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-21000 | Reciprocal of Complex Number in terms of Conjugate and Modulus | Let $z \in \C$ be a complex number.
The reciprocal of $z$ can be expressed as:
:$\dfrac 1 z = \dfrac {\overline z} {\cmod z^2}$
where:
:$\overline z$ denotes the complex conjugate of $z$
:$\cmod z^2$ denotes the modulus of $z$. | Let $z$ be defined as:
:$z = a = i b$
Then:
{{begin-eqn}}
{{eqn | l = \dfrac 1 z
| r = \dfrac {a - i b} {a^2 + b^2}
| c = Reciprocal of Complex Number
}}
{{eqn | r = \dfrac {a - i b} {\paren {\sqrt {a^2 + b^2} }^2}
| c = rearranging
}}
{{eqn | r = \dfrac {\overline z} {\paren {\sqrt {a^2 + b^2} }^2}
... | Let $z \in \C$ be a [[Definition:Complex Number|complex number]].
The [[Definition:Reciprocal|reciprocal]] of $z$ can be expressed as:
:$\dfrac 1 z = \dfrac {\overline z} {\cmod z^2}$
where:
:$\overline z$ denotes the [[Definition:Complex Conjugate|complex conjugate]] of $z$
:$\cmod z^2$ denotes the [[Definition:Compl... | Let $z$ be defined as:
:$z = a = i b$
Then:
{{begin-eqn}}
{{eqn | l = \dfrac 1 z
| r = \dfrac {a - i b} {a^2 + b^2}
| c = [[Reciprocal of Complex Number]]
}}
{{eqn | r = \dfrac {a - i b} {\paren {\sqrt {a^2 + b^2} }^2}
| c = rearranging
}}
{{eqn | r = \dfrac {\overline z} {\paren {\sqrt {a^2 + b^2}... | Reciprocal of Complex Number in terms of Conjugate and Modulus | https://proofwiki.org/wiki/Reciprocal_of_Complex_Number_in_terms_of_Conjugate_and_Modulus | https://proofwiki.org/wiki/Reciprocal_of_Complex_Number_in_terms_of_Conjugate_and_Modulus | [
"Complex Modulus",
"Complex Conjugates",
"Reciprocals"
] | [
"Definition:Complex Number",
"Definition:Reciprocal",
"Definition:Complex Conjugate",
"Definition:Complex Modulus"
] | [
"Reciprocal of Complex Number"
] |
proofwiki-21001 | Reciprocal of Complex Number | Let $z = a + i b$ be a complex number.
The reciprocal of $z$ is:
:$\dfrac 1 z = \dfrac {a - i b} {a^2 + b^2}$ | {{begin-eqn}}
{{eqn | l = \dfrac 1 z
| r = \dfrac 1 {a + i b}
| c = Definition of $z$
}}
{{eqn | r = \dfrac {a - i b} {\paren {a + i b} \paren {a - i b} }
| c = multiplying top and bottom by $a - i b$
}}
{{eqn | r = \dfrac {a - i b} {a^2 + b^2}
| c = Product of Complex Number with Conjugate
}}
{... | Let $z = a + i b$ be a [[Definition:Complex Number|complex number]].
The [[Definition:Reciprocal|reciprocal]] of $z$ is:
:$\dfrac 1 z = \dfrac {a - i b} {a^2 + b^2}$ | {{begin-eqn}}
{{eqn | l = \dfrac 1 z
| r = \dfrac 1 {a + i b}
| c = Definition of $z$
}}
{{eqn | r = \dfrac {a - i b} {\paren {a + i b} \paren {a - i b} }
| c = multiplying [[Definition:Numerator|top]] and [[Definition:Denominator|bottom]] by $a - i b$
}}
{{eqn | r = \dfrac {a - i b} {a^2 + b^2}
... | Reciprocal of Complex Number | https://proofwiki.org/wiki/Reciprocal_of_Complex_Number | https://proofwiki.org/wiki/Reciprocal_of_Complex_Number | [
"Complex Numbers",
"Reciprocals"
] | [
"Definition:Complex Number",
"Definition:Reciprocal"
] | [
"Definition:Fraction/Numerator",
"Definition:Fraction/Denominator",
"Product of Complex Number with Conjugate"
] |
proofwiki-21002 | Reciprocal of Power of Complex Number | Let $z \in \C$ be a complex number.
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $z^n$ denote $z$ raised to the $n$th power.
The reciprocal of $z^n$ can be expressed as:
:$\dfrac 1 {z^n} = \dfrac {\overline z^n} {\cmod z^{2 n} }$
where:
:$\overline z$ denotes the complex conjugate of $z$
:$\cmod z^2$ denot... | {{begin-eqn}}
{{eqn | l = \dfrac 1 {z^n}
| r = \paren {\dfrac 1 z}^n
| c =
}}
{{eqn | r = \paren {\dfrac {\overline z} {\cmod z^2} }^n
| c =
}}
{{eqn | r = \dfrac {\paren {\overline z}^n} {\paren {\cmod z^2}^n}
| c =
}}
{{eqn | r = \dfrac {\overline z^n} {\cmod z^{2 n} }
| c =
}}
{{end... | Let $z \in \C$ be a [[Definition:Complex Number|complex number]].
Let $n \in \Z_{>0}$ be a [[Definition:Strictly Positive Integer|(strictly) positive integer]].
Let $z^n$ denote $z$ raised to the [[Definition:Integer Power|$n$th power]].
The [[Definition:Reciprocal|reciprocal]] of $z^n$ can be expressed as:
:$\dfra... | {{begin-eqn}}
{{eqn | l = \dfrac 1 {z^n}
| r = \paren {\dfrac 1 z}^n
| c =
}}
{{eqn | r = \paren {\dfrac {\overline z} {\cmod z^2} }^n
| c =
}}
{{eqn | r = \dfrac {\paren {\overline z}^n} {\paren {\cmod z^2}^n}
| c =
}}
{{eqn | r = \dfrac {\overline z^n} {\cmod z^{2 n} }
| c =
}}
{{end... | Reciprocal of Power of Complex Number | https://proofwiki.org/wiki/Reciprocal_of_Power_of_Complex_Number | https://proofwiki.org/wiki/Reciprocal_of_Power_of_Complex_Number | [
"Reciprocals",
"Complex Powers"
] | [
"Definition:Complex Number",
"Definition:Strictly Positive/Integer",
"Definition:Power (Algebra)/Integer",
"Definition:Reciprocal",
"Definition:Complex Conjugate",
"Definition:Complex Modulus"
] | [] |
proofwiki-21003 | Bounded Linear Transformation is Into Linear Isomorphism iff Dual Operator is Surjective | Let $\GF \in \set {\R, \C}$.
Let $X$ and $Y$ be Banach spaces over $\GF$.
Let $X^\ast$ and $Y^\ast$ be the normed duals of $X$ and $Y$ respectively.
Let $T : X \to Y$ be a bounded linear transformation.
Let $T^\ast : Y^\ast \to X^\ast$ be the dual operator of $T$.
Then $T$ is an into linear isomorphism {{iff}} $T^\a... | Let $B_{X^\ast}^-$ and $B_{Y^\ast}^-$ be the closed unit balls of $X^\ast$ and $Y^\ast$ respectively. | Let $\GF \in \set {\R, \C}$.
Let $X$ and $Y$ be [[Definition:Banach Space|Banach spaces]] over $\GF$.
Let $X^\ast$ and $Y^\ast$ be the [[Definition:Normed Dual Space|normed duals]] of $X$ and $Y$ respectively.
Let $T : X \to Y$ be a [[Definition:Bounded Linear Transformation|bounded linear transformation]].
Let $... | Let $B_{X^\ast}^-$ and $B_{Y^\ast}^-$ be the [[Definition:Closed Unit Ball|closed unit balls]] of $X^\ast$ and $Y^\ast$ respectively. | Bounded Linear Transformation is Into Linear Isomorphism iff Dual Operator is Surjective | https://proofwiki.org/wiki/Bounded_Linear_Transformation_is_Into_Linear_Isomorphism_iff_Dual_Operator_is_Surjective | https://proofwiki.org/wiki/Bounded_Linear_Transformation_is_Into_Linear_Isomorphism_iff_Dual_Operator_is_Surjective | [
"Linear Isomorphisms",
"Dual Operators"
] | [
"Definition:Banach Space",
"Definition:Normed Dual Space",
"Definition:Bounded Linear Transformation",
"Definition:Dual Operator",
"Definition:Into Linear Isomorphism",
"Definition:Surjection"
] | [
"Definition:Closed Unit Ball"
] |
proofwiki-21004 | Kernel of Bounded Linear Transformation is Annihilator of Image of Dual Operator | Let $\GF \in \set {\R, \C}$.
Let $X$ and $Y$ be normed vector spaces over $\GF$.
Let $X^\ast$ and $Y^\ast$ be the normed dual spaces of $X$ and $Y$ respectively.
Let $T : X \to Y$ be a Bounded linear transformation.
Let $T^\ast : Y^\ast \to X^\ast$ be the dual operator of $T$.
Then we have:
:$\map \ker T = {}^\bot... | Let $x \in \map \ker T$.
Let $\phi \in T^\ast \sqbrk {Y^\ast}$.
Then from the definition of the dual operator, there exists $f \in Y^\ast$ such that $f \circ T = \phi$.
Then we have:
{{begin-eqn}}
{{eqn | l = \map \phi x
| r = \map f {T x}
}}
{{eqn | r = 0
| c = since $T x = {\mathbf 0}_X$
}}
{{end-eqn}}
Since th... | Let $\GF \in \set {\R, \C}$.
Let $X$ and $Y$ be [[Definition:Normed Vector Space|normed vector spaces]] over $\GF$.
Let $X^\ast$ and $Y^\ast$ be the [[Definition:Normed Dual Space|normed dual spaces]] of $X$ and $Y$ respectively.
Let $T : X \to Y$ be a [[Definition:Bounded Linear Transformation|Bounded linear tra... | Let $x \in \map \ker T$.
Let $\phi \in T^\ast \sqbrk {Y^\ast}$.
Then from the definition of the [[Definition:Dual Operator|dual operator]], there exists $f \in Y^\ast$ such that $f \circ T = \phi$.
