d]^*+{p_2}
&
A
\ar[d]^*+{f}
\\
&
B
\ar[r]_*+{g}
&
C
}\end{xy}$</nowiki>
with the equalizer UMP:
:::<nowiki>$\begin{xy}\xymatrix{
P
\ar[r]^*{\family{p_1, p_2}}
&
A\time... | Let $\mathbf C$ be a [[Definition:Metacategory|metacategory]] [[Definition:Category with Binary Products|with binary products]].
Suppose that the following diagram [[Definition:Commutative Diagram|commutes]] (possibly except for the square):
$\quad\quad \begin{xy}\xymatrix@+1em{
P
\ar@/^/[rrd]^*+{p_1}
\ar@/_/[ddr]_*+... | Compare the [[Definition:Pullback UMP|pullback UMP]]:
:::<nowiki>$\begin{xy}\xymatrix{
Q
\ar@/^/[rrd]^*+{q_1}
\ar@/_/[ddr]_*+{q_2}
\ar@{.>}[dr]^*+{u}
\\
&
P
\ar[r]_*+{p_1}
\ar[d]^*+{p_2}
&
A
\ar[d]^*+{f}
\\
&
B
\ar[r]_*+{g}
&
C
}\end{xy}$</nowiki>
with the [[Definition:Equalizer UMP|equalizer UMP]]:
:::<nowiki>$\beg... | Pullback as Equalizer | https://proofwiki.org/wiki/Pullback_as_Equalizer | https://proofwiki.org/wiki/Pullback_as_Equalizer | [
"Pullbacks"
] | [
"Definition:Metacategory",
"Definition:Category with Products/Binary",
"Definition:Commutative Diagram",
"Definition:Product (Category Theory)/Binary Product",
"Definition:Product (Category Theory)/Binary Product Projection",
"Definition:Pullback (Category Theory)",
"Definition:Equalizer"
] | [
"Definition:Pullback UMP",
"Definition:Equalizer UMP"
] |
proofwiki-6352 | Minkowski's Inequality for Sums/Index 2 | Let $a_1, a_2, \ldots, a_n, b_1, b_2, \ldots, b_n \ge 0$ be non-negative real numbers.
Then:
:$\ds \paren {\sum_{k \mathop = 1}^n \paren {a_k + b_k}^2}^{1 / 2} \le \paren {\sum_{k \mathop = 1}^n a_k^2}^{1 / 2} + \paren {\sum_{k \mathop = 1}^n b_k^2}^{1 / 2}$ | {{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 1}^n \paren {a_k + b_k}^2
| r = \sum_{k \mathop = 1}^n \paren {a_k^2 + 2 a_k b_k + b_k^2}
}}
{{eqn | r = \sum_{k \mathop = 1}^n a_k^2 + 2 \sum_{k \mathop = 1}^n a_k b_k + \sum_{k \mathop = 1}^n b_k^2
}}
{{eqn | o = \le
| r = \sum_{k \mathop = 1}^n a_k^2 + 2 \paren... | Let $a_1, a_2, \ldots, a_n, b_1, b_2, \ldots, b_n \ge 0$ be [[Definition:Non-Negative Real Number|non-negative real numbers]].
Then:
:$\ds \paren {\sum_{k \mathop = 1}^n \paren {a_k + b_k}^2}^{1 / 2} \le \paren {\sum_{k \mathop = 1}^n a_k^2}^{1 / 2} + \paren {\sum_{k \mathop = 1}^n b_k^2}^{1 / 2}$ | {{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 1}^n \paren {a_k + b_k}^2
| r = \sum_{k \mathop = 1}^n \paren {a_k^2 + 2 a_k b_k + b_k^2}
}}
{{eqn | r = \sum_{k \mathop = 1}^n a_k^2 + 2 \sum_{k \mathop = 1}^n a_k b_k + \sum_{k \mathop = 1}^n b_k^2
}}
{{eqn | o = \le
| r = \sum_{k \mathop = 1}^n a_k^2 + 2 \paren... | Minkowski's Inequality for Sums/Index 2 | https://proofwiki.org/wiki/Minkowski's_Inequality_for_Sums/Index_2 | https://proofwiki.org/wiki/Minkowski's_Inequality_for_Sums/Index_2 | [
"Minkowski's Inequality for Sums"
] | [
"Definition:Positive/Real Number"
] | [
"Cauchy's Inequality",
"Order is Preserved on Positive Reals by Squaring"
] |
proofwiki-6353 | Banach-Steinhaus Theorem/Normed Vector Space | Let $\struct {X, \norm {\,\cdot\,}_X}$ be a Banach space.
Let $\struct {Y, \norm {\,\cdot\,}_Y}$ be a normed vector space.
Let $\family {T_\alpha: X \to Y}_{\alpha \mathop \in A}$ be an $A$-indexed family of bounded linear transformations from $X$ to $Y$.
Suppose that:
:$\ds \forall x \in X: \sup_{\alpha \mathop \in A}... | From Banach Space is F-Space, $\struct {X, \norm {\, \cdot \,}_X}$ can be considered as an $F$-Space.
From Normed Vector Space is Hausdorff Topological Vector Space, $\struct {Y, \norm {\, \cdot \,}_Y}$ can be considered as a topological vector space.
Let $\Gamma = \set {T_\alpha : \alpha \in A}$ and:
:$\map \Gamma x =... | Let $\struct {X, \norm {\,\cdot\,}_X}$ be a [[Definition:Banach Space|Banach space]].
Let $\struct {Y, \norm {\,\cdot\,}_Y}$ be a [[Definition:Normed Vector Space|normed vector space]].
Let $\family {T_\alpha: X \to Y}_{\alpha \mathop \in A}$ be an [[Definition:Indexed Family|$A$-indexed family]] of [[Definition:Boun... | From [[Banach Space is F-Space]], $\struct {X, \norm {\, \cdot \,}_X}$ can be considered as an [[Definition:F-Space|$F$-Space]].
From [[Normed Vector Space is Hausdorff Topological Vector Space]], $\struct {Y, \norm {\, \cdot \,}_Y}$ can be considered as a [[Definition:Topological Vector Space|topological vector space... | Banach-Steinhaus Theorem/Normed Vector Space/Proof 2 | https://proofwiki.org/wiki/Banach-Steinhaus_Theorem/Normed_Vector_Space | https://proofwiki.org/wiki/Banach-Steinhaus_Theorem/Normed_Vector_Space/Proof_2 | [
"Linear Transformations on Banach Spaces",
"Banach-Steinhaus Theorem"
] | [
"Definition:Banach Space",
"Definition:Normed Vector Space",
"Definition:Indexing Set/Family",
"Definition:Bounded Linear Transformation",
"Definition:Finite Extended Real Number",
"Definition:Finite Extended Real Number",
"Definition:Norm/Bounded Linear Transformation"
] | [
"Banach Space is F-Space",
"Definition:F-Space",
"Normed Vector Space is Hausdorff Topological Vector Space",
"Definition:Topological Vector Space",
"Characterization of von Neumann-Boundedness in Normed Vector Space",
"Definition:Von Neumann-Bounded Subset of Topological Vector Space",
"Banach-Steinhau... |
proofwiki-6354 | Banach-Steinhaus Theorem/Normed Vector Space | Let $\struct {X, \norm {\,\cdot\,}_X}$ be a Banach space.
Let $\struct {Y, \norm {\,\cdot\,}_Y}$ be a normed vector space.
Let $\family {T_\alpha: X \to Y}_{\alpha \mathop \in A}$ be an $A$-indexed family of bounded linear transformations from $X$ to $Y$.
Suppose that:
:$\ds \forall x \in X: \sup_{\alpha \mathop \in A}... | It suffices to show that there exist an $x_0 \in X$ and an $r \in \R_{>0}$ such that:
:$\ds K : = \sup_{x \mathop \in \map {B_r} {x_0} } \sup_{\alpha \mathop \in A} \norm {T_\alpha x}_Y$ is finite
where $\map {B_r} {x_0}$ is the open $r$-ball of $x_0$.
Indeed, then we have for all $x \in X \setminus \set 0$:
{{begin-eq... | Let $\struct {X, \norm {\,\cdot\,}_X}$ be a [[Definition:Banach Space|Banach space]].
Let $\struct {Y, \norm {\,\cdot\,}_Y}$ be a [[Definition:Normed Vector Space|normed vector space]].
Let $\family {T_\alpha: X \to Y}_{\alpha \mathop \in A}$ be an [[Definition:Indexed Family|$A$-indexed family]] of [[Definition:Boun... | It suffices to show that there exist an $x_0 \in X$ and an $r \in \R_{>0}$ such that:
:$\ds K : = \sup_{x \mathop \in \map {B_r} {x_0} } \sup_{\alpha \mathop \in A} \norm {T_\alpha x}_Y$ is [[Definition:Finite Extended Real Number|finite]]
where $\map {B_r} {x_0}$ is the [[Definition:Open Ball in Normed Vector Space|op... | Banach-Steinhaus Theorem/Normed Vector Space/Proof 3 | https://proofwiki.org/wiki/Banach-Steinhaus_Theorem/Normed_Vector_Space | https://proofwiki.org/wiki/Banach-Steinhaus_Theorem/Normed_Vector_Space/Proof_3 | [
"Linear Transformations on Banach Spaces",
"Banach-Steinhaus Theorem"
] | [
"Definition:Banach Space",
"Definition:Normed Vector Space",
"Definition:Indexing Set/Family",
"Definition:Bounded Linear Transformation",
"Definition:Finite Extended Real Number",
"Definition:Finite Extended Real Number",
"Definition:Norm/Bounded Linear Transformation"
] | [
"Definition:Finite Extended Real Number",
"Definition:Open Ball/Normed Vector Space",
"Definition:Continuous Mapping (Metric Space)",
"Definition:Cauchy Sequence",
"Definition:Banach Space",
"Definition:Closed Ball/Normed Vector Space",
"Definition:Contradiction"
] |
proofwiki-6355 | Minkowski's Inequality for Sums/Index Greater than 1 | Let $a_1, a_2, \ldots, a_n, b_1, b_2, \ldots, b_n \in \R_{\ge 0}$ be non-negative real numbers.
Let $p \in \R$ be a real number such that $p > 1$.
Then:
:$\ds \paren {\sum_{k \mathop = 1}^n \paren {a_k + b_k}^p}^{1 / p} \le \paren {\sum_{k \mathop = 1}^n {a_k}^p}^{1 / p} + \paren {\sum_{k \mathop = 1}^n {b_k}^p}^{1 / p... | Define:
:$q = \dfrac p {p - 1}$
Then:
:$\dfrac 1 p + \dfrac 1 q = \dfrac 1 p + \dfrac {p - 1} p = 1$
It follows that:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 1}^n \paren {a_k + b_k}^p
| r = \sum_{k \mathop = 1}^n a_k \paren {a_k + b_k}^{p - 1} + \sum_{k \mathop = 1}^n b_k \paren {a_k + b_k}^{p - 1}
| c ... | Let $a_1, a_2, \ldots, a_n, b_1, b_2, \ldots, b_n \in \R_{\ge 0}$ be [[Definition:Non-Negative Real Number|non-negative real numbers]].
Let $p \in \R$ be a [[Definition:Real Number|real number]] such that $p > 1$.
Then:
:$\ds \paren {\sum_{k \mathop = 1}^n \paren {a_k + b_k}^p}^{1 / p} \le \paren {\sum_{k \mathop = ... | Define:
:$q = \dfrac p {p - 1}$
Then:
:$\dfrac 1 p + \dfrac 1 q = \dfrac 1 p + \dfrac {p - 1} p = 1$
It follows that:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 1}^n \paren {a_k + b_k}^p
| r = \sum_{k \mathop = 1}^n a_k \paren {a_k + b_k}^{p - 1} + \sum_{k \mathop = 1}^n b_k \paren {a_k + b_k}^{p - 1}
|... | Minkowski's Inequality for Sums/Index Greater than 1/Proof 1 | https://proofwiki.org/wiki/Minkowski's_Inequality_for_Sums/Index_Greater_than_1 | https://proofwiki.org/wiki/Minkowski's_Inequality_for_Sums/Index_Greater_than_1/Proof_1 | [
"Minkowski's Inequality for Sums"
] | [
"Definition:Positive/Real Number",
"Definition:Real Number"
] | [
"Hölder's Inequality for Sums",
"Exponent Combination Laws/Power of Power"
] |
proofwiki-6356 | Minkowski's Inequality for Sums/Index Greater than 1 | Let $a_1, a_2, \ldots, a_n, b_1, b_2, \ldots, b_n \in \R_{\ge 0}$ be non-negative real numbers.