Then we have:
{{begin-eqn}}
{{eqn | l = \map \phi x
| r = \map f {T x}
}}
{{eqn | r = 0
| c = since $T x = {\math... | Kernel of Bounded Linear Transformation is Annihilator of Image of Dual Operator | https://proofwiki.org/wiki/Kernel_of_Bounded_Linear_Transformation_is_Annihilator_of_Image_of_Dual_Operator | https://proofwiki.org/wiki/Kernel_of_Bounded_Linear_Transformation_is_Annihilator_of_Image_of_Dual_Operator | [
"Annihilators of Subspaces of Normed Dual Spaces",
"Dual Operators"
] | [
"Definition:Normed Vector Space",
"Definition:Normed Dual Space",
"Definition:Bounded Linear Transformation",
"Definition:Dual Operator",
"Definition:Annihilator of Subspace of Normed Dual Space",
"Definition:Subspace"
] | [
"Definition:Dual Operator",
"Definition:Dual Operator",
"Normed Dual Space Separates Points"
] |
proofwiki-21005 | Annihilator of Subspace of Banach Space is Weak-* Closed | Let $X$ be a Banach space.
Let $M$ be a vector subspace of $X$.
Let $X^\ast$ be the normed dual space of $X$.
Let $w^\ast$ be the weak-$\ast$ topology on $X^\ast$.
Let $M^\bot$ be the annihilator of $M$.
Then $M^\bot$ is closed in $\struct {X^\ast, w^\ast}$. | From Set is Closed iff Equals Topological Closure, we aim to show:
:$M^\bot = \map {\cl_{w^\ast} } {M^\bot}$
From Set is Subset of its Topological Closure, we have:
:$M^\bot \subseteq \map {\cl_{w^\ast} } {M^\bot}$
Now let:
:$f \in \map {\cl_{w^\ast} } {M^\bot}$
From Point in Set Closure iff Limit of Net, there exist... | Let $X$ be a [[Definition:Banach Space|Banach space]].
Let $M$ be a [[Definition:Vector Subspace|vector subspace]] of $X$.
Let $X^\ast$ be the [[Definition:Normed Dual Space|normed dual space]] of $X$.
Let $w^\ast$ be the [[Definition:Weak-* Topology|weak-$\ast$ topology]] on $X^\ast$.
Let $M^\bot$ be the [[Defin... | From [[Set is Closed iff Equals Topological Closure]], we aim to show:
:$M^\bot = \map {\cl_{w^\ast} } {M^\bot}$
From [[Set is Subset of its Topological Closure]], we have:
:$M^\bot \subseteq \map {\cl_{w^\ast} } {M^\bot}$
Now let:
:$f \in \map {\cl_{w^\ast} } {M^\bot}$
From [[Point in Set Closure iff Limit of Net... | Annihilator of Subspace of Banach Space is Weak-* Closed/Proof 1 | https://proofwiki.org/wiki/Annihilator_of_Subspace_of_Banach_Space_is_Weak-*_Closed | https://proofwiki.org/wiki/Annihilator_of_Subspace_of_Banach_Space_is_Weak-*_Closed/Proof_1 | [
"Annihilator of Subspace of Banach Space is Weak-* Closed",
"Annihilators of Subspaces of Banach Spaces",
"Weak-* Topologies"
] | [
"Definition:Banach Space",
"Definition:Vector Subspace",
"Definition:Normed Dual Space",
"Definition:Weak-* Topology",
"Definition:Annihilator of Subspace of Banach Space",
"Definition:Closed Set"
] | [
"Set is Closed iff Equals Topological Closure",
"Set is Subset of its Topological Closure",
"Point in Set Closure iff Limit of Net",
"Definition:Directed Preordering",
"Definition:Net (Set Theory)",
"Definition:Convergent Net",
"Characterization of Convergent Net in Weak-* Topology",
"Definition:Net (... |
proofwiki-21006 | Annihilator of Subspace of Banach Space is Weak-* Closed | Let $X$ be a Banach space.
Let $M$ be a vector subspace of $X$.
Let $X^\ast$ be the normed dual space of $X$.
Let $w^\ast$ be the weak-$\ast$ topology on $X^\ast$.
Let $M^\bot$ be the annihilator of $M$.
Then $M^\bot$ is closed in $\struct {X^\ast, w^\ast}$. | From Annihilator of Subspace of Banach Space as Intersection of Kernels, we have:
:$\ds M^\bot = \bigcap_{x \in M} \map \ker {x^\wedge}$
From Characterization of Continuity of Linear Functional in Weak-* Topology:
:the linear functional $x^\wedge : \struct {X^\ast, w^\ast} \to \GF$ is continuous.
From Characterization... | Let $X$ be a [[Definition:Banach Space|Banach space]].
Let $M$ be a [[Definition:Vector Subspace|vector subspace]] of $X$.
Let $X^\ast$ be the [[Definition:Normed Dual Space|normed dual space]] of $X$.
Let $w^\ast$ be the [[Definition:Weak-* Topology|weak-$\ast$ topology]] on $X^\ast$.
Let $M^\bot$ be the [[Defin... | From [[Annihilator of Subspace of Banach Space as Intersection of Kernels]], we have:
:$\ds M^\bot = \bigcap_{x \in M} \map \ker {x^\wedge}$
From [[Characterization of Continuity of Linear Functional in Weak-* Topology]]:
:the [[Definition:Linear Functional|linear functional]] $x^\wedge : \struct {X^\ast, w^\ast} \to... | Annihilator of Subspace of Banach Space is Weak-* Closed/Proof 2 | https://proofwiki.org/wiki/Annihilator_of_Subspace_of_Banach_Space_is_Weak-*_Closed | https://proofwiki.org/wiki/Annihilator_of_Subspace_of_Banach_Space_is_Weak-*_Closed/Proof_2 | [
"Annihilator of Subspace of Banach Space is Weak-* Closed",
"Annihilators of Subspaces of Banach Spaces",
"Weak-* Topologies"
] | [
"Definition:Banach Space",
"Definition:Vector Subspace",
"Definition:Normed Dual Space",
"Definition:Weak-* Topology",
"Definition:Annihilator of Subspace of Banach Space",
"Definition:Closed Set"
] | [
"Annihilator of Subspace of Banach Space as Intersection of Kernels",
"Characterization of Continuity of Linear Functional in Weak-* Topology",
"Definition:Linear Functional",
"Definition:Continuous Mapping",
"Characterization of Continuous Linear Functionals on Topological Vector Space",
"Definition:Clos... |
proofwiki-21007 | Constant Net is Convergent | Let $\struct {X, \tau}$ be a topological space.
Let $\struct {\Lambda, \preceq}$ be a directed set.
Let $x \in X$.
Define a net $\family {x_\lambda}_{\lambda \in \Lambda}$ by:
:$x_\lambda = x$ for each $\lambda \in \Lambda$.
Then $\family {x_\lambda}_{\lambda \in \Lambda}$ converges to $x$. | Let $U$ be an open neighborhood of $x$.
Let $\lambda_0 \in \Lambda$.
Then we have $x_\lambda \in U$ for all $\lambda \in \Lambda$, and in particular all $\lambda \in \Lambda$ with $\lambda_0 \preceq \lambda$.
So $\family {x_\lambda}_{\lambda \in \Lambda}$ converges to $x$.
{{qed}}
Category:Nets (Set Theory)
f6k99d3i4... | Let $\struct {X, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $\struct {\Lambda, \preceq}$ be a [[Definition:Directed Set|directed set]].
Let $x \in X$.
Define a [[Definition:Net (Set Theory)|net]] $\family {x_\lambda}_{\lambda \in \Lambda}$ by:
:$x_\lambda = x$ for each $\lambda \in \Lambda... | Let $U$ be an [[Definition:Open Neighborhood|open neighborhood]] of $x$.
Let $\lambda_0 \in \Lambda$.
Then we have $x_\lambda \in U$ for all $\lambda \in \Lambda$, and in particular all $\lambda \in \Lambda$ with $\lambda_0 \preceq \lambda$.
So $\family {x_\lambda}_{\lambda \in \Lambda}$ [[Definition:Convergent Ne... | Constant Net is Convergent | https://proofwiki.org/wiki/Constant_Net_is_Convergent | https://proofwiki.org/wiki/Constant_Net_is_Convergent | [
"Moore-Smith Sequences",
"Nets (Set Theory)",
"Nets (Set Theory)"
] | [
"Definition:Topological Space",
"Definition:Directed Preordering",
"Definition:Net (Set Theory)",
"Definition:Convergent Net"
] | [
"Definition:Open Neighborhood",
"Definition:Convergent Net",
"Category:Nets (Set Theory)"
] |
proofwiki-21008 | Annihilator of Subspace of Banach Space as Intersection of Kernels | Let $\GF \in \set {\R, \C}$.
Let $X$ be a Banach space over $\GF$.
Let $M$ be a vector subspace of $X$.
Let $X^\ast$ be the normed dual space of $X$.
Let $M^\bot$ be the annihilator of $M$.
Then:
:$\ds M^\bot = \bigcap_{x \mathop \in M} \map \ker {x^\wedge}$
where $x^\wedge$ denotes the evaluation linear transform... | We have:
{{begin-eqn}}
{{eqn | l = M^\bot
| r = \set {g \in X^\ast : \map g x = 0 \text { for all } x \in M}
| c = {{Defof|Annihilator of Subspace of Banach Space}}
}}
{{eqn | r = \bigcap_{x \mathop \in M} \set {g \in X^\ast : \map g x = 0}
}}
{{eqn | r = \bigcap_{x \mathop \in M} \set {g \in X^\ast : \map... | Let $\GF \in \set {\R, \C}$.
Let $X$ be a [[Definition:Banach Space|Banach space]] over $\GF$.
Let $M$ be a [[Definition:Vector Subspace|vector subspace]] of $X$.
Let $X^\ast$ be the [[Definition:Normed Dual Space|normed dual space]] of $X$.
Let $M^\bot$ be the [[Definition:Annihilator of Subspace of Banach Spac... | We have:
{{begin-eqn}}
{{eqn | l = M^\bot
| r = \set {g \in X^\ast : \map g x = 0 \text { for all } x \in M}
| c = {{Defof|Annihilator of Subspace of Banach Space}}
}}
{{eqn | r = \bigcap_{x \mathop \in M} \set {g \in X^\ast : \map g x = 0}
}}
{{eqn | r = \bigcap_{x \mathop \in M} \set {g \in X^\ast : \map... | Annihilator of Subspace of Banach Space as Intersection of Kernels | https://proofwiki.org/wiki/Annihilator_of_Subspace_of_Banach_Space_as_Intersection_of_Kernels | https://proofwiki.org/wiki/Annihilator_of_Subspace_of_Banach_Space_as_Intersection_of_Kernels | [
"Annihilators of Subspaces of Banach Spaces"
] | [
"Definition:Banach Space",
"Definition:Vector Subspace",
"Definition:Normed Dual Space",
"Definition:Annihilator of Subspace of Banach Space",
"Definition:Evaluation Linear Transformation/Normed Vector Space"
] | [
"Category:Annihilators of Subspaces of Banach Spaces"
] |
proofwiki-21009 | Closure in Weak-* Topology in terms of Annihilators | Let $X$ be a Banach space.
Let $X^\ast$ be the normed dual space of $X$.
Let $w^\ast$ be the weak-$\ast$ topology on $X^\ast$.
Let $N$ be a linear subspace of $X^\ast$.
Then:
:$\map {\cl_{w^\ast} } N = \paren { {}^\bot N}^\bot$
where:
:$\cl_{w^\ast}$ denotes closure in the weak-$\ast$ topology
:${}^\bot N$ denotes ... | From Linear Subspace is Subset of Double Annihilator, we have:
:$N \subseteq \paren { {}^\bot N}^\bot$
So, from Set Closure Preserves Set Inclusion:
:$\map {\cl_{w^\ast} } N \subseteq \map {\cl_{w^\ast} } {\paren { {}^\bot N}^\bot}$
From Annihilator of Subspace of Banach Space is Weak-* Closed and Set is Closed iff Equ... | Let $X$ be a [[Definition:Banach Space|Banach space]].
Let $X^\ast$ be the [[Definition:Normed Dual Space|normed dual space]] of $X$.
Let $w^\ast$ be the [[Definition:Weak-* Topology|weak-$\ast$ topology on $X^\ast$]].
Let $N$ be a [[Definition:Linear Subspace|linear subspace]] of $X^\ast$.
Then:
:$\map {\cl_{... | From [[Linear Subspace is Subset of Double Annihilator]], we have:
:$N \subseteq \paren { {}^\bot N}^\bot$
So, from [[Set Closure Preserves Set Inclusion]]:
:$\map {\cl_{w^\ast} } N \subseteq \map {\cl_{w^\ast} } {\paren { {}^\bot N}^\bot}$
From [[Annihilator of Subspace of Banach Space is Weak-* Closed]] and [[Set i... | Closure in Weak-* Topology in terms of Annihilators | https://proofwiki.org/wiki/Closure_in_Weak-*_Topology_in_terms_of_Annihilators | https://proofwiki.org/wiki/Closure_in_Weak-*_Topology_in_terms_of_Annihilators | [
"Annihilators of Subspaces of Banach Spaces",
"Annihilators of Subspaces of Normed Dual Spaces",
"Weak-* Topologies"
] | [
"Definition:Banach Space",
"Definition:Normed Dual Space",
"Definition:Weak-* Topology",
"Definition:Linear Subspace",
"Definition:Closure (Topology)",
"Definition:Weak-* Topology",
"Definition:Annihilator of Subspace of Normed Dual Space",
"Definition:Annihilator of Subspace of Banach Space"
] | [
"Linear Subspace is Subset of Double Annihilator",
"Set Closure Preserves Set Inclusion",
"Annihilator of Subspace of Banach Space is Weak-* Closed",
"Set is Closed iff Equals Topological Closure",
"Set Complement inverts Subsets",
"Existence of Non-Zero Continuous Linear Functional vanishing on Proper Cl... |
proofwiki-21010 | Linear Subspace is Subset of Double Annihilator | Let $X$ be a Banach space.
Let $X^\ast$ be the normed dual space of $X$.
Let $N$ be a linear subspace of $X^\ast$.
Then:
:$N \subseteq \paren { {}^\bot N}^\bot$
where:
:${}^\bot N$ denotes the annihilator of $N \subseteq X^\ast$
:$\paren { {}^\bot N}^\bot$ denotes the annihilator of ${}^\bot N \subseteq X$. | From Set Complement inverts Subsets, we can equivalently show:
:$X^\ast \setminus \paren { {}^\bot N}^\bot \subseteq X^\ast \setminus N$
Let $f \in X^\ast \setminus \paren { {}^\bot N}^\bot$, then:
:$\map f x \ne 0$ for some $x \in {}^\bot N$.
By the definition of the annihilator of a subspace of $X$, we have:
:$\map ... | Let $X$ be a [[Definition:Banach Space|Banach space]].
Let $X^\ast$ be the [[Definition:Normed Dual Space|normed dual space]] of $X$.
Let $N$ be a [[Definition:Linear Subspace|linear subspace]] of $X^\ast$.
Then:
:$N \subseteq \paren { {}^\bot N}^\bot$
where:
:${}^\bot N$ denotes the [[Definition:Annihilator of ... | From [[Set Complement inverts Subsets]], we can equivalently show:
:$X^\ast \setminus \paren { {}^\bot N}^\bot \subseteq X^\ast \setminus N$
Let $f \in X^\ast \setminus \paren { {}^\bot N}^\bot$, then:
:$\map f x \ne 0$ for some $x \in {}^\bot N$.
By the definition of the [[Definition:Annihilator of Subspace of Banac... | Linear Subspace is Subset of Double Annihilator | https://proofwiki.org/wiki/Linear_Subspace_is_Subset_of_Double_Annihilator | https://proofwiki.org/wiki/Linear_Subspace_is_Subset_of_Double_Annihilator | [
"Annihilators of Subspaces of Banach Spaces",
"Annihilators of Subspaces of Normed Dual Spaces"
] | [
"Definition:Banach Space",
"Definition:Normed Dual Space",
"Definition:Linear Subspace",
"Definition:Annihilator of Subspace of Normed Dual Space",
"Definition:Annihilator of Subspace of Banach Space"
] | [
"Set Complement inverts Subsets",
"Definition:Annihilator of Subspace of Banach Space",
"Set Complement inverts Subsets",
"Category:Annihilators of Subspaces of Banach Spaces",
"Category:Annihilators of Subspaces of Normed Dual Spaces"
] |
proofwiki-21011 | Annihilator of Image of Bounded Linear Transformation is Kernel of Dual Operator | Let $\GF \in \set {\R, \C}$.
Let $X$ and $Y$ be normed vector spaces over $\GF$.
Let $T : X \to Y$ be a bounded linear transformation.
Let $X^\ast$ and $Y^\ast$ be the normed duals of $X$ and $Y$ respectively.
Let $T^\ast : Y^\ast \to X^\ast$ be the dual operator of $T$.
Let $T \sqbrk X^\bot$ be the annihilator of $T... | We have:
:$f \in \map \ker {T^\ast}$
{{iff}}:
:$\map {\paren {T^\ast f} } x = \map f {T x} = 0$ for each $x \in X$.
That is, {{iff}}:
:$\map f y = 0$ for all $y \in T \sqbrk X$
To conclude, we have $f \in \map \ker {T^\ast}$ {{iff}}:
:$f \in T \sqbrk X^\bot$
{{qed}} | Let $\GF \in \set {\R, \C}$.
Let $X$ and $Y$ be [[Definition:Normed Vector Space|normed vector spaces]] over $\GF$.
Let $T : X \to Y$ be a [[Definition:Bounded Linear Transformation|bounded linear transformation]].
Let $X^\ast$ and $Y^\ast$ be the [[Definition:Normed Dual Space|normed duals]] of $X$ and $Y$ respect... | We have:
:$f \in \map \ker {T^\ast}$
{{iff}}:
:$\map {\paren {T^\ast f} } x = \map f {T x} = 0$ for each $x \in X$.
That is, {{iff}}:
:$\map f y = 0$ for all $y \in T \sqbrk X$
To conclude, we have $f \in \map \ker {T^\ast}$ {{iff}}:
:$f \in T \sqbrk X^\bot$
{{qed}} | Annihilator of Image of Bounded Linear Transformation is Kernel of Dual Operator | https://proofwiki.org/wiki/Annihilator_of_Image_of_Bounded_Linear_Transformation_is_Kernel_of_Dual_Operator | https://proofwiki.org/wiki/Annihilator_of_Image_of_Bounded_Linear_Transformation_is_Kernel_of_Dual_Operator | [
"Annihilators of Subspaces of Banach Spaces",
"Dual Operators"
] | [
"Definition:Normed Vector Space",
"Definition:Bounded Linear Transformation",
"Definition:Normed Dual Space",
"Definition:Dual Operator",
"Definition:Annihilator of Subspace of Banach Space"
] | [] |
proofwiki-21012 | Existence of Complementary Subspace | Let $X$ be a vector space.
Let $N \subseteq X$ be a subspace.
Then $N$ has a complementary subspace.
That is, there exists a subspace $Y \subseteq X$ such that:
:$X = N \oplus Y$ | Let $S$ be the set of all subspaces $V \subseteq X$ such that:
:$N \cap V = \set 0$
By Zorn's Lemma, $\struct {S, \subseteq}$ has a maximal element $Y$.
We claim:
:$X = N \oplus Y$
Indeed, if there would be an $x \in X \setminus \paren {N \oplus Y}$, then:
:$Y \subsetneq \map \span {Y \cup \set x} \in S$
where $\span$ ... | Let $X$ be a [[Definition:Vector Space|vector space]].
Let $N \subseteq X$ be a [[Definition:Vector Subspace|subspace]].
Then $N$ has a [[Definition:Complementary Subspace|complementary subspace]].
That is, there exists a [[Definition:Vector Subspace|subspace]] $Y \subseteq X$ such that:
:$X = N \oplus Y$ | Let $S$ be the [[Definition:Set|set]] of all [[Definition:Vector Subspace|subspaces]] $V \subseteq X$ such that:
:$N \cap V = \set 0$
By [[Zorn's Lemma]], $\struct {S, \subseteq}$ has a [[Definition:Maximal Element|maximal element]] $Y$.