Let $p \in \R$ be a real number such that $p > 1$.
Then:
:$\ds \paren {\sum_{k \mathop = 1}^n \paren {a_k + b_k}^p}^{1 / p} \le \paren {\sum_{k \mathop = 1}^n {a_k}^p}^{1 / p} + \paren {\sum_{k \mathop = 1}^n {b_k}^p}^{1 / p... | Let $\mathbf a$ and $\mathbf b$ be real finite sequences:
{{begin-eqn}}
{{eqn | l = \mathbf a
| r = \sequence {a_k}_{1 \mathop \le k \mathop \le n}
}}
{{eqn | l = \mathbf b
| r = \sequence {b_k}_{1 \mathop \le k \mathop \le n}
}}
{{end-eqn}}
Let $\norm {\mathbf a}_p$ denote the $p$-norm of $\mathbf a$:
:$\n... | Let $a_1, a_2, \ldots, a_n, b_1, b_2, \ldots, b_n \in \R_{\ge 0}$ be [[Definition:Non-Negative Real Number|non-negative real numbers]].
Let $p \in \R$ be a [[Definition:Real Number|real number]] such that $p > 1$.
Then:
:$\ds \paren {\sum_{k \mathop = 1}^n \paren {a_k + b_k}^p}^{1 / p} \le \paren {\sum_{k \mathop = ... | Let $\mathbf a$ and $\mathbf b$ be [[Definition:Real Sequence|real]] [[Definition:Finite Sequence|finite sequences]]:
{{begin-eqn}}
{{eqn | l = \mathbf a
| r = \sequence {a_k}_{1 \mathop \le k \mathop \le n}
}}
{{eqn | l = \mathbf b
| r = \sequence {b_k}_{1 \mathop \le k \mathop \le n}
}}
{{end-eqn}}
Let ... | Minkowski's Inequality for Sums/Index Greater than 1/Proof 2 | https://proofwiki.org/wiki/Minkowski's_Inequality_for_Sums/Index_Greater_than_1 | https://proofwiki.org/wiki/Minkowski's_Inequality_for_Sums/Index_Greater_than_1/Proof_2 | [
"Minkowski's Inequality for Sums"
] | [
"Definition:Positive/Real Number",
"Definition:Real Number"
] | [
"Definition:Real Sequence",
"Definition:Finite Sequence",
"Definition:P-Norm/Real",
"Hölder's Inequality for Sums",
"Transformation of P-Norm"
] |
proofwiki-6357 | Minkowski's Inequality for Sums/Index Less than 1 | Let $a_1, a_2, \ldots, a_n, b_1, b_2, \ldots, b_n \in \R_{\ge 0}$ be non-negative real numbers.
Let $p \in \R$ be a real number.
If $p < 0$, then we require that $a_1, a_2, \ldots, a_n, b_1, b_2, \ldots, b_n$ be strictly positive.
If $p < 1$, $p \ne 0$, then:
:$\ds \paren {\sum_{k \mathop = 1}^n \paren {a_k + b_k}^p}^{... | In this case, $p$ and $q$ have opposite sign.
The proof then follows the same lines as the proof for $p > 1$, except that the Reverse Hölder's Inequality for Sums is applied instead. | Let $a_1, a_2, \ldots, a_n, b_1, b_2, \ldots, b_n \in \R_{\ge 0}$ be [[Definition:Non-Negative Real Number|non-negative real numbers]].
Let $p \in \R$ be a [[Definition:Real Number|real number]].
If $p < 0$, then we require that $a_1, a_2, \ldots, a_n, b_1, b_2, \ldots, b_n$ be [[Definition:Strictly Positive/Real Num... | In this case, $p$ and $q$ have opposite sign.
The proof then follows the same lines as [[Minkowski's Inequality for Sums/Index Greater than 1|the proof for $p > 1$]], except that the [[Reverse Hölder's Inequality for Sums]] is applied instead. | Minkowski's Inequality for Sums/Index Less than 1 | https://proofwiki.org/wiki/Minkowski's_Inequality_for_Sums/Index_Less_than_1 | https://proofwiki.org/wiki/Minkowski's_Inequality_for_Sums/Index_Less_than_1 | [
"Algebra",
"Analysis"
] | [
"Definition:Positive/Real Number",
"Definition:Real Number",
"Definition:Strictly Positive/Real Number"
] | [
"Minkowski's Inequality for Sums/Index Greater than 1",
"Reverse Hölder's Inequality for Sums"
] |
proofwiki-6358 | Monotone Real Function is Darboux Integrable | Let $\closedint a b$ be a closed real interval, where $a < b$.
Let $f: \closedint a b \to \R$ be a monotone real function.
Then $f$ is Darboux integrable over $\closedint a b$. | We consider the case where $f$ is increasing; the case where $f$ is decreasing is handled similarly.
Let $\epsilon \in \R_{>0}$ be a (strictly) positive real number.
By the Axiom of Archimedes, there exists a natural number $n$ such that:
:$n > \dfrac {\paren {b - a} \paren {\map f b - \map f a} } \epsilon$
For $k \in ... | Let $\closedint a b$ be a [[Definition:Closed Real Interval|closed real interval]], where $a < b$.
Let $f: \closedint a b \to \R$ be a [[Definition:Monotone Real Function|monotone real function]].
Then $f$ is [[Definition:Darboux Integrable Function|Darboux integrable]] over $\closedint a b$. | We consider the case where $f$ is [[Definition:Increasing Real Function|increasing]]; the case where $f$ is [[Definition:Decreasing Real Function|decreasing]] is handled similarly.
Let $\epsilon \in \R_{>0}$ be a [[Definition:Strictly Positive Real Number|(strictly) positive real number]].
By the [[Axiom of Archimed... | Monotone Real Function is Darboux Integrable | https://proofwiki.org/wiki/Monotone_Real_Function_is_Darboux_Integrable | https://proofwiki.org/wiki/Monotone_Real_Function_is_Darboux_Integrable | [
"Integral Calculus",
"Monotone Real Functions",
"Darboux Integrable Functions"
] | [
"Definition:Real Interval/Closed",
"Definition:Monotone (Order Theory)/Real Function",
"Definition:Darboux Integrable Function"
] | [
"Definition:Increasing/Real Function",
"Definition:Decreasing/Real Function",
"Definition:Strictly Positive/Real Number",
"Axiom of Archimedes",
"Definition:Natural Numbers",
"Definition:Subdivision of Interval",
"Definition:Lower Darboux Sum",
"Definition:Upper Darboux Sum",
"Definition:Subdivision... |
proofwiki-6359 | Topologies are not necessarily Comparable by Coarseness | Let $S$ be a set with at least $2$ elements.
Let $\mathbb T$ be the set of all topologies on $S$.
For two topologies $\tau_a, \tau_b \in \mathbb T$, let $\tau_a \le \tau_b$ denote that $\tau_a$ is coarser than $\tau_b$.
Then there exist $\tau_1, \tau_2 \in \mathbb T$ such that neither:
:$\tau_1 \le \tau_2$
nor:
:$\tau_... | Let $a, b \in S$.
Let:
:$\tau_a$ be the particular point topology with respect to $a$ on $S$
:$\tau_b$ be the particular point topology with respect to $b$ on $S$
From Homeomorphic Non-Comparable Particular Point Topologies:
:neither $\tau_a \le \tau_b$ nor $\tau_b \le \tau_a$.
Hence the result.
{{qed}} | Let $S$ be a [[Definition:Set|set]] with at least $2$ [[Definition:Element|elements]].
Let $\mathbb T$ be the [[Definition:Set|set]] of all [[Definition:Topology|topologies]] on $S$.
For two [[Definition:Topology|topologies]] $\tau_a, \tau_b \in \mathbb T$, let $\tau_a \le \tau_b$ denote that $\tau_a$ is [[Definition... | Let $a, b \in S$.
Let:
:$\tau_a$ be the [[Definition:Particular Point Topology|particular point topology]] with respect to $a$ on $S$
:$\tau_b$ be the [[Definition:Particular Point Topology|particular point topology]] with respect to $b$ on $S$
From [[Homeomorphic Non-Comparable Particular Point Topologies]]:
:neithe... | Topologies are not necessarily Comparable by Coarseness | https://proofwiki.org/wiki/Topologies_are_not_necessarily_Comparable_by_Coarseness | https://proofwiki.org/wiki/Topologies_are_not_necessarily_Comparable_by_Coarseness | [
"Coarser Topology"
] | [
"Definition:Set",
"Definition:Element",
"Definition:Set",
"Definition:Topology",
"Definition:Topology",
"Definition:Coarser Topology",
"Definition:Topology",
"Definition:Comparable Topologies"
] | [
"Definition:Particular Point Topology",
"Definition:Particular Point Topology",
"Homeomorphic Non-Comparable Particular Point Topologies"
] |
proofwiki-6360 | Euclidean Plane is Abstract Geometry | The Euclidean plane $\struct {\R^2, L_E}$ is an abstract geometry. | We will show that the axioms for an abstract geometry hold. | The [[Definition:Euclidean Plane|Euclidean plane]] $\struct {\R^2, L_E}$ is an [[Definition:Abstract Geometry|abstract geometry]]. | We will show that the axioms for an [[Definition:Abstract Geometry|abstract geometry]] hold. | Euclidean Plane is Abstract Geometry | https://proofwiki.org/wiki/Euclidean_Plane_is_Abstract_Geometry | https://proofwiki.org/wiki/Euclidean_Plane_is_Abstract_Geometry | [
"Abstract Geometries",
"Euclidean Geometry"
] | [
"Definition:Euclidean Plane",
"Definition:Abstract Geometry"
] | [
"Definition:Abstract Geometry",
"Definition:Abstract Geometry"
] |
proofwiki-6361 | Pullback of Subset Inclusion | Denote with $\mathbf{Set}$ the category of sets.
Let $A, B$ be sets, and let $f: A \to B$ be a mapping.
Let $V \subseteq B$ be a subset of $B$.
Denote with $i: V \to B$ the inclusion mapping.
Let $f^{-1} \left({V}\right) \subseteq A$ be the preimage of $V$ under $f$.
Denote with $j: f^{-1} \left({V}\right) \to A$ the i... | From the definition of pullback, given a commutative diagram:
$\quad\quad \begin{xy}\xymatrix{
Q
\ar[r]^*+{q_1}
\ar[d]_*+{q_2}
&
V
\ar[d]^*+{i}
\\
A
\ar[r]_*+{f}
&
B
}\end{xy}$
we need to find a mapping $u: Q \to \map {f^{-1}} V$ such that $\bar f \circ u = q_1$ and $j \circ u = q_2$.
Let $x \in Q$; then $\map {q_1} x ... | Denote with $\mathbf{Set}$ the [[Definition:Category of Sets|category of sets]].
Let $A, B$ be [[Definition:Set|sets]], and let $f: A \to B$ be a [[Definition:Mapping|mapping]].
Let $V \subseteq B$ be a [[Definition:Subset|subset]] of $B$.
Denote with $i: V \to B$ the [[Definition:Inclusion Mapping|inclusion mapping... | From the definition of [[Definition:Pullback (Category Theory)|pullback]], given a [[Definition:Commutative Diagram|commutative diagram]]:
$\quad\quad \begin{xy}\xymatrix{
Q
\ar[r]^*+{q_1}
\ar[d]_*+{q_2}
&
V
\ar[d]^*+{i}
\\
A
\ar[r]_*+{f}
&
B
}\end{xy}$
we need to find a [[Definition:Mapping|mapping]] $u: Q \to \map ... | Pullback of Subset Inclusion | https://proofwiki.org/wiki/Pullback_of_Subset_Inclusion | https://proofwiki.org/wiki/Pullback_of_Subset_Inclusion | [
"Pullbacks",
"Category of Sets",
"Subsets",
"Inclusion Mappings"
] | [
"Definition:Category of Sets",
"Definition:Set",
"Definition:Mapping",
"Definition:Subset",
"Definition:Inclusion Mapping",
"Definition:Preimage/Mapping/Subset",
"Definition:Inclusion Mapping",
"Definition:Restriction/Mapping",
"Definition:Pullback (Category Theory)"
] | [
"Definition:Pullback (Category Theory)",
"Definition:Commutative Diagram",
"Definition:Mapping",
"Definition:Inclusion Mapping",
"Definition:Preimage/Mapping/Subset",
"Definition:Restriction/Mapping",
"Definition:Inclusion Mapping",
"Equality of Mappings",
"Definition:Restriction/Mapping",
"Equali... |
proofwiki-6362 | Poincaré Plane is Abstract Geometry | The Poincaré plane $\struct {\H, L_H}$ is an abstract geometry. | We will show that the axioms for an abstract geometry hold. | The [[Definition:Poincaré Plane|Poincaré plane]] $\struct {\H, L_H}$ is an [[Definition:Abstract Geometry|abstract geometry]]. | We will show that the axioms for an [[Definition:Abstract Geometry|abstract geometry]] hold. | Poincaré Plane is Abstract Geometry | https://proofwiki.org/wiki/Poincaré_Plane_is_Abstract_Geometry | https://proofwiki.org/wiki/Poincaré_Plane_is_Abstract_Geometry | [
"Poincaré Plane",
"Abstract Geometries"
] | [
"Definition:Poincaré Plane",
"Definition:Abstract Geometry"
] | [
"Definition:Abstract Geometry",
"Definition:Abstract Geometry"
] |
proofwiki-6363 | Strong Separation Theorem | Let $C \subset \R^\ell$ be closed and convex.