We claim:
:$X = N \oplus Y$
Indeed, if there would be an $x \in X \setminus \p... | Existence of Complementary Subspace | https://proofwiki.org/wiki/Existence_of_Complementary_Subspace | https://proofwiki.org/wiki/Existence_of_Complementary_Subspace | [
"Linear Algebra"
] | [
"Definition:Vector Space",
"Definition:Vector Subspace",
"Definition:Complementary Subspace",
"Definition:Vector Subspace"
] | [
"Definition:Set",
"Definition:Vector Subspace",
"Zorn's Lemma",
"Definition:Maximal/Element",
"Definition:Generated Submodule/Linear Span",
"Definition:Contradiction",
"Definition:Maximal/Element"
] |
proofwiki-21013 | Cauchy-Riemann Equations/Polar Form | The Cauchy-Riemann equations can be expressed in polar form as:
{{begin-eqn}}
{{eqn | n = 1
| l = \dfrac {\partial u} {\partial r}
| r = \dfrac 1 r \dfrac {\partial v} {\partial \theta}
}}
{{eqn | n = 2
| l = \dfrac 1 r \dfrac {\partial u} {\partial \theta}
| r = -\dfrac {\partial v} {\partial r... | {{ProofWanted}}
{{Namedfor|Augustin Louis Cauchy|name2 = Georg Friedrich Bernhard Riemann}} | The [[Definition:Cauchy-Riemann Equations|Cauchy-Riemann equations]] can be expressed in [[Definition:Polar Form of Complex Number|polar form]] as:
{{begin-eqn}}
{{eqn | n = 1
| l = \dfrac {\partial u} {\partial r}
| r = \dfrac 1 r \dfrac {\partial v} {\partial \theta}
}}
{{eqn | n = 2
| l = \dfrac 1... | {{ProofWanted}}
{{Namedfor|Augustin Louis Cauchy|name2 = Georg Friedrich Bernhard Riemann}} | Cauchy-Riemann Equations/Polar Form | https://proofwiki.org/wiki/Cauchy-Riemann_Equations/Polar_Form | https://proofwiki.org/wiki/Cauchy-Riemann_Equations/Polar_Form | [
"Cauchy-Riemann Equations"
] | [
"Definition:Cauchy-Riemann Equations",
"Definition:Complex Number/Polar Form",
"Definition:Complex Number/Polar Form/Exponential Form"
] | [] |
proofwiki-21014 | Degenerate Linear Operator Plus Identity is Fredholm Operator | Let $U$ be a vector space.
Let $T : U \to U$ be a degenerate linear operator.
Let $I_U : U \to U$ be the identity operator.
Then:
:$T + I_U$
is a Fredholm operator. | We need to show that both:
:$\map \ker {T + I_U}$
and:
:$U / {\Img {T + I_U} }$
are finite-dimensional. | Let $U$ be a [[Definition:Vector Space|vector space]].
Let $T : U \to U$ be a [[Definition:Degenerate Linear Transformation|degenerate]] [[Definition:Linear Operator on Vector Space|linear operator]].
Let $I_U : U \to U$ be the [[Definition:Identity Operator|identity operator]].
Then:
:$T + I_U$
is a [[Definition:F... | We need to show that both:
:$\map \ker {T + I_U}$
and:
:$U / {\Img {T + I_U} }$
are [[Definition:Finite Dimensional Vector Space|finite-dimensional]]. | Degenerate Linear Operator Plus Identity is Fredholm Operator | https://proofwiki.org/wiki/Degenerate_Linear_Operator_Plus_Identity_is_Fredholm_Operator | https://proofwiki.org/wiki/Degenerate_Linear_Operator_Plus_Identity_is_Fredholm_Operator | [
"Fredholm Operators",
"Vector Spaces",
"Functional Analysis"
] | [
"Definition:Vector Space",
"Definition:Degenerate Linear Transformation",
"Definition:Linear Operator/Vector Space",
"Definition:Identity Mapping",
"Definition:Fredholm Operator"
] | [
"Definition:Dimension of Vector Space/Finite"
] |
proofwiki-21015 | Recursively Enumerable Set is Image of Primitive Recursive Function | Let $S \subseteq \N$ be recursively enumerable.
Suppose $S$ is non-empty.
Then there exists a primitive recursive function $f : \N^k \to \N$ such that:
:$\Img f = S$ | {{Proofread}}
{{tidy|minor stuff, I'll be on the case in due course}}
By definition of recursively enumerable, there exists a recursive function $g : \N^\ell \to \N$ such that:
:$\Img g = S$
By definition of non-empty, there exists some $x \in S$.
By Kleene's Normal Form Theorem, there exist:
* A primitive recursive re... | Let $S \subseteq \N$ be [[Definition:Recursively Enumerable Set|recursively enumerable]].
Suppose $S$ is [[Definition:Non-Empty Set|non-empty]].
Then there exists a [[Definition:Primitive Recursive Function|primitive recursive function]] $f : \N^k \to \N$ such that:
:$\Img f = S$ | {{Proofread}}
{{tidy|minor stuff, I'll be on the case in due course}}
By definition of [[Definition:Recursively Enumerable Set|recursively enumerable]], there exists a [[Definition:Recursive Function|recursive function]] $g : \N^\ell \to \N$ such that:
:$\Img g = S$
By definition of [[Definition:Non-Empty Set|non-emp... | Recursively Enumerable Set is Image of Primitive Recursive Function | https://proofwiki.org/wiki/Recursively_Enumerable_Set_is_Image_of_Primitive_Recursive_Function | https://proofwiki.org/wiki/Recursively_Enumerable_Set_is_Image_of_Primitive_Recursive_Function | [
"Recursion Theory"
] | [
"Definition:Recursively Enumerable Set",
"Definition:Non-Empty Set",
"Definition:Primitive Recursive/Function"
] | [
"Definition:Recursively Enumerable Set",
"Definition:Recursive/Function",
"Definition:Non-Empty Set",
"Kleene's Normal Form Theorem",
"Definition:Primitive Recursive/Relation",
"Definition:Primitive Recursive/Function",
"Definition:Partial Function",
"Definition:Recursive/Function",
"Definition:Recu... |
proofwiki-21016 | Recursively Enumerable Set is Image of Primitive Recursive Function/Corollary | Let $S \subseteq \N$ be recursively enumerable.
Suppose $S$ is non-empty.
Then there exists a primitive recursive function $f : \N \to \N$ such that:
:$\Img f = S$ | By Recursively Enumerable Set is Image of Primitive Recursive Function, there is some primitive recursive $g : \N^k \to \N$ such that:
:$\Img g = S$
As $S$ is non-empty, let $n \in S$.
Define:
:$\map f x = \begin{cases}
\map g {\map {\operatorname{pred}} {\paren{x}_1}, \dotsc, \map {\operatorname{pred}} {\paren{x}_k}} ... | Let $S \subseteq \N$ be [[Definition:Recursively Enumerable Set|recursively enumerable]].
Suppose $S$ is [[Definition:Non-Empty Set|non-empty]].
Then there exists a [[Definition:Primitive Recursive Function|primitive recursive function]] $f : \N \to \N$ such that:
:$\Img f = S$ | By [[Recursively Enumerable Set is Image of Primitive Recursive Function]], there is some [[Definition:Primitive Recursive Function|primitive recursive]] $g : \N^k \to \N$ such that:
:$\Img g = S$
As $S$ is [[Definition:Non-Empty Set|non-empty]], let $n \in S$.
Define:
:$\map f x = \begin{cases}
\map g {\map {\operat... | Recursively Enumerable Set is Image of Primitive Recursive Function/Corollary | https://proofwiki.org/wiki/Recursively_Enumerable_Set_is_Image_of_Primitive_Recursive_Function/Corollary | https://proofwiki.org/wiki/Recursively_Enumerable_Set_is_Image_of_Primitive_Recursive_Function/Corollary | [
"Recursion Theory"
] | [
"Definition:Recursively Enumerable Set",
"Definition:Non-Empty Set",
"Definition:Primitive Recursive/Function"
] | [
"Recursively Enumerable Set is Image of Primitive Recursive Function",
"Definition:Primitive Recursive/Function",
"Definition:Non-Empty Set",
"Definition:Primitive Recursive/Function",
"Definition by Cases is Primitive Recursive/Corollary",
"Predecessor Function is Primitive Recursive",
"Set of Sequence... |
proofwiki-21017 | Set is Recursively Enumerable iff Domain of Recursive Function | Let $S \subseteq \N$ be a set of natural numbers.
Then:
:$S$ is recursively enumerable
{{iff}}:
:There exists a recursive function $f : \N \to \N$ such that $S = \Dom f$
where $\Dom f$ is the domain of $f$. | === Necessary Condition ===
Suppose $S$ is the empty set.
Define $\map f x = \map {\mu z} {z \ne z}$.
Then, $f$ is undefined for every $x \in \N$, and thus is defined on the empty set.
Additionally, $f$ is recursive by:
* Equality Relation is Primitive Recursive
* Set Operations on Primitive Recursive Relations
* Primi... | Let $S \subseteq \N$ be a [[Definition:Set|set]] of [[Definition:Natural Number|natural numbers]].
Then:
:$S$ is [[Definition:Recursively Enumerable Set|recursively enumerable]]
{{iff}}:
:There exists a [[Definition:Recursive Function|recursive function]] $f : \N \to \N$ such that $S = \Dom f$
where $\Dom f$ is the [[... | === Necessary Condition ===
Suppose $S$ is the [[Definition:Empty Set|empty set]].
Define $\map f x = \map {\mu z} {z \ne z}$.
Then, $f$ is [[Definition:Undefined|undefined]] for every $x \in \N$, and thus is [[Definition:Defined (Partial Mapping)|defined]] on the [[Definition:Empty Set|empty set]].
Additionally, $... | Set is Recursively Enumerable iff Domain of Recursive Function | https://proofwiki.org/wiki/Set_is_Recursively_Enumerable_iff_Domain_of_Recursive_Function | https://proofwiki.org/wiki/Set_is_Recursively_Enumerable_iff_Domain_of_Recursive_Function | [
"Recursion Theory"
] | [
"Definition:Set",
"Definition:Natural Numbers",
"Definition:Recursively Enumerable Set",
"Definition:Recursive/Function",
"Definition:Domain (Set Theory)/Relation"
] | [
"Definition:Empty Set",
"Definition:Undefined",
"Definition:Many-to-One Relation/Defined",
"Definition:Empty Set",
"Definition:Recursive/Function",
"Equality Relation is Primitive Recursive",
"Set Operations on Primitive Recursive Relations",
"Primitive Recursive Function is Total Recursive Function",... |
proofwiki-21018 | Set is Recursive iff Set and Complement are Recursively Enumerable | Let $S \subseteq \N$ be a set of natural numbers.
Then:
:$S$ is a recursive set
{{iff}}:
:$S$ and $\N \setminus S$ are recursively enumerable sets. | === Necessary Condition ===
{{ExtractTheorem|Complement of Recursive Set. More generally, we should show that constructions with primitive recursive ''things'' work on recursive ones as well, such as Definition by Cases is Primitive Recursive, etc.}}
By Complement of Primitive Recursive Set, $\N \setminus S$ is a recur... | Let $S \subseteq \N$ be a [[Definition:Set|set]] of [[Definition:Natural Number|natural numbers]].
Then:
:$S$ is a [[Definition:Recursive Set|recursive set]]
{{iff}}:
:$S$ and $\N \setminus S$ are [[Definition:Recursively Enumerable Set|recursively enumerable sets]]. | === Necessary Condition ===
{{ExtractTheorem|[[Complement of Recursive Set]]. More generally, we should show that constructions with primitive recursive ''things'' work on recursive ones as well, such as [[Definition by Cases is Primitive Recursive]], etc.}}
By [[Complement of Primitive Recursive Set]], $\N \setminus... | Set is Recursive iff Set and Complement are Recursively Enumerable | https://proofwiki.org/wiki/Set_is_Recursive_iff_Set_and_Complement_are_Recursively_Enumerable | https://proofwiki.org/wiki/Set_is_Recursive_iff_Set_and_Complement_are_Recursively_Enumerable | [
"Recursion Theory"
] | [
"Definition:Set",
"Definition:Natural Numbers",
"Definition:Recursive/Set",
"Definition:Recursively Enumerable Set"
] | [
"Complement of Recursive Set",
"Definition by Cases is Primitive Recursive",
"Complement of Primitive Recursive Set",
"Definition:Recursive/Set",
"Definition:Recursive/Set",
"Definition:Recursively Enumerable Set",
"Definition:Recursive/Set",
"Definition:Characteristic Function (Set Theory)/Set",
"D... |
proofwiki-21019 | Union of Recursively Enumerable Sets | Let $S, T \subseteq \N$ be recursively enumerable sets.
Then $S \cup T$ is recursively enumerable. | Suppose $S = \O$.
Then by Union with Empty Set:
:$S \cup T = T$
which is recursively enumerable by assumption.
{{qed|lemma}}
Suppose $T = \O$.
For the same reason:
:$S \cup T = S$
which is also recursively enumerable by assumption.
{{qed|lemma}}
Now, suppose neither $S$ nor $T$ is empty.
By {{Corollary|Recursively Enum... | Let $S, T \subseteq \N$ be [[Definition:Recursively Enumerable Set|recursively enumerable sets]].
Then $S \cup T$ is [[Definition:Recursively Enumerable Set|recursively enumerable]]. | Suppose $S = \O$.
Then by [[Union with Empty Set]]:
:$S \cup T = T$
which is [[Definition:Recursively Enumerable Set|recursively enumerable]] by assumption.
{{qed|lemma}}
Suppose $T = \O$.
For the same reason:
:$S \cup T = S$
which is also [[Definition:Recursively Enumerable Set|recursively enumerable]] by assumpti... | Union of Recursively Enumerable Sets | https://proofwiki.org/wiki/Union_of_Recursively_Enumerable_Sets | https://proofwiki.org/wiki/Union_of_Recursively_Enumerable_Sets | [
"Recursion Theory"
] | [
"Definition:Recursively Enumerable Set",
"Definition:Recursively Enumerable Set"
] | [
"Union with Empty Set",
"Definition:Recursively Enumerable Set",
"Definition:Recursively Enumerable Set",
"Definition:Empty Set",
"Definition:Primitive Recursive/Function",
"Equality Relation is Primitive Recursive",
"Definition:Primitive Recursive/Function",
"Primitive Recursive Function is Total Rec... |
proofwiki-21020 | Intersection of Recursively Enumerable Sets | Let $S, T \subseteq \N$ be recursively enumerable.
Then $S \cap T$ is recursively enumerable. | By Set is Recursively Enumerable iff Domain of Recursive Function, there exist recursive $f, g : \N \to \N$ such that:
:$S = \Dom f$
:$T = \Dom g$
Define:
:$\map h x = \map f x + \map g x$
By:
* Addition is Primitive Recursive
* Primitive Recursive Function is Total Recursive Function
it follows that $h$ is recursive.
... | Let $S, T \subseteq \N$ be [[Definition:Recursively Enumerable Set|recursively enumerable]].
Then $S \cap T$ is [[Definition:Recursively Enumerable Set|recursively enumerable]]. | By [[Set is Recursively Enumerable iff Domain of Recursive Function]], there exist [[Definition:Recursive Function|recursive]] $f, g : \N \to \N$ such that:
:$S = \Dom f$
:$T = \Dom g$
Define:
:$\map h x = \map f x + \map g x$
By:
* [[Addition is Primitive Recursive]]
* [[Primitive Recursive Function is Total Recursi... | Intersection of Recursively Enumerable Sets | https://proofwiki.org/wiki/Intersection_of_Recursively_Enumerable_Sets | https://proofwiki.org/wiki/Intersection_of_Recursively_Enumerable_Sets | [
"Recursion Theory"
] | [
"Definition:Recursively Enumerable Set",
"Definition:Recursively Enumerable Set"
] | [
"Set is Recursively Enumerable iff Domain of Recursive Function",
"Definition:Recursive/Function",
"Addition is Primitive Recursive",
"Primitive Recursive Function is Total Recursive Function",
"Definition:Recursive/Function",
"Definition:Subset",
"Definition:Substitution (Mathematical Logic)",
"Defin... |
proofwiki-21021 | Generalized Sum with Finite Non-zero Summands | Let $G$ be a commutative topological semigroup with identity $0_G$.
Let $\family{g }_{i \in I}$ be an indexed family of elements of $G$.
Let $\set{i \in I : g_i \ne 0_G}$ be finite.
Let $\set{i_1, i_2, \cdots, i_n}$ be a finite enumeration of $\set{i \in I : g_i \ne 0_G}$.
Then the generalized sum $\ds \sum_{i \mathop ... | Let $J = \set{i \in I : g_i \ne 0_G}$.
From Finite Generalized Sum Converges to Summation:
:the generalized sum $\ds \sum_{j \mathop \in J} g_j$ converges to the summation $\ds \sum_{k \mathop = 1}^n g_{i_k}$
From Generalized Sum Restricted to Non-zero Summands:
:the generalized sum $\ds \sum_{i \mathop \in I} g_i$ co... | Let $G$ be a [[Definition:Commutative Semigroup|commutative]] [[Definition:Topological Semigroup|topological semigroup]] with [[Definition:Identity Element|identity]] $0_G$.
Let $\family{g }_{i \in I}$ be an [[Definition:Indexed Family|indexed family]] of [[Definition:Element|elements]] of $G$.
Let $\set{i \in I : ... | Let $J = \set{i \in I : g_i \ne 0_G}$.
From [[Finite Generalized Sum Converges to Summation]]:
:the [[Definition:Generalized Sum|generalized sum]] $\ds \sum_{j \mathop \in J} g_j$ [[Definition:Convergent Net|converges]] to the [[Definition:Summation|summation]] $\ds \sum_{k \mathop = 1}^n g_{i_k}$
From [[Generalized... | Generalized Sum with Finite Non-zero Summands | https://proofwiki.org/wiki/Generalized_Sum_with_Finite_Non-zero_Summands | https://proofwiki.org/wiki/Generalized_Sum_with_Finite_Non-zero_Summands | [
"Generalized Sums"
] | [
"Definition:Commutative Semigroup",
"Definition:Topological Semigroup",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Indexing Set/Family",
"Definition:Element",
"Definition:Finite Set",
"Definition:Enumeration/Finite",
"Definition:Generalized Sum",
"Definition:Convergent ... | [
"Finite Generalized Sum Converges to Summation",
"Definition:Generalized Sum",
"Definition:Convergent Net",
"Definition:Summation",
"Generalized Sum Restricted to Non-zero Summands",
"Definition:Generalized Sum",
"Definition:Convergent Net",
"Definition:Summation",
"Category:Generalized Sums"
] |
proofwiki-21022 | Generalized Sum with Countable Non-zero Summands | Let $V$ be a Banach space.
Let $\norm {\, \cdot \,}$ denote the norm on $V$.
Let $\family {v_i}_{i \mathop \in I}$ be an indexed family of elements of $V$.
Let $J$ be a countably infinite subset of $I$ such that $\set{i \in I : v_i \ne 0} \subseteq J$.
Let $\set {j_0, j_1, j_2, \ldots}$ be a countably infinite enumerat... | From {{Corollary|Generalized Sum Restricted to Non-zero Summands}}:
:the generalized sum $\ds \sum_{i \mathop \in I} v_i$ converges absolutely to $r$
{{iff}}
:the generalized sum $\ds \sum_{j \mathop \in J} v_j$ converges absolutely to $r$
From Absolute Net Convergence Equivalent to Absolute Convergence:
:the genera... | Let $V$ be a [[Definition:Banach Space|Banach space]].
Let $\norm {\, \cdot \,}$ denote the [[Definition:Norm on Vector Space|norm]] on $V$.
Let $\family {v_i}_{i \mathop \in I}$ be an [[Definition:Indexed Family|indexed family]] of [[Definition:Element|elements]] of $V$.