Let $D = \set {\mathbf v} \subset C^c$.
Then $C$ and $D$ can be strongly separated. | === Lemma 1 ===
For any $\mathbf y \ne 0$ and any $\mathbf z \in \map {H_{\mathbf y}^<} r$, $r = \mathbf y \mathbf y$, the line segment joining $\mathbf y$ and $\mathbf z$ contains points with lengths strictly less than $\mathbf y$.
{{explain|Because of the loose use of commas to separate clauses, it is unclear what th... | Let $C \subset \R^\ell$ be closed and convex.
Let $D = \set {\mathbf v} \subset C^c$.
Then $C$ and $D$ can be strongly separated. | === Lemma 1 ===
For any $\mathbf y \ne 0$ and any $\mathbf z \in \map {H_{\mathbf y}^<} r$, $r = \mathbf y \mathbf y$, the line segment joining $\mathbf y$ and $\mathbf z$ contains points with lengths strictly less than $\mathbf y$.
{{explain|Because of the loose use of commas to separate clauses, it is unclear what ... | Strong Separation Theorem | https://proofwiki.org/wiki/Strong_Separation_Theorem | https://proofwiki.org/wiki/Strong_Separation_Theorem | [
"Convex Analysis",
"Named Theorems"
] | [] | [] |
proofwiki-6364 | Carathéodory's Theorem (Convex Analysis) | Let $E \subset \R^l$.
Let $\mathbb x \in \map {\operatorname {conv} } E$, where $\map {\operatorname {conv} } E$ denotes the convex hull of $E$.
Then $\mathbf x$ is a convex combination of affinely independent points of $E$.
In particular, $\mathbf x$ is a convex combination of at most $l + 1$ points of $E$. | Since $\mathbb x \in \map {\operatorname {conv} } E$, $\mathbf x$ is a convex combination of points in $E$.
From the definition of convex combination:
:$\ds \mathbf x = \sum_{i \mathop = 1}^k \gamma_i \mathbf y_i$
with:
:$\gamma_i \ge 0$
:$\ds \sum_{i \mathop = 1}^k \gamma_i = 1$
and:
:$\mathbf y_i \in E$
Let $K \subse... | Let $E \subset \R^l$.
Let $\mathbb x \in \map {\operatorname {conv} } E$, where $\map {\operatorname {conv} } E$ denotes the [[Definition:Convex Hull|convex hull]] of $E$.
Then $\mathbf x$ is a convex combination of [[Definition:Affinely Independent|affinely independent]] points of $E$.
In particular, $\mathbf x$ ... | Since $\mathbb x \in \map {\operatorname {conv} } E$, $\mathbf x$ is a convex combination of points in $E$.
From the definition of convex combination:
:$\ds \mathbf x = \sum_{i \mathop = 1}^k \gamma_i \mathbf y_i$
with:
:$\gamma_i \ge 0$
:$\ds \sum_{i \mathop = 1}^k \gamma_i = 1$
and:
:$\mathbf y_i \in E$
Let $K \s... | Carathéodory's Theorem (Convex Analysis) | https://proofwiki.org/wiki/Carathéodory's_Theorem_(Convex_Analysis) | https://proofwiki.org/wiki/Carathéodory's_Theorem_(Convex_Analysis) | [
"Convex Analysis"
] | [
"Definition:Convex Hull",
"Definition:Affinely Dependent/Independent"
] | [
"Well-Ordering Principle",
"Definition:Well-Ordering",
"Definition:Well-Ordering",
"Definition:Affinely Dependent",
"Definition:Affinely Dependent",
"Proof by Contradiction"
] |
proofwiki-6365 | Limit of Sinc Function at Zero/Corollary | :$\ds \lim_{x \mathop \to 0} \frac x {\sin x} = 1$ | From the Sine Inequality, we have:
:$\size {\sin x} \le \size x$
From Sine of Zero is Zero, equality occurs at $x = 0$.
From the Tangent Inequality, we have:
:$x < \tan x$
From Tangent of Zero, equality occurs at $x = 0$.
Putting the two inequalities together, we obtain:
:$\sin x \le x \le \tan x$
Therefore, dividing t... | :$\ds \lim_{x \mathop \to 0} \frac x {\sin x} = 1$ | From the [[Sine Inequality]], we have:
:$\size {\sin x} \le \size x$
From [[Sine of Zero is Zero]], [[Definition:Equality|equality]] occurs at $x = 0$.
From the [[Tangent Inequality]], we have:
:$x < \tan x$
From [[Tangent of Zero]], [[Definition:Equality|equality]] occurs at $x = 0$.
Putting the two [[Definition:I... | Limit of Sinc Function at Zero/Corollary | https://proofwiki.org/wiki/Limit_of_Sinc_Function_at_Zero/Corollary | https://proofwiki.org/wiki/Limit_of_Sinc_Function_at_Zero/Corollary | [
"Limit of Sinc Function at Zero",
"Sinc Function"
] | [] | [
"Sine Inequality",
"Sine of Zero is Zero",
"Definition:Equals",
"Tangent Inequality",
"Tangent of Zero",
"Definition:Equals",
"Definition:Inequality",
"Definition:Limit of Real Function",
"Definition:Term",
"Definition:Term",
"Squeeze Theorem",
"Category:Limit of Sinc Function at Zero",
"Cat... |
proofwiki-6366 | Dicyclic Group is Non-Abelian Group/Corollary | The quaternion group $Q_4$ is a non-abelian group. | The quaternion group $Q_4$ is an example of the dicyclic group with $4$ elements.
The result follows from Dicyclic Group is Non-Abelian Group.
{{qed}} | The [[Definition:Quaternion Group|quaternion group]] $Q_4$ is a non-[[Definition:Abelian Group|abelian]] [[Definition:Group|group]]. | The [[Definition:Quaternion Group|quaternion group]] $Q_4$ is an example of the [[Definition:Dicyclic Group|dicyclic group]] with $4$ elements.
The result follows from [[Dicyclic Group is Non-Abelian Group]].
{{qed}} | Dicyclic Group is Non-Abelian Group/Corollary | https://proofwiki.org/wiki/Dicyclic_Group_is_Non-Abelian_Group/Corollary | https://proofwiki.org/wiki/Dicyclic_Group_is_Non-Abelian_Group/Corollary | [
"Dicyclic Groups"
] | [
"Definition:Dicyclic Group/Quaternion Group",
"Definition:Abelian Group",
"Definition:Group"
] | [
"Definition:Dicyclic Group/Quaternion Group",
"Definition:Dicyclic Group",
"Dicyclic Group is Non-Abelian Group"
] |
proofwiki-6367 | Integral of Constant/Definite | :$\ds \int_a^b c \rd x = c \paren {b - a}$ | Let $f_c: \R \to \R$ be the constant function.
By definition:
:$\forall x \in \R: \map {f_c} x = c$
Thus:
:$\map \sup {f_c} = \map \inf {f_c} = c$
So from Darboux's Theorem, we have:
:$\ds c \paren {b - a} \le \int_a^b c \rd x \le c \paren {b - a}$
Hence the result.
{{qed}} | :$\ds \int_a^b c \rd x = c \paren {b - a}$ | Let $f_c: \R \to \R$ be the [[Definition:Constant Mapping|constant function]].
By definition:
:$\forall x \in \R: \map {f_c} x = c$
Thus:
:$\map \sup {f_c} = \map \inf {f_c} = c$
So from [[Darboux's Theorem]], we have:
:$\ds c \paren {b - a} \le \int_a^b c \rd x \le c \paren {b - a}$
Hence the result.
{{qed}} | Integral of Constant/Definite | https://proofwiki.org/wiki/Integral_of_Constant/Definite | https://proofwiki.org/wiki/Integral_of_Constant/Definite | [
"Definite Integrals"
] | [] | [
"Definition:Constant Mapping",
"Darboux's Theorem"
] |
proofwiki-6368 | Primitive of Constant | :$\ds \int c \rd x = c x + C$ where $C$ is an arbitrary constant. | Let:
:$\ds \map F x = \int c \rd x$
From the definition of primitive:
:$\map {F'} x = c$
From Derivative of Function of Constant Multiple:
:$\map {\dfrac \d {\d x} } {c x} = c$
From Primitives which Differ by Constant:
:$\map {\dfrac \d {\d x} } {c x + C} = c$
Hence the result.
{{qed}} | :$\ds \int c \rd x = c x + C$ where $C$ is an [[Definition:Arbitrary Constant (Calculus)|arbitrary constant]]. | Let:
:$\ds \map F x = \int c \rd x$
From the definition of [[Definition:Primitive (Calculus)|primitive]]:
:$\map {F'} x = c$
From [[Derivative of Function of Constant Multiple]]:
:$\map {\dfrac \d {\d x} } {c x} = c$
From [[Primitives which Differ by Constant]]:
:$\map {\dfrac \d {\d x} } {c x + C} = c$
Hence the r... | Primitive of Constant | https://proofwiki.org/wiki/Primitive_of_Constant | https://proofwiki.org/wiki/Primitive_of_Constant | [
"Primitives"
] | [
"Definition:Primitive (Calculus)/Constant of Integration"
] | [
"Definition:Primitive (Calculus)",
"Derivative of Function of Constant Multiple",
"Primitives which Differ by Constant"
] |
proofwiki-6369 | Primitive of Constant | :$\ds \int c \rd x = c x + C$ where $C$ is an arbitrary constant. | From Linear Combination of Primitives:
:$\ds \int \paren {\lambda \map f x + \mu \map g x} \rd x = \lambda \int \map f x \rd x + \mu \int \map g x \rd x$
The result follows by setting $\lambda = c$ and $\mu = 0$.
{{qed}} | :$\ds \int c \rd x = c x + C$ where $C$ is an [[Definition:Arbitrary Constant (Calculus)|arbitrary constant]]. | From [[Linear Combination of Primitives]]:
:$\ds \int \paren {\lambda \map f x + \mu \map g x} \rd x = \lambda \int \map f x \rd x + \mu \int \map g x \rd x$
The result follows by setting $\lambda = c$ and $\mu = 0$.
{{qed}} | Primitive of Constant Multiple of Function/Proof 1 | https://proofwiki.org/wiki/Primitive_of_Constant | https://proofwiki.org/wiki/Primitive_of_Constant_Multiple_of_Function/Proof_1 | [
"Primitives"
] | [
"Definition:Primitive (Calculus)/Constant of Integration"
] | [
"Linear Combination of Integrals/Indefinite"
] |
proofwiki-6370 | Primitive of Constant | :$\ds \int c \rd x = c x + C$ where $C$ is an arbitrary constant. | From Derivative of Constant Multiple:
:$\map {\dfrac \d {\d x} } {c \map f x} = c \map {\dfrac \d {\d x} } {\map f x}$
The result follows from the definition of primitive. | :$\ds \int c \rd x = c x + C$ where $C$ is an [[Definition:Arbitrary Constant (Calculus)|arbitrary constant]]. | From [[Derivative of Constant Multiple]]:
:$\map {\dfrac \d {\d x} } {c \map f x} = c \map {\dfrac \d {\d x} } {\map f x}$
The result follows from the definition of [[Definition:Primitive (Calculus)|primitive]]. | Primitive of Constant Multiple of Function/Proof 2 | https://proofwiki.org/wiki/Primitive_of_Constant | https://proofwiki.org/wiki/Primitive_of_Constant_Multiple_of_Function/Proof_2 | [
"Primitives"
] | [
"Definition:Primitive (Calculus)/Constant of Integration"
] | [
"Derivative of Constant Multiple",
"Definition:Primitive (Calculus)"
] |
proofwiki-6371 | Unbounded Set of Real Numbers is not Compact | Let $\R$ be the set of real numbers considered as a Euclidean space.