Let $J$ be a [[Definition:Countably Infini... | From {{Corollary|Generalized Sum Restricted to Non-zero Summands}}:
:the [[Definition:Generalized Sum|generalized sum]] $\ds \sum_{i \mathop \in I} v_i$ [[Definition:Absolute Net Convergence|converges absolutely]] to $r$
{{iff}}
:the [[Definition:Generalized Sum|generalized sum]] $\ds \sum_{j \mathop \in J} v_j$ [[Def... | Generalized Sum with Countable Non-zero Summands | https://proofwiki.org/wiki/Generalized_Sum_with_Countable_Non-zero_Summands | https://proofwiki.org/wiki/Generalized_Sum_with_Countable_Non-zero_Summands | [
"Generalized Sums",
"Banach Spaces",
"Generalized Sum with Countable Non-zero Summands"
] | [
"Definition:Banach Space",
"Definition:Norm/Vector Space",
"Definition:Indexing Set/Family",
"Definition:Element",
"Definition:Countably Infinite",
"Definition:Subset",
"Definition:Enumeration/Countably Infinite",
"Definition:Generalized Sum",
"Definition:Generalized Sum/Absolute Net Convergence",
... | [
"Definition:Generalized Sum",
"Definition:Generalized Sum/Absolute Net Convergence",
"Definition:Generalized Sum",
"Definition:Generalized Sum/Absolute Net Convergence",
"Absolute Net Convergence Equivalent to Absolute Convergence",
"Definition:Generalized Sum",
"Definition:Generalized Sum/Absolute Net ... |
proofwiki-21023 | Neumann Series Theorem/Corollary 2 | The mapping $\paren {I - A}^{-1} : X \to X$ is continuous. | {{ProofWanted|Obviously, we have invertibility in $CL(X)$, but the source mentions $\norm x \le \frac 1 {1 - \norm A} \norm y$. A chance to have two proofs?}} | The [[Definition:Mapping|mapping]] $\paren {I - A}^{-1} : X \to X$ is [[Definition:Continuous Mapping|continuous]]. | {{ProofWanted|Obviously, we have invertibility in $CL(X)$, but the source mentions $\norm x \le \frac 1 {1 - \norm A} \norm y$. A chance to have two proofs?}} | Neumann Series Theorem/Corollary 2 | https://proofwiki.org/wiki/Neumann_Series_Theorem/Corollary_2 | https://proofwiki.org/wiki/Neumann_Series_Theorem/Corollary_2 | [
"Operators",
"Banach Spaces",
"Bijections"
] | [
"Definition:Mapping",
"Definition:Continuous Mapping"
] | [] |
proofwiki-21024 | System of Linear Equations as Continuous Linear Transformation | Let $x_1, x_2 \in \R$ be real numbers.
Consider the following system of simultaneous linear equations $\paren S$:
{{begin-eqn}}
{{eqn | l = x_1
| r = \frac 1 2 x_1 + \frac 1 3 x_2 + 1
}}
{{eqn | l = x_2
| r = \frac 1 3 x_1 + \frac 1 4 x_2 + 2
}}
{{end-eqn}}
Let $\norm {\, \cdot \,}_2$ be the Euclidean norm.... | === $\paren S$ is expressible as $\paren {I - K} x = y$ ===
We have that:
{{begin-eqn}}
{{eqn | l = x_1
| r = \frac 1 2 x_1 + \frac 1 3 x_2 + 1
}}
{{eqn | ll = \leadsto
| l = \frac 1 2 x_1 - \frac 1 3 x_2
| r = 1
}}
{{end-eqn}}
Furthermore:
{{begin-eqn}}
{{eqn | l = x_1
| r = \frac 1 2 x_1 + \frac... | Let $x_1, x_2 \in \R$ be [[Definition:Real Number|real numbers]].
Consider the following system of [[Definition:Simultaneous Linear Equations|simultaneous linear equations]] $\paren S$:
{{begin-eqn}}
{{eqn | l = x_1
| r = \frac 1 2 x_1 + \frac 1 3 x_2 + 1
}}
{{eqn | l = x_2
| r = \frac 1 3 x_1 + \frac 1 4... | === $\paren S$ is expressible as $\paren {I - K} x = y$ ===
We have that:
{{begin-eqn}}
{{eqn | l = x_1
| r = \frac 1 2 x_1 + \frac 1 3 x_2 + 1
}}
{{eqn | ll = \leadsto
| l = \frac 1 2 x_1 - \frac 1 3 x_2
| r = 1
}}
{{end-eqn}}
Furthermore:
{{begin-eqn}}
{{eqn | l = x_1
| r = \frac 1 2 x_1 + \... | System of Linear Equations as Continuous Linear Transformation | https://proofwiki.org/wiki/System_of_Linear_Equations_as_Continuous_Linear_Transformation | https://proofwiki.org/wiki/System_of_Linear_Equations_as_Continuous_Linear_Transformation | [
"Simultaneous Equations",
"Neumann Series"
] | [
"Definition:Real Number",
"Definition:Simultaneous Equations/Linear Equations",
"Definition:Euclidean Norm",
"Definition:Supremum Operator Norm",
"Definition:Unit Matrix",
"Definition:Unique",
"Definition:Simultaneous Equations/Solution"
] | [] |
proofwiki-21025 | Continuous Linear Transformation between Topological Vector Spaces is Bounded | Let $\GF \in \set {\R, \C}$.
Let $X$ and $Y$ be topological vector spaces over $\GF$.
Let $T : X \to Y$ be a continuous linear transformation.
Then $T$ is bounded. | Let $E$ be a von Neumann-bounded subset of $X$.
We want to show that $T \sqbrk E$ is a von Neumann-bounded subset of $Y$.
Let $W$ be a open neighborhood of ${\mathbf 0}_Y$ in $Y$.
Since $T$ is continuous, it is continuous at $\mathbf 0_X$.
Since $\map T {\mathbf 0_X} = \mathbf 0_Y$, there exists an open neighborhood ... | Let $\GF \in \set {\R, \C}$.
Let $X$ and $Y$ be [[Definition:Topological Vector Space|topological vector spaces]] over $\GF$.
Let $T : X \to Y$ be a [[Definition:Continuous Mapping|continuous]] [[Definition:Linear Transformation|linear transformation]].
Then $T$ is [[Definition:Bounded Linear Transformation/Topol... | Let $E$ be a [[Definition:Von Neumann-Bounded Subset of Topological Vector Space|von Neumann-bounded subset]] of $X$.
We want to show that $T \sqbrk E$ is a [[Definition:Von Neumann-Bounded Subset of Topological Vector Space|von Neumann-bounded subset]] of $Y$.
Let $W$ be a [[Definition:Open Neighborhood|open neigh... | Continuous Linear Transformation between Topological Vector Spaces is Bounded | https://proofwiki.org/wiki/Continuous_Linear_Transformation_between_Topological_Vector_Spaces_is_Bounded | https://proofwiki.org/wiki/Continuous_Linear_Transformation_between_Topological_Vector_Spaces_is_Bounded | [
"Bounded Linear Transformations (Topological Vector Spaces)"
] | [
"Definition:Topological Vector Space",
"Definition:Continuous Mapping",
"Definition:Linear Transformation",
"Definition:Bounded Linear Transformation/Topological Vector Space"
] | [
"Definition:Von Neumann-Bounded Subset of Topological Vector Space",
"Definition:Von Neumann-Bounded Subset of Topological Vector Space",
"Definition:Open Neighborhood",
"Definition:Continuous Mapping",
"Definition:Continuous Mapping (Topology)/Point",
"Definition:Open Neighborhood",
"Definition:Von Neu... |
proofwiki-21026 | Image of Convergent Sequence in Topological Vector Space under Bounded Linear Transformation is von Neumann-Bounded | Let $\GF \in \set {\R, \C}$.
Let $X$ and $Y$ be topological vector spaces over $\GF$.
Let $T : X \to Y$ be a bounded linear transformation.
Let $\sequence {x_n}_{n \in \N}$ be a convergent sequence in $X$.
Then $\set {T x_n : n \in \N}$ is a von Neumann-bounded subset of $Y$. | Let:
:$E = \set {x_n : n \in \N}$
From Image of Convergent Sequence in Topological Vector Space is von Neumann-Bounded, $E$ is von Neumann-bounded.
So, from the definition of a bounded linear transformation:
:$T \sqbrk E = \set {T x_n : n \in \N}$ is von Neumann-bounded.
{{qed}} | Let $\GF \in \set {\R, \C}$.
Let $X$ and $Y$ be [[Definition:Topological Vector Space|topological vector spaces]] over $\GF$.
Let $T : X \to Y$ be a [[Definition:Bounded Linear Transformation/Topological Vector Space|bounded linear transformation]].
Let $\sequence {x_n}_{n \in \N}$ be a [[Definition:Convergent Seq... | Let:
:$E = \set {x_n : n \in \N}$
From [[Image of Convergent Sequence in Topological Vector Space is von Neumann-Bounded]], $E$ is [[Definition:Von Neumann-Bounded Subset of Topological Vector Space|von Neumann-bounded]].
So, from the definition of a [[Definition:Bounded Linear Transformation/Topological Vector Space... | Image of Convergent Sequence in Topological Vector Space under Bounded Linear Transformation is von Neumann-Bounded | https://proofwiki.org/wiki/Image_of_Convergent_Sequence_in_Topological_Vector_Space_under_Bounded_Linear_Transformation_is_von_Neumann-Bounded | https://proofwiki.org/wiki/Image_of_Convergent_Sequence_in_Topological_Vector_Space_under_Bounded_Linear_Transformation_is_von_Neumann-Bounded | [
"Bounded Linear Transformations (Topological Vector Spaces)"
] | [
"Definition:Topological Vector Space",
"Definition:Bounded Linear Transformation/Topological Vector Space",
"Definition:Convergent Sequence",
"Definition:Von Neumann-Bounded Subset of Topological Vector Space"
] | [
"Image of Convergent Sequence in Topological Vector Space is von Neumann-Bounded",
"Definition:Von Neumann-Bounded Subset of Topological Vector Space",
"Definition:Bounded Linear Transformation/Topological Vector Space",
"Definition:Von Neumann-Bounded Subset of Topological Vector Space"
] |
proofwiki-21027 | Uniform Mean Ergodic Theorem | Let $\GF \in \set {\R, \C}$.
Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space over $\mathbb F$.
Let $U : \HH \to \HH$ be a bounded linear operator such that:
:$\forall f \in \HH : \norm {\map U f} \le \norm f$
Then for each $f \in \HH$:
:$\ds \lim_{N - M \mathop \to \infty} \dfrac 1 {N - M} \sum_{n \matho... | {{begin-eqn}}
{{eqn | l = \norm {\frac 1 {N - M} \sum_{n \mathop = M}^{N - 1} \map {U^n} f - \map P f}
| r = \norm {\frac 1 {N - M} \sum_{n \mathop = 0}^{N - M - 1} \map {U^{M+n} } f - \map P f}
}}
{{eqn | r = \norm {\frac 1 {N - M} \sum_{n \mathop = M}^{N - 1} \map {U^{M+n} } f - \map {U^M} {\map P f} }
| ... | Let $\GF \in \set {\R, \C}$.
Let $\struct {\HH, \innerprod \cdot \cdot}$ be a [[Definition:Hilbert Space|Hilbert space]] over $\mathbb F$.
Let $U : \HH \to \HH$ be a [[Definition:Bounded Linear Operator/Inner Product Space|bounded linear operator]] such that:
:$\forall f \in \HH : \norm {\map U f} \le \norm f$
Then... | {{begin-eqn}}
{{eqn | l = \norm {\frac 1 {N - M} \sum_{n \mathop = M}^{N - 1} \map {U^n} f - \map P f}
| r = \norm {\frac 1 {N - M} \sum_{n \mathop = 0}^{N - M - 1} \map {U^{M+n} } f - \map P f}
}}
{{eqn | r = \norm {\frac 1 {N - M} \sum_{n \mathop = M}^{N - 1} \map {U^{M+n} } f - \map {U^M} {\map P f} }
| ... | Uniform Mean Ergodic Theorem | https://proofwiki.org/wiki/Uniform_Mean_Ergodic_Theorem | https://proofwiki.org/wiki/Uniform_Mean_Ergodic_Theorem | [
"Mean Ergodic Theorem"
] | [
"Definition:Hilbert Space",
"Definition:Bounded Linear Operator/Inner Product Space",
"Definition:Composition of Mappings",
"Definition:Orthogonal Projection"
] | [
"Mean Ergodic Theorem (Hilbert Space)"
] |
proofwiki-21028 | Cantor Pairing Function is Well-Defined | The Cantor pairing function is well-defined. | By definition, the Cantor pairing function $\pi : \N^2 \to \N$ is:
:$\ds \map \pi {m, n} = \frac 1 2 \paren {m + n} \paren {m + n + 1} + m$
It suffices to show that, for every $m, n \in \N$:
:$\paren {m + n} \paren {m + n + 1}$
is divisible by $2$.
Suppose that $m + n$ is even.
Then, by definition, $2 \divides \paren {... | The [[Definition:Cantor Pairing Function|Cantor pairing function]] is [[Definition:Well-Defined|well-defined]]. | By definition, the [[Definition:Cantor Pairing Function|Cantor pairing function]] $\pi : \N^2 \to \N$ is:
:$\ds \map \pi {m, n} = \frac 1 2 \paren {m + n} \paren {m + n + 1} + m$
It suffices to show that, for every $m, n \in \N$:
:$\paren {m + n} \paren {m + n + 1}$
is [[Definition:Divisor in Natural Numbers|divisible... | Cantor Pairing Function is Well-Defined | https://proofwiki.org/wiki/Cantor_Pairing_Function_is_Well-Defined | https://proofwiki.org/wiki/Cantor_Pairing_Function_is_Well-Defined | [
"Cantor Pairing Function"
] | [
"Definition:Cantor Pairing Function",
"Definition:Well-Defined"
] | [
"Definition:Cantor Pairing Function",
"Definition:Divisor (Algebra)/Natural Numbers",
"Definition:Even Integer",
"Divisor Divides Multiple",
"Definition:Even Integer",
"Definition:Odd Integer/Definition 1",
"Definition:Odd Integer/Definition 2",
"Definition:Even Integer/Definition 2",
"Definition:Ev... |
proofwiki-21029 | Inverse of Cantor Pairing Function | Let $\pi : \N^2 \to \N$ be the Cantor pairing function.
Define $k : \N \to \N$ as:
:$\map k z$ is the largest $k$ such that $T_k \le z$
where $T_k$ is the $k$-th triangular number.
Let $\pi_1 : \N \to \N$ be defined as:
:$\ds \map {\pi_1} z = z - T_{\map k z}$
Let $\pi_2 : \N \to \N$ be defined as:
:$\map {\pi_2} z = \... | By definition of $\map k z$, we have $T_{\map k z} \le z$.
Thus, $z - T_{\map k z} \ge 0$.
Therefore, $\pi_1$ is well-defined.
{{AimForCont}} $\map {\pi_1} z > \map k z$.
That is:
:$z > T_{\map k z} + \map k z$
Or:
:$z \ge T_{\map k z} + \map k z + 1 = T_{\map k z + 1}$
which contradicts the maximality of $\map k z$.
T... | Let $\pi : \N^2 \to \N$ be the [[Definition:Cantor Pairing Function|Cantor pairing function]].
Define $k : \N \to \N$ as:
:$\map k z$ is the largest $k$ such that $T_k \le z$
where $T_k$ is the $k$-th [[Definition:Triangular Number|triangular number]].
Let $\pi_1 : \N \to \N$ be defined as:
:$\ds \map {\pi_1} z = z ... | By definition of $\map k z$, we have $T_{\map k z} \le z$.
Thus, $z - T_{\map k z} \ge 0$.
Therefore, $\pi_1$ is [[Definition:Well-Defined|well-defined]].
{{AimForCont}} $\map {\pi_1} z > \map k z$.
That is:
:$z > T_{\map k z} + \map k z$
Or:
:$z \ge T_{\map k z} + \map k z + 1 = T_{\map k z + 1}$
which [[Definit... | Inverse of Cantor Pairing Function | https://proofwiki.org/wiki/Inverse_of_Cantor_Pairing_Function | https://proofwiki.org/wiki/Inverse_of_Cantor_Pairing_Function | [
"Cantor Pairing Function"
] | [
"Definition:Cantor Pairing Function",
"Definition:Triangular Number",
"Definition:Well-Defined"
] | [
"Definition:Well-Defined",
"Definition:Contradiction",
"Definition:Maximal",
"Proof by Contradiction",
"Definition:Well-Defined",
"Closed Form for Triangular Numbers",
"Closed Form for Triangular Numbers",
"Category:Cantor Pairing Function"
] |
proofwiki-21030 | Expression for Closure of Set in Topological Vector Space/Corollary | Let $\BB$ be a local basis for ${\mathbf 0}_X$ in $\struct {X, \tau}$.
Then:
:$\ds A^- = \bigcap_{V \in \BB} \paren {A + V}$
where $A^-$ is the closure of $A$. | Let $\VV$ be the set of open neighborhoods of ${\mathbf 0}_X$ in $\struct {X, \tau}$.
From Intersection is Decreasing, we have:
:$\ds \bigcap_{V \in \mathcal V} \paren {A + V} \subseteq \bigcap_{V \in \BB} \paren {A + V}$
Conversely, suppose that:
:$\ds x \in \bigcap_{V \in \BB} \paren {A + V}$
Let $O \in \VV$.
Since ... | Let $\BB$ be a [[Definition:Local Basis|local basis]] for ${\mathbf 0}_X$ in $\struct {X, \tau}$.
Then:
:$\ds A^- = \bigcap_{V \in \BB} \paren {A + V}$
where $A^-$ is the [[Definition:Closure (Topology)|closure]] of $A$. | Let $\VV$ be the [[Definition:Set|set]] of [[Definition:Open Neighborhood|open neighborhoods]] of ${\mathbf 0}_X$ in $\struct {X, \tau}$.
From [[Intersection is Decreasing]], we have:
:$\ds \bigcap_{V \in \mathcal V} \paren {A + V} \subseteq \bigcap_{V \in \BB} \paren {A + V}$
Conversely, suppose that:
:$\ds x \in \... | Expression for Closure of Set in Topological Vector Space/Corollary | https://proofwiki.org/wiki/Expression_for_Closure_of_Set_in_Topological_Vector_Space/Corollary | https://proofwiki.org/wiki/Expression_for_Closure_of_Set_in_Topological_Vector_Space/Corollary | [
"Expression for Closure of Set in Topological Vector Space"
] | [
"Definition:Local Basis",
"Definition:Closure (Topology)"
] | [
"Definition:Set",
"Definition:Open Neighborhood",
"Intersection is Decreasing",
"Definition:Local Basis",
"Category:Expression for Closure of Set in Topological Vector Space"
] |
proofwiki-21031 | Locally Compact Hausdorff Topological Vector Space has Finite Dimension | Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \tau}$ be a locally compact Hausdorff topological vector space over $\GF$.
Then $X$ is a finite dimensional vector space. | Since $X$ is locally compact, there exists a von Neumann-bounded open neighborhood $V$ of ${\mathbf 0}_X$ such that:
:$\map \cl V$ is compact.
From Dilations of von Neumann-Bounded Neighborhood of Origin in Topological Vector Space form Local Basis for Origin:
:$\BB = \set {2^{-n} V : n \in \N}$ is a local basis for $... | Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \tau}$ be a [[Definition:Locally Compact Topological Vector Space|locally compact]] [[Definition:Hausdorff Topological Vector Space|Hausdorff topological vector space]] over $\GF$.