Let $S \subseteq \R$ be unbounded in $\R$.
Then $S$ is not a compact subspace of $\R$. | By the rule of transposition, it suffices to show that if $S$ is a compact subspace of $\R$, then $S$ is bounded.
Let $\CC$ be the set of all open $\epsilon$-balls of $0$ in $\R$:
:$\CC = \set {\map {B_\epsilon} 0: \epsilon \in \R_{>0}}$
We have that:
:$\ds \bigcup \CC = \R \supseteq S$
From Open Ball of Metric Space i... | Let $\R$ be the [[Definition:Real Number|set of real numbers]] considered as a [[Definition:Euclidean Space|Euclidean space]].
Let $S \subseteq \R$ be [[Definition:Unbounded Metric Space|unbounded]] in $\R$.
Then $S$ is not a [[Definition:Compact Subspace|compact subspace]] of $\R$. | By the [[Rule of Transposition|rule of transposition]], it suffices to show that if $S$ is a [[Definition:Compact Subspace|compact subspace]] of $\R$, then $S$ is [[Definition:Bounded Metric Space|bounded]].
Let $\CC$ be the [[Definition:Set|set]] of all [[Definition:Open Ball of Metric Space|open $\epsilon$-balls]] ... | Unbounded Set of Real Numbers is not Compact/Proof 1 | https://proofwiki.org/wiki/Unbounded_Set_of_Real_Numbers_is_not_Compact | https://proofwiki.org/wiki/Unbounded_Set_of_Real_Numbers_is_not_Compact/Proof_1 | [
"Unbounded Set of Real Numbers is not Compact",
"Compact Spaces (Real Analysis)",
"Real Analysis"
] | [
"Definition:Real Number",
"Definition:Euclidean Space",
"Definition:Bounded Metric Space/Unbounded",
"Definition:Compact Topological Space/Subspace"
] | [
"Rule of Transposition",
"Definition:Compact Topological Space/Subspace",
"Definition:Bounded Metric Space",
"Definition:Set",
"Definition:Open Ball",
"Open Ball is Open Set/Pseudometric Space",
"Definition:Open Cover",
"Definition:Subcover/Finite",
"Definition:Open Ball",
"Definition:Bounded Metr... |
proofwiki-6372 | Unbounded Set of Real Numbers is not Compact | Let $\R$ be the set of real numbers considered as a Euclidean space.
Let $S \subseteq \R$ be unbounded in $\R$.
Then $S$ is not a compact subspace of $\R$. | From:
* Real Number Line is Metric Space
* Compact Metric Space is Totally Bounded
* Totally Bounded Metric Space is Bounded
the result follows by the rule of transposition.
{{qed}} | Let $\R$ be the [[Definition:Real Number|set of real numbers]] considered as a [[Definition:Euclidean Space|Euclidean space]].
Let $S \subseteq \R$ be [[Definition:Unbounded Metric Space|unbounded]] in $\R$.
Then $S$ is not a [[Definition:Compact Subspace|compact subspace]] of $\R$. | From:
* [[Real Number Line is Metric Space]]
* [[Compact Metric Space is Totally Bounded]]
* [[Totally Bounded Metric Space is Bounded]]
the result follows by the [[Rule of Transposition|rule of transposition]].
{{qed}} | Unbounded Set of Real Numbers is not Compact/Proof 2 | https://proofwiki.org/wiki/Unbounded_Set_of_Real_Numbers_is_not_Compact | https://proofwiki.org/wiki/Unbounded_Set_of_Real_Numbers_is_not_Compact/Proof_2 | [
"Unbounded Set of Real Numbers is not Compact",
"Compact Spaces (Real Analysis)",
"Real Analysis"
] | [
"Definition:Real Number",
"Definition:Euclidean Space",
"Definition:Bounded Metric Space/Unbounded",
"Definition:Compact Topological Space/Subspace"
] | [
"Real Number Line is Metric Space",
"Compact Metric Space is Totally Bounded",
"Totally Bounded Metric Space is Bounded",
"Rule of Transposition"
] |
proofwiki-6373 | Continuous Mapping is Sequentially Continuous | Let $X$ and $Y$ be topological spaces.
Let $x \in X$.
Let $f: X \to Y$ be a mapping that is continuous at $x$.
Then $f$ is sequentially continuous at $x$. | Let $\sequence {x_n}_{n \mathop \ge 1}$ be a sequence in $X$ converging to $x$.
Let $V$ be a neighborhood of $\map f x$ in $Y$.
We are required to show that there exists $N \in \N$ such that $\map f {x_n} \in V$ for all $n \ge N$.
By continuity of $f$, choose a neighborhood $U$ of $x$ in $X$ such that $\map f U \subset... | Let $X$ and $Y$ be [[Definition:Topological Space|topological spaces]].
Let $x \in X$.
Let $f: X \to Y$ be a [[Definition:Mapping|mapping]] that is [[Definition:Continuous Mapping at Point (Topology)|continuous at $x$]].
Then $f$ is [[Definition:Sequential Continuity|sequentially continuous at $x$]]. | Let $\sequence {x_n}_{n \mathop \ge 1}$ be a [[Definition:Sequence|sequence]] in $X$ [[Definition:Convergent Sequence (Topology)|converging]] to $x$.
Let $V$ be a [[Definition:Neighborhood of Point|neighborhood]] of $\map f x$ in $Y$.
We are required to show that there exists $N \in \N$ such that $\map f {x_n} \in V$... | Continuous Mapping is Sequentially Continuous | https://proofwiki.org/wiki/Continuous_Mapping_is_Sequentially_Continuous | https://proofwiki.org/wiki/Continuous_Mapping_is_Sequentially_Continuous | [
"Continuous Mappings",
"Sequential Continuity"
] | [
"Definition:Topological Space",
"Definition:Mapping",
"Definition:Continuous Mapping (Topology)/Point",
"Definition:Sequential Continuity"
] | [
"Definition:Sequence",
"Definition:Convergent Sequence/Topology",
"Definition:Neighborhood (Topology)/Point",
"Definition:Continuous Mapping (Topology)/Everywhere",
"Definition:Neighborhood (Topology)/Point",
"Category:Continuous Mappings",
"Category:Sequential Continuity"
] |
proofwiki-6374 | Sequential Continuity is Equivalent to Continuity in Metric Space | Let $\struct {X, d}$ and $\struct {Y, e}$ be metric spaces.
Let $f: X \to Y$ be a mapping.
Let $x \in X$.
Then $f$ is continuous at $x$ {{iff}} $f$ is sequentially continuous at $x$. | We have that a Continuous Mapping is Sequentially Continuous.
To prove the converse, by the Rule of Transposition we may prove the contrapositive:
:If $f$ is not continuous at $x$, then $f$ is not sequentially continuous at $x$.
We suppose therefore that there exists $\epsilon_0 > 0$ such that for all $\delta > 0$ the... | Let $\struct {X, d}$ and $\struct {Y, e}$ be [[Definition:Metric Space|metric spaces]].
Let $f: X \to Y$ be a [[Definition:Mapping|mapping]].
Let $x \in X$.
Then $f$ is [[Definition:Continuous at Point of Metric Space|continuous at $x$]] {{iff}} $f$ is [[Definition:Sequential Continuity|sequentially continuous at $... | We have that a [[Continuous Mapping is Sequentially Continuous]].
To prove the [[Definition:Converse Statement|converse]], by the [[Rule of Transposition]] we may prove the [[Definition:Contrapositive Statement|contrapositive]]:
:If $f$ is not [[Definition:Continuous at Point of Metric Space|continuous at $x$]], then ... | Sequential Continuity is Equivalent to Continuity in Metric Space | https://proofwiki.org/wiki/Sequential_Continuity_is_Equivalent_to_Continuity_in_Metric_Space | https://proofwiki.org/wiki/Sequential_Continuity_is_Equivalent_to_Continuity_in_Metric_Space | [
"Metric Spaces",
"Sequential Continuity",
"Continuous Mappings on Metric Spaces"
] | [
"Definition:Metric Space",
"Definition:Mapping",
"Definition:Continuous Mapping (Metric Space)/Point",
"Definition:Sequential Continuity"
] | [
"Continuous Mapping is Sequentially Continuous",
"Definition:Converse Statement",
"Rule of Transposition",
"Definition:Contrapositive Statement",
"Definition:Continuous Mapping (Metric Space)/Point",
"Definition:Sequential Continuity",
"Definition:Sequence",
"Definition:Convergent Sequence/Metric Spac... |
proofwiki-6375 | Pullback of Commutative Triangle | Let $\mathbf C$ be a metacategory.
Suppose that the following is a commutative diagram in $\mathbf C$:
::<nowiki>$\begin{xy}\xymatrix@+1em@L+2px{
A'
\ar[rr]^*{h_\alpha}
\ar[dd]_*{\alpha'}
& &
A
\ar[rd]^*{\gamma}
\ar[dd]^(.4)*{\alpha}
\\ &
B'
\ar[ld]^*{\beta'}
\ar[rr] |{\hole} ^(.3)*{h_\beta}
& &
B
\ar... | The first diagram above can be distorted into:
::<nowiki>$\begin{xy}\xymatrix@+.5em@L+2px{
A'
\ar@/^/[rrd]^*{\gamma \circ h_\alpha}
\ar@/_/[ddr]_*{\alpha'}
\\ &
B'
\ar[r]^*{h_\beta}
\ar[d]^*{\beta'}
&
B
\ar[d]^*{\beta}
\\ &
C'
\ar[r]_*{h}
&
C
}\end{xy}$</nowiki>
and since the square is a pullback, ther... | Let $\mathbf C$ be a [[Definition:Metacategory|metacategory]].
Suppose that the following is a [[Definition:Commutative Diagram|commutative diagram]] in $\mathbf C$:
::<nowiki>$\begin{xy}\xymatrix@+1em@L+2px{
A'
\ar[rr]^*{h_\alpha}
\ar[dd]_*{\alpha'}
& &
A
\ar[rd]^*{\gamma}
\ar[dd]^(.4)*{\alpha}
\\ &
B'
... | The first diagram above can be distorted into:
::<nowiki>$\begin{xy}\xymatrix@+.5em@L+2px{
A'
\ar@/^/[rrd]^*{\gamma \circ h_\alpha}
\ar@/_/[ddr]_*{\alpha'}
\\ &
B'
\ar[r]^*{h_\beta}
\ar[d]^*{\beta'}
&
B
\ar[d]^*{\beta}
\\ &
C'
\ar[r]_*{h}
&
C
}\end{xy}$</nowiki>
and since the square is a [[Definiti... | Pullback of Commutative Triangle | https://proofwiki.org/wiki/Pullback_of_Commutative_Triangle | https://proofwiki.org/wiki/Pullback_of_Commutative_Triangle | [
"Pullbacks"
] | [
"Definition:Metacategory",
"Definition:Commutative Diagram",
"Definition:Pullback Diagram",
"Definition:Unique",
"Definition:Morphism",
"Definition:Pullback (Category Theory)"
] | [
"Definition:Pullback (Category Theory)",
"Definition:Unique",
"Definition:Pullback (Category Theory)",
"Pullback Lemma"
] |
proofwiki-6376 | Associative Commutative Idempotent Operation is Self-Distributive | Let $\struct {S, \circ}$ be an algebraic structure, such that:
{{begin-itemize}}
{{item|(1):|$\circ$ is associative}}
{{item|(2):|$\circ$ is commutative}}
{{item|(3):|$\circ$ is idempotent}}
{{end-itemize}}
Then $\circ$ is self-distributive.
That is:
:$\forall a, b, c \in S: \paren {a \circ b} \circ \paren {a \circ c} ... | {{begin-eqn}}
{{eqn | l = \paren {a \circ b} \circ \paren {a \circ c}
| r = a \circ \paren {b \circ a} \circ c
| c = $\circ$ is associative
}}
{{eqn | r = a \circ \paren {a \circ b} \circ c
| c = $\circ$ is commutative
}}
{{eqn | r = \paren {a \circ a} \circ b \circ c
| c = $\circ$ is associativ... | Let $\struct {S, \circ}$ be an [[Definition:Algebraic Structure with One Operation|algebraic structure]], such that:
{{begin-itemize}}
{{item|(1):|$\circ$ is [[Definition:Associative Operation|associative]]}}
{{item|(2):|$\circ$ is [[Definition:Commutative Operation|commutative]]}}
{{item|(3):|$\circ$ is [[Definition:I... | {{begin-eqn}}
{{eqn | l = \paren {a \circ b} \circ \paren {a \circ c}
| r = a \circ \paren {b \circ a} \circ c
| c = $\circ$ is [[Definition:Associative Operation|associative]]
}}
{{eqn | r = a \circ \paren {a \circ b} \circ c
| c = $\circ$ is [[Definition:Commutative Operation|commutative]]
}}
{{eqn ... | Associative Commutative Idempotent Operation is Self-Distributive | https://proofwiki.org/wiki/Associative_Commutative_Idempotent_Operation_is_Self-Distributive | https://proofwiki.org/wiki/Associative_Commutative_Idempotent_Operation_is_Self-Distributive | [
"Abstract Algebra",
"Commutativity",
"Associativity",
"Idempotence",
"Self-Distributive Operations",
"Semilattices"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Associative Operation",
"Definition:Commutative/Operation",
"Definition:Idempotence/Operation",
"Definition:Self-Distributive Operation"
] | [
"Definition:Associative Operation",
"Definition:Commutative/Operation",
"Definition:Associative Operation",
"Definition:Idempotence/Operation",
"Definition:Associative Operation",
"Definition:Commutative/Operation",
"Definition:Associative Operation",
"Definition:Idempotence/Operation",
"Category:Ab... |
proofwiki-6377 | Set Intersection is Self-Distributive | Set intersection is self-distributive:
:$\forall A, B, C: \paren {A \cap B} \cap \paren {A \cap C} = A \cap B \cap C = \paren {A \cap C} \cap \paren {B \cap C}$
where $A, B, C$ are sets. | We have:
:Intersection is Associative
:Intersection is Commutative
:Set Intersection is Idempotent
The result follows from Associative Commutative Idempotent Operation is Self-Distributive.
{{qed}}
Category:Set Intersection
Category:Examples of Self-Distributive Operations
Category:Set Intersection is Self-Distributive... | [[Definition:Set Intersection|Set intersection]] is [[Definition:Self-Distributive Operation|self-distributive]]:
:$\forall A, B, C: \paren {A \cap B} \cap \paren {A \cap C} = A \cap B \cap C = \paren {A \cap C} \cap \paren {B \cap C}$
where $A, B, C$ are [[Definition:Set|sets]]. | We have:
:[[Intersection is Associative]]
:[[Intersection is Commutative]]
:[[Set Intersection is Idempotent]]
The result follows from [[Associative Commutative Idempotent Operation is Self-Distributive]].
{{qed}}
[[Category:Set Intersection]]
[[Category:Examples of Self-Distributive Operations]]
[[Category:Set Inter... | Set Intersection is Self-Distributive | https://proofwiki.org/wiki/Set_Intersection_is_Self-Distributive | https://proofwiki.org/wiki/Set_Intersection_is_Self-Distributive | [
"Set Intersection",
"Examples of Self-Distributive Operations",
"Set Intersection is Self-Distributive"
] | [
"Definition:Set Intersection",
"Definition:Self-Distributive Operation",
"Definition:Set"
] | [
"Intersection is Associative",
"Intersection is Commutative",
"Set Intersection is Idempotent",
"Associative Commutative Idempotent Operation is Self-Distributive",
"Category:Set Intersection",
"Category:Examples of Self-Distributive Operations",
"Category:Set Intersection is Self-Distributive"
] |
proofwiki-6378 | Inclusion Mapping is Restriction of Identity | Let $T$ be a set.
Let $S \subseteq T$ be a subset of $T$.
Let $i_S: S \to T$ be the inclusion mapping on $S$.
Then $i_S$ is the restriction of the identity mapping $I_T: T \to T$ on $T$. | By definition of inclusion mapping:
:$i_S: S \to T: \forall x \in S: \map {i_S} x = x$
By definition of identity mapping:
:$I_T: T \to T: \forall x \in T: \map {I_T} x = x$
The result follows by definition of restriction of mapping.
{{qed}} | Let $T$ be a [[Definition:Set|set]].
Let $S \subseteq T$ be a [[Definition:Subset|subset]] of $T$.
Let $i_S: S \to T$ be the [[Definition:Inclusion Mapping|inclusion mapping]] on $S$.
Then $i_S$ is the [[Definition:Restriction of Mapping|restriction]] of the [[Definition:Identity Mapping|identity mapping]] $I_T: T ... | By definition of [[Definition:Inclusion Mapping|inclusion mapping]]:
:$i_S: S \to T: \forall x \in S: \map {i_S} x = x$
By definition of [[Definition:Identity Mapping|identity mapping]]:
:$I_T: T \to T: \forall x \in T: \map {I_T} x = x$
The result follows by definition of [[Definition:Restriction of Mapping|restrict... | Inclusion Mapping is Restriction of Identity | https://proofwiki.org/wiki/Inclusion_Mapping_is_Restriction_of_Identity | https://proofwiki.org/wiki/Inclusion_Mapping_is_Restriction_of_Identity | [
"Inclusion Mappings",
"Identity Mappings"
] | [
"Definition:Set",
"Definition:Subset",
"Definition:Inclusion Mapping",
"Definition:Restriction/Mapping",
"Definition:Identity Mapping"
] | [
"Definition:Inclusion Mapping",
"Definition:Identity Mapping",
"Definition:Restriction/Mapping"
] |
proofwiki-6379 | Preimage of Subset under Inclusion Mapping | Let $S$ be a set.
Let $H \subseteq S$ be a subset of $S$.
Let $i_H: H \to S$ be the inclusion mapping on $H$.
Let $T \subseteq S$.
Then:
:$i_H^{-1} \sqbrk T = T \cap H$
where $i_H^{-1} \sqbrk T$ is the preimage of $T$ under $i_H$. | By definition of preimage:
:$i_H^{-1} \sqbrk T = \set {h \in H: \exists t \in T: \map {i_H} h = t}$
Let $T \cap H = \O$.
Then:
:$\nexists h \in H: \exists t \in T: \map {i_H} h = t$
That is:
:$i_H^{-1} \sqbrk T = \O$
That is:
:$i_H^{-1} \sqbrk T = T \cap H$
{{qed|lemma}}
Let $T \cap H \ne \O$.
From Intersection is Subs... | Let $S$ be a [[Definition:Set|set]].
Let $H \subseteq S$ be a [[Definition:Subset|subset]] of $S$.
Let $i_H: H \to S$ be the [[Definition:Inclusion Mapping|inclusion mapping]] on $H$.
Let $T \subseteq S$.
Then:
:$i_H^{-1} \sqbrk T = T \cap H$
where $i_H^{-1} \sqbrk T$ is the [[Definition:Preimage of Subset under M... | By definition of [[Definition:Preimage of Subset under Mapping|preimage]]:
:$i_H^{-1} \sqbrk T = \set {h \in H: \exists t \in T: \map {i_H} h = t}$
Let $T \cap H = \O$.
Then:
:$\nexists h \in H: \exists t \in T: \map {i_H} h = t$
That is:
:$i_H^{-1} \sqbrk T = \O$
That is:
:$i_H^{-1} \sqbrk T = T \cap H$
{{qed|lem... | Preimage of Subset under Inclusion Mapping | https://proofwiki.org/wiki/Preimage_of_Subset_under_Inclusion_Mapping | https://proofwiki.org/wiki/Preimage_of_Subset_under_Inclusion_Mapping | [
"Inclusion Mappings",
"Subsets"
] | [
"Definition:Set",
"Definition:Subset",
"Definition:Inclusion Mapping",
"Definition:Preimage/Mapping/Subset"
] | [
"Definition:Preimage/Mapping/Subset",
"Intersection is Subset",
"Subset of Empty Set",
"Empty Set is Subset of All Sets",
"Definition:Inclusion Mapping",
"Definition:Preimage/Mapping/Subset",
"Definition:Inclusion Mapping",
"Definition:Set Intersection",
"Definition:Subset",
"Definition:Set Inters... |
proofwiki-6380 | Inclusion Mapping is Continuous | Let $T = \struct {S, \tau}$ be a topological space.
Let $T_H = \struct {H, \tau_H}$ be a topological subspace of $T$ where $H \subseteq S$.
Let $i_H: H \to S$ be the inclusion mapping on $H$.
Then $i_H$ is a $\struct {\tau_H, \tau}$-continuous mapping. | Let $I_H$ be the identity mapping on $H$.
From Continuity of Composite with Inclusion: Inclusion on Mapping, $I_H$ is $\tuple {\tau_H , \tau_H}$-continuous {{iff}} $i_H \circ I_H$ is $\tuple {\tau_H , \tau}$-continuous.
From Identity Mapping is Right Identity, it follows that:
:$i_H \circ I_H = i_H$
Therefore, $I_H$ is... | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $T_H = \struct {H, \tau_H}$ be a [[Definition:Topological Subspace|topological subspace]] of $T$ where $H \subseteq S$.
Let $i_H: H \to S$ be the [[Definition:Inclusion Mapping|inclusion mapping]] on $H$.
Then $i_H$ is a [[Defi... | Let $I_H$ be the [[Definition:Identity Mapping|identity mapping]] on $H$.
From [[Continuity of Composite with Inclusion/Inclusion on Mapping|Continuity of Composite with Inclusion: Inclusion on Mapping]], $I_H$ is [[Definition:Everywhere Continuous Mapping (Topology)|$\tuple {\tau_H , \tau_H}$-continuous]] {{iff}} $i_... | Inclusion Mapping is Continuous/Proof 2 | https://proofwiki.org/wiki/Inclusion_Mapping_is_Continuous | https://proofwiki.org/wiki/Inclusion_Mapping_is_Continuous/Proof_2 | [
"Inclusion Mapping is Continuous"
] | [
"Definition:Topological Space",
"Definition:Topological Subspace",
"Definition:Inclusion Mapping",
"Definition:Continuous Mapping (Topology)/Everywhere"
] | [
"Definition:Identity Mapping",
"Continuity of Composite with Inclusion/Inclusion on Mapping",
"Definition:Continuous Mapping (Topology)/Everywhere",
"Definition:Continuous Mapping (Topology)/Everywhere",
"Identity Mapping is Right Identity",
"Definition:Continuous Mapping (Topology)/Everywhere",
"Defini... |
proofwiki-6381 | Continuity of Composite with Inclusion/Mapping on Inclusion | If $f$ is $\tuple {\tau, \tau'}$-continuous, then $f \circ i$ is $\tuple {\tau_H, \tau'}$-continuous | From Inclusion Mapping is Continuous, $i$ is $\tuple {\tau_H, \tau}$-continuous.
It follows from Composite of Continuous Mappings is Continuous that $f \circ i$ is $\tuple {\tau_H, \tau'}$-continuous.
{{qed}} | If $f$ is [[Definition:Continuous Mapping (Topology)|$\tuple {\tau, \tau'}$-continuous]], then $f \circ i$ is [[Definition:Continuous Mapping (Topology)|$\tuple {\tau_H, \tau'}$-continuous]] | From [[Inclusion Mapping is Continuous]], $i$ is [[Definition:Continuous Mapping (Topology)|$\tuple {\tau_H, \tau}$-continuous]].
It follows from [[Composite of Continuous Mappings is Continuous]] that $f \circ i$ is [[Definition:Continuous Mapping (Topology)|$\tuple {\tau_H, \tau'}$-continuous]].
{{qed}} | Continuity of Composite with Inclusion/Mapping on Inclusion | https://proofwiki.org/wiki/Continuity_of_Composite_with_Inclusion/Mapping_on_Inclusion | https://proofwiki.org/wiki/Continuity_of_Composite_with_Inclusion/Mapping_on_Inclusion | [
"Continuity of Composite with Inclusion"
] | [
"Definition:Continuous Mapping (Topology)",
"Definition:Continuous Mapping (Topology)"
] | [
"Inclusion Mapping is Continuous",
"Definition:Continuous Mapping (Topology)",
"Composite of Continuous Mappings is Continuous",
"Definition:Continuous Mapping (Topology)"
] |
proofwiki-6382 | Continuity of Composite with Inclusion/Inclusion on Mapping | $g$ is $\tuple {\tau', \tau_H}$-continuous {{iff}} $i \circ g$ is $\tuple {\tau', \tau}$-continuous. | === Necessary Condition ===
Suppose $g$ is $\tuple {\tau', \tau_H}$-continuous.
From Inclusion Mapping is Continuous, $i$ is $\tuple {\tau_H, \tau}$-continuous.
It follows from Composite of Continuous Mappings is Continuous that $i \circ g$ is $\tuple {\tau', \tau}$-continuous. | $g$ is [[Definition:Continuous Mapping (Topology)|$\tuple {\tau', \tau_H}$-continuous]] {{iff}} $i \circ g$ is [[Definition:Continuous Mapping (Topology)|$\tuple {\tau', \tau}$-continuous]]. | === Necessary Condition ===
Suppose $g$ is [[Definition:Continuous Mapping (Topology)|$\tuple {\tau', \tau_H}$-continuous]].
From [[Inclusion Mapping is Continuous]], $i$ is [[Definition:Continuous Mapping (Topology)|$\tuple {\tau_H, \tau}$-continuous]].
It follows from [[Composite of Continuous Mappings is Continuo... | Continuity of Composite with Inclusion/Inclusion on Mapping | https://proofwiki.org/wiki/Continuity_of_Composite_with_Inclusion/Inclusion_on_Mapping | https://proofwiki.org/wiki/Continuity_of_Composite_with_Inclusion/Inclusion_on_Mapping | [
"Continuity of Composite with Inclusion"
] | [
"Definition:Continuous Mapping (Topology)",
"Definition:Continuous Mapping (Topology)"
] | [
"Definition:Continuous Mapping (Topology)",
"Inclusion Mapping is Continuous",
"Definition:Continuous Mapping (Topology)",
"Composite of Continuous Mappings is Continuous",
"Definition:Continuous Mapping (Topology)",
"Definition:Continuous Mapping (Topology)",
"Definition:Continuous Mapping (Topology)"
... |
proofwiki-6383 | Continuity of Composite with Inclusion/Uniqueness of Induced Topology | The induced topology $\tau_H$ is the ''only'' topology on $H$ satisfying Continuity of Composite with Inclusion: Inclusion on Mapping for all possible $g$. | Suppose $\tau' '$ is a topology on $H$ such that:
:$(1) \quad$ For any topological space $T' = \struct {A', \tau'}$, and
:$(2) \quad$ For any mapping $g: A' \to H$:
$g$ is $\tuple {\tau', \tau' '}$-continuous {{iff}} $i \circ g$ is $\tuple {\tau', \tau}$-continuous.
It needs to be shown that $\tau' '$ must be the same ... | The [[Definition:Topological Subspace|induced topology]] $\tau_H$ is the ''only'' [[Definition:Topology|topology]] on $H$ satisfying [[Continuity of Composite with Inclusion/Inclusion on Mapping|Continuity of Composite with Inclusion: Inclusion on Mapping]] for all possible $g$. | Suppose $\tau' '$ is a [[Definition:Topology|topology]] on $H$ such that:
:$(1) \quad$ For any [[Definition:Topological Space|topological space]] $T' = \struct {A', \tau'}$, and
:$(2) \quad$ For any [[Definition:Mapping|mapping]] $g: A' \to H$:
$g$ is [[Definition:Continuous Mapping (Topology)|$\tuple {\tau', \tau' '}... | Continuity of Composite with Inclusion/Uniqueness of Induced Topology | https://proofwiki.org/wiki/Continuity_of_Composite_with_Inclusion/Uniqueness_of_Induced_Topology | https://proofwiki.org/wiki/Continuity_of_Composite_with_Inclusion/Uniqueness_of_Induced_Topology | [
"Continuity of Composite with Inclusion"
] | [
"Definition:Topological Subspace",
"Definition:Topology",
"Continuity of Composite with Inclusion/Inclusion on Mapping"
] | [
"Definition:Topology",
"Definition:Topological Space",
"Definition:Mapping",
"Definition:Continuous Mapping (Topology)",
"Definition:Continuous Mapping (Topology)",
"Definition:Identity Mapping",
"Identity Mapping is Continuous",
"Definition:Continuous Mapping (Topology)",
"Composite of Continuous M... |
proofwiki-6384 | Products of Open Sets form Local Basis in Product Space | Let $T_1 = \struct {A_1, \tau_1}$ and $T_2 = \struct {A_2, \tau_2}$ be topological spaces.
Let $\struct {T, \tau} = T_1 \times T_2$ be the product space of $T_1$ and $T_2$.
Let $\tuple {x, y} \in A_1 \times A_2$.
Let $W \in \tau$ be an open set of $T$ such that $\tuple {x, y} \in W$.
Then:
:$\exists U_1 \in \tau_1, U_2... | Let $W \in \tau$ such that $\tuple {x, y} \in W$.
From Natural Basis of Product Topology of Finite Product, $\tau$ is the topology with basis:
:$\BB = \set{U_1 \times U_2: U_1 \in \tau_1, U_2 \in \tau_2}$
Thus $W$ is the union of sets of the form $U_1 \times U_2$ where $U_1 \in \tau_1$ and $U_2 \in \tau_2$.
That is:
:$... | Let $T_1 = \struct {A_1, \tau_1}$ and $T_2 = \struct {A_2, \tau_2}$ be [[Definition:Topological Space|topological spaces]].
Let $\struct {T, \tau} = T_1 \times T_2$ be the [[Definition:Product Topology|product space]] of $T_1$ and $T_2$.
Let $\tuple {x, y} \in A_1 \times A_2$.
Let $W \in \tau$ be an [[Definition:Ope... | Let $W \in \tau$ such that $\tuple {x, y} \in W$.
From [[Natural Basis of Product Topology of Finite Product]], $\tau$ is the [[Definition:Topology|topology]] with [[Definition:Basis (Topology)|basis]]:
:$\BB = \set{U_1 \times U_2: U_1 \in \tau_1, U_2 \in \tau_2}$
Thus $W$ is the [[Definition:Set Union|union]] of set... | Products of Open Sets form Local Basis in Product Space | https://proofwiki.org/wiki/Products_of_Open_Sets_form_Local_Basis_in_Product_Space | https://proofwiki.org/wiki/Products_of_Open_Sets_form_Local_Basis_in_Product_Space | [
"Product Topology"
] | [
"Definition:Topological Space",
"Definition:Product Topology",
"Definition:Open Set/Topology",
"Definition:Cartesian Product",
"Definition:Open Set/Topology",
"Definition:Local Basis"
] | [
"Natural Basis of Product Topology/Finite Product",
"Definition:Topology",
"Definition:Basis (Topology)",
"Definition:Set Union"
] |
proofwiki-6385 | Open Real Intervals are Homeomorphic | Let $\R$ be the real number line with the Euclidean topology.
Let $I_1 := \openint a b$ and $I_2 := \openint c d$ be non-empty open real intervals.
Then $I_1$ and $I_2$ are homeomorphic. | By definition of open real interval, for $I_1$ and $I_2$ to be non-empty it must be the case that $a < b$ and $c < d$.
In particular it is noted that $a \ne b$ and $c \ne d$.
Thus $a - b \ne 0$ and $c - d \ne 0$.
Consider the real function $f: I_1 \to I_2$ defined as:
:$\forall x \in I_1: \map f x = c + \dfrac {\paren ... | Let $\R$ be the [[Definition:Real Number Line|real number line]] with the [[Definition:Real Number Line with Euclidean Topology|Euclidean topology]].
Let $I_1 := \openint a b$ and $I_2 := \openint c d$ be [[Definition:Non-Empty Set|non-empty]] [[Definition:Open Real Interval|open real intervals]].
Then $I_1$ and $I_... | By definition of [[Definition:Open Real Interval|open real interval]], for $I_1$ and $I_2$ to be [[Definition:Non-Empty Set|non-empty]] it must be the case that $a < b$ and $c < d$.
In particular it is noted that $a \ne b$ and $c \ne d$.
Thus $a - b \ne 0$ and $c - d \ne 0$.
Consider the [[Definition:Real Function|r... | Open Real Intervals are Homeomorphic | https://proofwiki.org/wiki/Open_Real_Intervals_are_Homeomorphic | https://proofwiki.org/wiki/Open_Real_Intervals_are_Homeomorphic | [
"Examples of Homeomorphisms"
] | [
"Definition:Real Number/Real Number Line",
"Definition:Euclidean Space/Euclidean Topology/Real Number Line",
"Definition:Non-Empty Set",
"Definition:Real Interval/Open",
"Definition:Homeomorphism/Topological Spaces"
] | [
"Definition:Real Interval/Open",
"Definition:Non-Empty Set",
"Definition:Real Function",
"Combination Theorem for Continuous Functions/Real",
"Definition:Continuous Real Function",
"Definition:Real Interval/Open",
"Definition:Homeomorphism/Topological Spaces"
] |
proofwiki-6386 | Half-Open Real Interval is not Closed in Real Number Line | Let $\R$ be the real number line considered as an Euclidean space.
Let $\hointr a b \subset \R$ be a right half-open interval of $\R$.
Then $\hointr a b$ is not a closed set of $\R$.
Similarly, the left half-open interval $\hointl a b \subset \R$ is not a closed set of $\R$. | Consider:
:$A := \R \setminus \hointr a b = \openint \gets a \cup \hointr b \to$
Let $A:= \openint \gets a$ and let $B := \hointr b \to$.
Let $\map {B_\epsilon} b$ be the open $\epsilon$-ball of $b$.
We have that $b - \epsilon < b$ and so $\map {B_\epsilon} b = \openint {b - \epsilon} {b + \epsilon}$ does not lie entir... | Let $\R$ be the [[Definition:Real Number Line|real number line]] considered as an [[Definition:Euclidean Space|Euclidean space]].
Let $\hointr a b \subset \R$ be a [[Definition:Right Half-Open Real Interval|right half-open interval]] of $\R$.
Then $\hointr a b$ is not a [[Definition:Closed Set (Metric Space)|closed ... | Consider:
:$A := \R \setminus \hointr a b = \openint \gets a \cup \hointr b \to$
Let $A:= \openint \gets a$ and let $B := \hointr b \to$.
Let $\map {B_\epsilon} b$ be the [[Definition:Open Ball|open $\epsilon$-ball]] of $b$.
We have that $b - \epsilon < b$ and so $\map {B_\epsilon} b = \openint {b - \epsilon} {b + \... | Half-Open Real Interval is not Closed in Real Number Line | https://proofwiki.org/wiki/Half-Open_Real_Interval_is_not_Closed_in_Real_Number_Line | https://proofwiki.org/wiki/Half-Open_Real_Interval_is_not_Closed_in_Real_Number_Line | [
"Real Intervals",
"Closed Sets"
] | [
"Definition:Real Number/Real Number Line",
"Definition:Euclidean Space",
"Definition:Real Interval/Half-Open/Right",
"Definition:Closed Set/Metric Space",
"Definition:Real Interval/Half-Open/Left",
"Definition:Closed Set/Metric Space"
] | [
"Definition:Open Ball",
"Definition:Open Set/Metric Space",
"Definition:Closed Set/Metric Space",
"Definition:Open Set/Metric Space",
"Definition:Closed Set/Metric Space"
] |
proofwiki-6387 | Half-Open Real Interval is Closed in some Open Intervals | Let $\R$ be the real number line considered as an Euclidean space.
Let $\hointr a b \subset \R$ be a half-open interval of $\R$.
Let $c < a$.
Then $\hointr a b$ is a closed set of $\openint c b$.
Similarly, let $d > b$.
Then the half-open interval $\hointl a b$ is a closed set of $\openint a d$. | Consider:
:$A := \openint c b \setminus \hointr a b = \openint c a$
Then $A$ is an open interval.
By Open Real Interval is Open Set, $A$ is open in $\R$.
By the definition of the subspace topology it follows that $A$ is open in $\openint c b$.
Thus by the definition of closed set, $\openint c b \setminus A = \hointr a ... | Let $\R$ be the [[Definition:Real Number Line|real number line]] considered as an [[Definition:Euclidean Space|Euclidean space]].
Let $\hointr a b \subset \R$ be a [[Definition:Half-Open Real Interval|half-open interval]] of $\R$.
Let $c < a$.
Then $\hointr a b$ is a [[Definition:Closed Set (Metric Space)|closed se... | Consider:
:$A := \openint c b \setminus \hointr a b = \openint c a$
Then $A$ is an [[Definition:Open Real Interval|open interval]].
By [[Open Real Interval is Open Set]], $A$ is [[Definition:Open Set (Metric Space)|open]] in $\R$.
By the definition of the [[Definition:Topological Subspace|subspace topology]] it foll... | Half-Open Real Interval is Closed in some Open Intervals | https://proofwiki.org/wiki/Half-Open_Real_Interval_is_Closed_in_some_Open_Intervals | https://proofwiki.org/wiki/Half-Open_Real_Interval_is_Closed_in_some_Open_Intervals | [
"Real Intervals",
"Closed Sets"
] | [
"Definition:Real Number/Real Number Line",
"Definition:Euclidean Space",
"Definition:Real Interval/Half-Open",
"Definition:Closed Set/Metric Space",
"Definition:Real Interval/Half-Open",
"Definition:Closed Set/Metric Space"
] | [
"Definition:Real Interval/Open",
"Open Real Interval is Open Set",
"Definition:Open Set/Metric Space",
"Definition:Topological Subspace",
"Definition:Open Set/Metric Space",
"Definition:Closed Set/Metric Space",
"Definition:Closed Set/Metric Space",
"Definition:Closed Set/Metric Space"
] |
proofwiki-6388 | Underlying Set of Topological Space is Clopen | Let $T = \struct {S, \tau}$ be a topological space.
Then the underlying set $S$ of $T$ is both open and closed in $T$. | From the definition of topology, $S$ is open in $T$.
From Underlying Set of Topological Space is Closed $S$ is closed in $T$.
Hence the result.
{{qed}} | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Then the [[Definition:Underlying Set of Topological Space|underlying set]] $S$ of $T$ is [[Definition:Clopen Set|both open and closed]] in $T$. | From the definition of [[Definition:Topology|topology]], $S$ is [[Definition:Open Set (Topology)|open]] in $T$.
From [[Underlying Set of Topological Space is Closed]] $S$ is [[Definition:Closed Set (Topology)|closed]] in $T$.
Hence the result.
{{qed}} | Underlying Set of Topological Space is Clopen | https://proofwiki.org/wiki/Underlying_Set_of_Topological_Space_is_Clopen | https://proofwiki.org/wiki/Underlying_Set_of_Topological_Space_is_Clopen | [
"Clopen Sets"
] | [
"Definition:Topological Space",
"Definition:Underlying Set/Topological Space",
"Definition:Clopen Set"
] | [
"Definition:Topology",
"Definition:Open Set/Topology",
"Underlying Set of Topological Space is Closed",
"Definition:Closed Set/Topology"
] |
proofwiki-6389 | Empty Set is Closed/Topological Space | Let $T = \struct {S, \tau}$ be a topological space.
Then $\O$ is closed in $T$. | From the definition of closed set, $U$ is open in $T = \struct {S, \tau}$ {{iff}} $S \setminus U$ is closed in $T$.
By definition of topological space, $S$ is open in $T$.
From Set Difference with Self is Empty Set:
:$S \setminus S = \O$
Hence $\O$ is closed in $T$.
{{qed}} | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Then $\O$ is [[Definition:Closed Set (Topology)|closed]] in $T$. | From the definition of [[Definition:Closed Set (Topology)|closed set]], $U$ is [[Definition:Open Set (Topology)|open]] in $T = \struct {S, \tau}$ {{iff}} $S \setminus U$ is [[Definition:Closed Set (Topology)|closed]] in $T$.
By definition of [[Definition:Topological Space|topological space]], $S$ is [[Definition:Open ... | Empty Set is Closed/Topological Space | https://proofwiki.org/wiki/Empty_Set_is_Closed/Topological_Space | https://proofwiki.org/wiki/Empty_Set_is_Closed/Topological_Space | [
"Closed Sets",
"Empty Set",
"Empty Set is Closed",
"Empty Set is Closed"
] | [
"Definition:Topological Space",
"Definition:Closed Set/Topology"
] | [
"Definition:Closed Set/Topology",
"Definition:Open Set/Topology",
"Definition:Closed Set/Topology",
"Definition:Topological Space",
"Definition:Open Set/Topology",
"Set Difference with Self is Empty Set",
"Definition:Closed Set/Topology"
] |
proofwiki-6390 | Basis induces Local Basis | Let $\struct {S, \tau}$ be a topological space.
Let $\BB$ be an analytic basis for $\tau$.
Let $x \in S$, and denote with $\BB_x$ the set $\set {B \in \BB: x \in B}$.
Then $\BB_x$ is a local basis at $x$. | Let $x \in U$, with $U$ an open set of $\tau$.
Since $\BB$ is an analytic basis for $\tau$, we have:
:$U = \ds\bigcup \AA$
for some $\AA \subseteq \BB$.
Since $x \in \ds \bigcup \AA$, there is a $B \in \AA$ such that $x \in B$, by definition of set union.
Hence, by definition of $\BB_x$, $B \in \BB_x$.
From Set is Subs... | Let $\struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $\BB$ be an [[Definition:Analytic Basis|analytic basis]] for $\tau$.
Let $x \in S$, and denote with $\BB_x$ the set $\set {B \in \BB: x \in B}$.
Then $\BB_x$ is a [[Definition:Local Basis|local basis]] at $x$. | Let $x \in U$, with $U$ an [[Definition:Open Set (Topology)|open set]] of $\tau$.
Since $\BB$ is an [[Definition:Analytic Basis|analytic basis]] for $\tau$, we have:
:$U = \ds\bigcup \AA$
for some $\AA \subseteq \BB$.
Since $x \in \ds \bigcup \AA$, there is a $B \in \AA$ such that $x \in B$, by definition of [[Defi... | Basis induces Local Basis | https://proofwiki.org/wiki/Basis_induces_Local_Basis | https://proofwiki.org/wiki/Basis_induces_Local_Basis | [
"Topology"
] | [
"Definition:Topological Space",
"Definition:Basis (Topology)/Analytic Basis",
"Definition:Local Basis"
] | [
"Definition:Open Set/Topology",
"Definition:Basis (Topology)/Analytic Basis",
"Definition:Set Union/Set of Sets",
"Set is Subset of Union/General Result",
"Definition:Local Basis"
] |
proofwiki-6391 | Fort Space is T0 | Let $T = \struct {S, \tau_p}$ be a Fort space on an infinite set $S$.
Then $T$ is a $T_0$ space. | Follows directly from:
:Fort Space is $T_1$
:$T_1$ Space is $T_0$ Space
{{qed}}
Category:Fort Spaces
Category:Examples of T0 Spaces
51k0w3v3pxvtbmwsndw2uf55mo5zzuo | Let $T = \struct {S, \tau_p}$ be a [[Definition:Fort Space|Fort space]] on an [[Definition:Infinite|infinite set]] $S$.
Then $T$ is a [[Definition:T0 Space|$T_0$ space]]. | Follows directly from:
:[[Fort Space is T1|Fort Space is $T_1$]]
:[[T1 Space is T0 Space|$T_1$ Space is $T_0$ Space]]
{{qed}}
[[Category:Fort Spaces]]
[[Category:Examples of T0 Spaces]]
51k0w3v3pxvtbmwsndw2uf55mo5zzuo | Fort Space is T0 | https://proofwiki.org/wiki/Fort_Space_is_T0 | https://proofwiki.org/wiki/Fort_Space_is_T0 | [
"Fort Spaces",
"Examples of T0 Spaces"
] | [
"Definition:Fort Space",
"Definition:Infinite",
"Definition:T0 Space"
] | [
"Fort Space is T1",
"T1 Space is T0",
"Category:Fort Spaces",
"Category:Examples of T0 Spaces"
] |
proofwiki-6392 | Pullback Functor is Functor | Let $\mathbf C$ be a metacategory having all pullbacks.
Let $f: C \to D$ be a morphism of $\mathbf C$.
Let $\mathbf C \mathbin / C$ and $\mathbf C \mathbin / D$ be the slice categories over $C$ and $D$, respectively.
Let $f^* : \mathbf C \mathbin / D \to \mathbf C \mathbin / C$ be the pullback functor defined by $f$.
T... | Let $\alpha: A \to D$ be an object of $\mathbf C \mathbin / D$.
Then the identity morphism $\operatorname{id}_\alpha: \alpha \to \alpha$ is by definition $\operatorname{id}_A: A \to A$.
Thus $f^* \operatorname{id}_\alpha$ is by definition the unique morphism fitting:
::<nowiki>$\begin{xy}\xymatrix@+1em{
A'
\ar[rr]^*... | Let $\mathbf C$ be a [[Definition:Metacategory|metacategory]] having all [[Definition:Pullback (Category Theory)|pullbacks]].
Let $f: C \to D$ be a [[Definition:Morphism (Category Theory)|morphism]] of $\mathbf C$.
Let $\mathbf C \mathbin / C$ and $\mathbf C \mathbin / D$ be the [[Definition:Slice Category|slice cate... | Let $\alpha: A \to D$ be an [[Definition:Object (Category Theory)|object]] of $\mathbf C \mathbin / D$.
Then the [[Definition:Identity Morphism|identity morphism]] $\operatorname{id}_\alpha: \alpha \to \alpha$ is by definition $\operatorname{id}_A: A \to A$.
Thus $f^* \operatorname{id}_\alpha$ is by definition the u... | Pullback Functor is Functor | https://proofwiki.org/wiki/Pullback_Functor_is_Functor | https://proofwiki.org/wiki/Pullback_Functor_is_Functor | [
"Functors",
"Pullbacks"
] | [
"Definition:Metacategory",
"Definition:Pullback (Category Theory)",
"Definition:Morphism",
"Definition:Slice Category",
"Definition:Pullback Functor",
"Definition:Functor/Covariant"
] | [
"Definition:Object (Category Theory)",
"Definition:Identity Morphism",
"Definition:Morphism",
"Definition:Morphism",
"Definition:Functor/Covariant"
] |
proofwiki-6393 | Supremum of Bounded Above Set of Reals is in Closure | Let $\R$ be the real number line under the Euclidean metric.
Let $H \subseteq \R$ be a bounded above subset of $\R$ such that $H \ne \O$.
Let $u = \sup H$ be the supremum of $H$.
Then:
:$u \in \map \cl H$
where $\map \cl H$ denotes the closure of $H$ in $\R$. | Let $\epsilon \in \R_{>0}$ be a strictly positive real number.
Let $\map {B_\epsilon} u$ be the open $\epsilon$-ball of $u$ in $\R$.
From Distance from Subset of Real Numbers:
:$\map d {u, H} = 0$
Thus by definition of distance from subset:
:$\exists x \in H: \map d {u, x} < \epsilon$
Thus $x \in \map {B_\epsilon} u$.
... | Let $\R$ be the [[Definition:Real Number Line|real number line]] under the [[Definition:Euclidean Metric on Real Number Line|Euclidean metric]].
Let $H \subseteq \R$ be a [[Definition:Bounded Above Set|bounded above subset]] of $\R$ such that $H \ne \O$.
Let $u = \sup H$ be the [[Definition:Supremum of Set|supremum]]... | Let $\epsilon \in \R_{>0}$ be a [[Definition:Strictly Positive Real Number|strictly positive real number]].
Let $\map {B_\epsilon} u$ be the [[Definition:Open Ball|open $\epsilon$-ball]] of $u$ in $\R$.
From [[Distance from Subset of Real Numbers]]:
:$\map d {u, H} = 0$
Thus by definition of [[Definition:Distance fr... | Supremum of Bounded Above Set of Reals is in Closure | https://proofwiki.org/wiki/Supremum_of_Bounded_Above_Set_of_Reals_is_in_Closure | https://proofwiki.org/wiki/Supremum_of_Bounded_Above_Set_of_Reals_is_in_Closure | [
"Set Closures",
"Real Analysis"
] | [
"Definition:Real Number/Real Number Line",
"Definition:Euclidean Metric/Real Number Line",
"Definition:Bounded Above Set",
"Definition:Supremum of Set",
"Definition:Closure (Topology)"
] | [
"Definition:Strictly Positive/Real Number",
"Definition:Open Ball",
"Distance from Subset of Real Numbers",
"Definition:Distance/Sets",
"Definition:Set Intersection",
"Condition for Point being in Closure"
] |
proofwiki-6394 | Topological Space is Open Neighborhood of Subset | Let $T = \left({S, \tau}\right)$ be a topological space.
Let $H \subseteq S$ be a subset of $S$.
Then $S$ is an open neighborhood of $H$. | From Underlying Set of Topological Space is Clopen, $S$ is open in $T$.
By hypothesis, $H \subseteq S$.
The result follows from Open Superset is Open Neighborhood.
{{qed}}
Category:Topology
Category:Neighborhoods
23fxtpbd2ihzx2sxohd59o0o3q2xfuf | Let $T = \left({S, \tau}\right)$ be a [[Definition:Topological Space|topological space]].
Let $H \subseteq S$ be a [[Definition:Subset|subset]] of $S$.
Then $S$ is an [[Definition:Open Neighborhood|open neighborhood]] of $H$. | From [[Underlying Set of Topological Space is Clopen]], $S$ is [[Definition:Open Set (Topology)|open]] in $T$.
[[Definition:By Hypothesis|By hypothesis]], $H \subseteq S$.
The result follows from [[Open Superset is Open Neighborhood]].
{{qed}}
[[Category:Topology]]
[[Category:Neighborhoods]]
23fxtpbd2ihzx2sxohd59o0o... | Topological Space is Open Neighborhood of Subset | https://proofwiki.org/wiki/Topological_Space_is_Open_Neighborhood_of_Subset | https://proofwiki.org/wiki/Topological_Space_is_Open_Neighborhood_of_Subset | [
"Topology",
"Neighborhoods"
] | [
"Definition:Topological Space",
"Definition:Subset",
"Definition:Open Neighborhood"
] | [
"Underlying Set of Topological Space is Clopen",
"Definition:Open Set/Topology",
"Definition:By Hypothesis",
"Open Superset is Open Neighborhood",
"Category:Topology",
"Category:Neighborhoods"
] |
proofwiki-6395 | Open Superset is Open Neighborhood | Let $T = \struct {S, \tau}$ be a topological space.
Let $U \in \tau$ be an open set of $T$.
Let $H \subseteq U$.
Then $U$ is an open neighborhood of $H$. | By hypothesis, $U$ is open in $T$.
Also {{hypothesis}}, $H \subseteq U$.
The result follows by definition of open neighborhood.
{{qed}}
Category:Topology
3ijzgu0t8urimrs49ae4gjsfrubijzl | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $U \in \tau$ be an [[Definition:Open Set (Topology)|open set]] of $T$.
Let $H \subseteq U$.
Then $U$ is an [[Definition:Open Neighborhood|open neighborhood]] of $H$. | [[Definition:By Hypothesis|By hypothesis]], $U$ is [[Definition:Open Set (Topology)|open]] in $T$.
Also {{hypothesis}}, $H \subseteq U$.
The result follows by definition of [[Definition:Open Neighborhood|open neighborhood]].
{{qed}}
[[Category:Topology]]
3ijzgu0t8urimrs49ae4gjsfrubijzl | Open Superset is Open Neighborhood | https://proofwiki.org/wiki/Open_Superset_is_Open_Neighborhood | https://proofwiki.org/wiki/Open_Superset_is_Open_Neighborhood | [
"Topology"
] | [
"Definition:Topological Space",
"Definition:Open Set/Topology",
"Definition:Open Neighborhood"
] | [
"Definition:By Hypothesis",
"Definition:Open Set/Topology",
"Definition:Open Neighborhood",
"Category:Topology"
] |
proofwiki-6396 | Category has Products and Equalizers iff Pullbacks and Terminal Object | Let $\mathbf C$ be a metacategory.
{{TFAE}}
{{begin-itemize}}
{{item|(1):|$\mathbf C$ has all finite products and equalizers}}
{{item|(2):|$\mathbf C$ has all pullbacks and a terminal object}}
{{end-itemize}} | === $(1)$ implies $(2)$ ===
Suppose $\mathbf C$ has all finite products and equalizers.
That $\mathbf C$ has pullbacks follows from Pullback as Equalizer.
That $\mathbf C$ has a terminal object follows by Empty Product is Terminal Object.
{{qed|lemma}} | Let $\mathbf C$ be a [[Definition:Metacategory|metacategory]].
{{TFAE}}
{{begin-itemize}}
{{item|(1):|$\mathbf C$ has all [[Definition:Finite Product (Category Theory)|finite products]] and [[Definition:Equalizer|equalizers]]}}
{{item|(2):|$\mathbf C$ has all [[Definition:Pullback (Category Theory)|pullbacks]] and a ... | === $(1)$ implies $(2)$ ===
Suppose $\mathbf C$ has all [[Definition:Product (Category Theory)|finite products]] and [[Definition:Equalizer|equalizers]].
That $\mathbf C$ has [[Definition:Pullback (Category Theory)|pullbacks]] follows from [[Pullback as Equalizer]].
That $\mathbf C$ has a [[Definition:Terminal Objec... | Category has Products and Equalizers iff Pullbacks and Terminal Object | https://proofwiki.org/wiki/Category_has_Products_and_Equalizers_iff_Pullbacks_and_Terminal_Object | https://proofwiki.org/wiki/Category_has_Products_and_Equalizers_iff_Pullbacks_and_Terminal_Object | [
"Pullbacks",
"Products (Category Theory)"
] | [
"Definition:Metacategory",
"Definition:Product (Category Theory)/General Definition/Finite Product",
"Definition:Equalizer",
"Definition:Pullback (Category Theory)",
"Definition:Terminal Object"
] | [
"Definition:Product (Category Theory)",
"Definition:Equalizer",
"Definition:Pullback (Category Theory)",
"Pullback as Equalizer",
"Definition:Terminal Object",
"Empty Product is Terminal Object",
"Definition:Pullback (Category Theory)",
"Definition:Terminal Object",
"Definition:Equalizer"
] |
proofwiki-6397 | Infimum of Bounded Below Set of Reals is in Closure | Let $\R$ be the real number line with the usual (Euclidean} metric.
Let $H \subseteq \R$ be a bounded below subset of $\R$ such that $H \ne \O$.
Let $l = \map \inf H$ be the infimum of $H$.
Then:
:$l \in \map \cl H$
where $\map \cl H$ denotes the closure of $H$ in $\R$. | Let $\epsilon \in \R_{>0}$ be a strictly positive real number.
Let $\map {B_\epsilon} l$ be the open $\epsilon$-ball of $l$ in $\R$.
From Distance from Subset of Real Numbers:
:$\map d {l, H} = 0$
Thus by definition of distance from subset:
:$\exists x \in H: \map d {l, x} < \epsilon$
Thus $x \in \map {B_\epsilon} l$.
... | Let $\R$ be the [[Definition:Real Number Line with Euclidean Metric|real number line with the usual (Euclidean} metric]].
Let $H \subseteq \R$ be a [[Definition:Bounded Below Set|bounded below subset]] of $\R$ such that $H \ne \O$.
Let $l = \map \inf H$ be the [[Definition:Infimum of Set|infimum]] of $H$.
Then:
:... | Let $\epsilon \in \R_{>0}$ be a [[Definition:Strictly Positive Real Number|strictly positive real number]].
Let $\map {B_\epsilon} l$ be the [[Definition:Open Ball|open $\epsilon$-ball]] of $l$ in $\R$.
From [[Distance from Subset of Real Numbers]]:
:$\map d {l, H} = 0$
Thus by definition of [[Definition:Distance f... | Infimum of Bounded Below Set of Reals is in Closure | https://proofwiki.org/wiki/Infimum_of_Bounded_Below_Set_of_Reals_is_in_Closure | https://proofwiki.org/wiki/Infimum_of_Bounded_Below_Set_of_Reals_is_in_Closure | [
"Set Closures",
"Real Number Line with Euclidean Metric"
] | [
"Definition:Euclidean Metric/Real Number Line",
"Definition:Bounded Below Set",
"Definition:Infimum of Set",
"Definition:Closure (Topology)"
] | [
"Definition:Strictly Positive/Real Number",
"Definition:Open Ball",
"Distance from Subset of Real Numbers",
"Definition:Distance/Sets",
"Definition:Set Intersection",
"Condition for Point being in Closure"
] |
proofwiki-6398 | Closure of Infinite Union may not equal Union of Closures | Let $T$ be a topological space.
Let $I$ be an infinite indexing set.
Let $\family {H_i}_{i \mathop \in I}$ be an indexed family of subsets of a set $S$.
Let $\ds H = \bigcup_{i \mathop \in I} H_i$ be the union of $\family {H_i}_{i \mathop \in I}$.
Then it is not always the case that:
:$\ds \bigcup_{i \mathop \in I} \ma... | Proof by Counterexample:
Consider the real number line $\struct {\R, \tau_d}$ with the usual (Euclidean) topology $\tau_d$.
Let:
:$H_n \subseteq \R: H_n = \closedint {\dfrac 1 n} 1$ for $n \ge 2$
where $\closedint {\dfrac 1 n} 1$ denotes the closed real interval from $\dfrac 1 n$ to $1$.
From Closed Real Interval is Cl... | Let $T$ be a [[Definition:Topological Space|topological space]].
Let $I$ be an [[Definition:Infinite Set|infinite]] [[Definition:Indexing Set|indexing set]].
Let $\family {H_i}_{i \mathop \in I}$ be an [[Definition:Indexed Family of Subsets|indexed family of subsets]] of a set $S$.
Let $\ds H = \bigcup_{i \mathop \... | [[Proof by Counterexample]]:
Consider the [[Definition:Real Number Line with Euclidean Topology|real number line $\struct {\R, \tau_d}$ with the usual (Euclidean) topology]] $\tau_d$.
Let:
:$H_n \subseteq \R: H_n = \closedint {\dfrac 1 n} 1$ for $n \ge 2$
where $\closedint {\dfrac 1 n} 1$ denotes the [[Definition:Clo... | Closure of Infinite Union may not equal Union of Closures/Proof 1 | https://proofwiki.org/wiki/Closure_of_Infinite_Union_may_not_equal_Union_of_Closures | https://proofwiki.org/wiki/Closure_of_Infinite_Union_may_not_equal_Union_of_Closures/Proof_1 | [
"Set Union",
"Set Closures",
"Closure of Infinite Union may not equal Union of Closures"
] | [
"Definition:Topological Space",
"Definition:Infinite Set",
"Definition:Indexing Set",
"Definition:Indexing Set/Family of Subsets",
"Definition:Set Union/Family of Sets",
"Definition:Closure (Topology)"
] | [
"Proof by Counterexample",
"Definition:Euclidean Space/Euclidean Topology/Real Number Line",
"Definition:Real Interval/Closed",
"Closed Real Interval is Closed Set",
"Definition:Closed Set/Topology",
"Set is Closed iff Equals Topological Closure"
] |
proofwiki-6399 | Closure of Infinite Union may not equal Union of Closures | Let $T$ be a topological space.
Let $I$ be an infinite indexing set.
Let $\family {H_i}_{i \mathop \in I}$ be an indexed family of subsets of a set $S$.
Let $\ds H = \bigcup_{i \mathop \in I} H_i$ be the union of $\family {H_i}_{i \mathop \in I}$.
Then it is not always the case that:
:$\ds \bigcup_{i \mathop \in I} \ma... | Proof by Counterexample:
Let $\struct {\R, \tau_d}$ denote the real number line with the usual (Euclidean) topology $\tau_d$.
Let $\struct {\Q, \tau_d}$ be the rational number space, also under the usual (Euclidean) topology $\tau_d$.
For a rational number $\alpha \in \Q$, let $B_\alpha$ denote the singleton containing... | Let $T$ be a [[Definition:Topological Space|topological space]].
Let $I$ be an [[Definition:Infinite Set|infinite]] [[Definition:Indexing Set|indexing set]].
Let $\family {H_i}_{i \mathop \in I}$ be an [[Definition:Indexed Family of Subsets|indexed family of subsets]] of a set $S$.
Let $\ds H = \bigcup_{i \mathop \... | [[Proof by Counterexample]]:
Let $\struct {\R, \tau_d}$ denote the [[Definition:Real Number Line with Euclidean Topology|real number line with the usual (Euclidean) topology]] $\tau_d$.
Let $\struct {\Q, \tau_d}$ be the [[Definition:Rational Number Space|rational number space]], also under the [[Definition:Euclidean ... | Closure of Infinite Union may not equal Union of Closures/Proof 2 | https://proofwiki.org/wiki/Closure_of_Infinite_Union_may_not_equal_Union_of_Closures | https://proofwiki.org/wiki/Closure_of_Infinite_Union_may_not_equal_Union_of_Closures/Proof_2 | [
"Set Union",
"Set Closures",
"Closure of Infinite Union may not equal Union of Closures"
] | [
"Definition:Topological Space",
"Definition:Infinite Set",
"Definition:Indexing Set",
"Definition:Indexing Set/Family of Subsets",
"Definition:Set Union/Family of Sets",
"Definition:Closure (Topology)"
] | [
"Proof by Counterexample",
"Definition:Euclidean Space/Euclidean Topology/Real Number Line",
"Definition:Rational Number Space",
"Definition:Euclidean Space/Euclidean Topology/Real Number Line",
"Definition:Rational Number",
"Definition:Singleton",
"Union of Closures of Singleton Rationals is Rational S... |
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