Then $X$ is a [[Definition:Finite Dimensional Vector Space|finite dimensional vector spa... | Since $X$ is [[Definition:Locally Compact Topological Vector Space|locally compact]], there exists a [[Definition:Von Neumann-Bounded Subset of Topological Vector Space|von Neumann-bounded]] [[Definition:Open Neighborhood|open neighborhood]] $V$ of ${\mathbf 0}_X$ such that:
:$\map \cl V$ is [[Definition:Compact Topol... | Locally Compact Hausdorff Topological Vector Space has Finite Dimension | https://proofwiki.org/wiki/Locally_Compact_Hausdorff_Topological_Vector_Space_has_Finite_Dimension | https://proofwiki.org/wiki/Locally_Compact_Hausdorff_Topological_Vector_Space_has_Finite_Dimension | [
"Locally Compact Hausdorff Topological Vector Space has Finite Dimension",
"Locally Compact Hausdorff Spaces",
"Hausdorff Topological Vector Spaces"
] | [
"Definition:Locally Compact Topological Vector Space",
"Definition:Hausdorff Topological Vector Space",
"Definition:Dimension of Vector Space/Finite"
] | [
"Definition:Locally Compact Topological Vector Space",
"Definition:Von Neumann-Bounded Subset of Topological Vector Space",
"Definition:Open Neighborhood",
"Definition:Compact Topological Space",
"Dilations of von Neumann-Bounded Neighborhood of Origin in Topological Vector Space form Local Basis for Origin... |
proofwiki-21032 | Locally Bounded Hausdorff Topological Vector Space with Heine-Borel Property has Finite Dimension | Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \tau}$ be a locally bounded Hausdorff topological vector space over $\GF$ with the Heine-Borel property.
Then $X$ is a finite dimensional vector space. | Since $\struct {X, \tau}$ is locally bounded, there exists a von Neumann-bounded open neighborhood $V$ of ${\mathbf 0}_X$.
From Closure of von Neumann-Bounded Subset of Topological Vector Space is von Neumann-Bounded, $\map \cl V$ is von Neumann-bounded.
From Topological Closure is Closed, $\map \cl V$ is closed.
So $... | Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \tau}$ be a [[Definition:Locally Bounded Topological Vector Space|locally bounded]] [[Definition:Hausdorff Topological Vector Space|Hausdorff topological vector space]] over $\GF$ with the [[Definition:Heine-Borel Property of Topological Vector Space|Heine-Borel property]... | Since $\struct {X, \tau}$ is [[Definition:Locally Bounded Topological Vector Space|locally bounded]], there exists a [[Definition:Von Neumann-Bounded Subset of Topological Vector Space|von Neumann-bounded]] [[Definition:Open Neighborhood|open neighborhood]] $V$ of ${\mathbf 0}_X$.
From [[Closure of von Neumann-Bounde... | Locally Bounded Hausdorff Topological Vector Space with Heine-Borel Property has Finite Dimension | https://proofwiki.org/wiki/Locally_Bounded_Hausdorff_Topological_Vector_Space_with_Heine-Borel_Property_has_Finite_Dimension | https://proofwiki.org/wiki/Locally_Bounded_Hausdorff_Topological_Vector_Space_with_Heine-Borel_Property_has_Finite_Dimension | [
"Heine-Borel Property of Topological Vector Spaces",
"Locally Bounded Topological Vector Spaces",
"Hausdorff Topological Vector Spaces"
] | [
"Definition:Locally Bounded Topological Vector Space",
"Definition:Hausdorff Topological Vector Space",
"Definition:Heine-Borel Property of Topological Vector Space",
"Definition:Dimension of Vector Space/Finite"
] | [
"Definition:Locally Bounded Topological Vector Space",
"Definition:Von Neumann-Bounded Subset of Topological Vector Space",
"Definition:Open Neighborhood",
"Closure of von Neumann-Bounded Subset of Topological Vector Space is von Neumann-Bounded",
"Definition:Von Neumann-Bounded Subset of Topological Vector... |
proofwiki-21033 | Union of Chain of Convex Sets in Vector Space is Convex | Let $\GF \in \set {\R, \C}$.
Let $X$ be a vector space over $\GF$.
Let $\Gamma$ be a chain of convex sets.
Let:
:$\ds C = \bigcup \Gamma$
Then $C$ is convex. | Let $t \in \closedint 0 1$ and $x, y \in C$.
Then there exists $C_1, C_2 \in \Gamma$ such that $x \in C_1$ and $y \in C_2$.
Since $\Gamma$ is a chain, we have $C_1 \subseteq C_2$ or $C_2 \subseteq C_1$.
{{WLOG}} suppose that $C_1 \subseteq C_2$.
Then $x, y \in C_2$.
Since $C_2$ is convex, we have $t x + \paren {1 - t}... | Let $\GF \in \set {\R, \C}$.
Let $X$ be a [[Definition:Vector Space|vector space]] over $\GF$.
Let $\Gamma$ be a [[Definition:Chain (Order Theory)|chain]] of [[Definition:Convex Set (Vector Space)|convex sets]].
Let:
:$\ds C = \bigcup \Gamma$
Then $C$ is [[Definition:Convex Set (Vector Space)|convex]]. | Let $t \in \closedint 0 1$ and $x, y \in C$.
Then there exists $C_1, C_2 \in \Gamma$ such that $x \in C_1$ and $y \in C_2$.
Since $\Gamma$ is a [[Definition:Chain (Order Theory)|chain]], we have $C_1 \subseteq C_2$ or $C_2 \subseteq C_1$.
{{WLOG}} suppose that $C_1 \subseteq C_2$.
Then $x, y \in C_2$.
Since $C_2$... | Union of Chain of Convex Sets in Vector Space is Convex | https://proofwiki.org/wiki/Union_of_Chain_of_Convex_Sets_in_Vector_Space_is_Convex | https://proofwiki.org/wiki/Union_of_Chain_of_Convex_Sets_in_Vector_Space_is_Convex | [
"Convex Sets (Vector Spaces)"
] | [
"Definition:Vector Space",
"Definition:Chain (Order Theory)",
"Definition:Convex Set (Vector Space)",
"Definition:Convex Set (Vector Space)"
] | [
"Definition:Chain (Order Theory)",
"Definition:Convex Set (Vector Space)",
"Definition:Convex Set (Vector Space)",
"Category:Convex Sets (Vector Spaces)"
] |
proofwiki-21034 | Least Fixed Point of Enumeration Operator | Let $\psi : \powerset \N \to \powerset \N$ be an enumeration operator.
Let $A_i$ be defined recursively as:
{{begin-eqn}}
{{eqn | l = A_0
| r = \O
}}
{{eqn | l = A_{n + 1}
| r = \map \psi {A_n}
}}
{{end-eqn}}
Let $A = \ds \bigcup_{i \mathop \in \N} A_i$.
Then:
* $A$ is a fixed point of $\psi$
* Every fixed ... | === Lemma ===
{{:Least Fixed Point of Enumeration Operator/Lemma}}{{qed|lemma}}
By definition of enumeration operator, let $\phi \subseteq \N$ be a recursively enumerable set that yields $\psi$. | Let $\psi : \powerset \N \to \powerset \N$ be an [[Definition:Enumeration Operator (Recursion Theory)|enumeration operator]].
Let $A_i$ be defined [[Definition:Recursively Defined Mapping|recursively]] as:
{{begin-eqn}}
{{eqn | l = A_0
| r = \O
}}
{{eqn | l = A_{n + 1}
| r = \map \psi {A_n}
}}
{{end-eqn}}
... | === [[Least Fixed Point of Enumeration Operator/Lemma|Lemma]] ===
{{:Least Fixed Point of Enumeration Operator/Lemma}}{{qed|lemma}}
By definition of [[Definition:Enumeration Operator (Recursion Theory)|enumeration operator]], let $\phi \subseteq \N$ be a [[Definition:Recursively Enumerable Set|recursively enumerable ... | Least Fixed Point of Enumeration Operator | https://proofwiki.org/wiki/Least_Fixed_Point_of_Enumeration_Operator | https://proofwiki.org/wiki/Least_Fixed_Point_of_Enumeration_Operator | [
"Least Fixed Point of Enumeration Operator",
"Recursion Theory"
] | [
"Definition:Enumeration Operator (Recursion Theory)",
"Definition:Recursively Defined Mapping",
"Definition:Fixed Point",
"Definition:Fixed Point",
"Definition:Subset/Superset"
] | [
"Least Fixed Point of Enumeration Operator/Lemma",
"Definition:Enumeration Operator (Recursion Theory)",
"Definition:Recursively Enumerable Set",
"Definition:Enumeration Operator (Recursion Theory)",
"Definition:Enumeration Operator (Recursion Theory)",
"Definition:Enumeration Operator (Recursion Theory)"... |
proofwiki-21035 | Image of Interior of Set under Homeomorphism is Interior of Image | Let $X$ and $Y$ be topological spaces.
Let $f : X \to Y$ be a homeomorphism.
Let $A \subseteq X$.
Then:
:$\paren {f \sqbrk A}^\circ = f \sqbrk {A^\circ}$
where:
:$A^\circ$ is the interior of $A$ in $X$
:$\paren {f \sqbrk A}^\circ$ is the interior of $f \sqbrk A$ in $Y$. | First, we show that $f \sqbrk {A^\circ} \subseteq \paren {f \sqbrk A}^\circ$.
Let $y \in f \sqbrk {A^\circ}$.
Then there exists $x \in A^\circ$ such that $y = \map f x$.
Since $x \in A^\circ$, there exists an open set $U \subseteq A$ in $X$ such that $x \in U$.
Then $y = \map f x \in f \sqbrk U \subseteq f \sqbrk A$.
... | Let $X$ and $Y$ be [[Definition:Topological Space|topological spaces]].
Let $f : X \to Y$ be a [[Definition:Homeomorphism (Topological Spaces)|homeomorphism]].
Let $A \subseteq X$.
Then:
:$\paren {f \sqbrk A}^\circ = f \sqbrk {A^\circ}$
where:
:$A^\circ$ is the [[Definition:Interior (Topology)|interior]] of $A$ in ... | First, we show that $f \sqbrk {A^\circ} \subseteq \paren {f \sqbrk A}^\circ$.
Let $y \in f \sqbrk {A^\circ}$.
Then there exists $x \in A^\circ$ such that $y = \map f x$.
Since $x \in A^\circ$, there exists an [[Definition:Open Set|open set]] $U \subseteq A$ in $X$ such that $x \in U$.
Then $y = \map f x \in f \sqb... | Image of Interior of Set under Homeomorphism is Interior of Image | https://proofwiki.org/wiki/Image_of_Interior_of_Set_under_Homeomorphism_is_Interior_of_Image | https://proofwiki.org/wiki/Image_of_Interior_of_Set_under_Homeomorphism_is_Interior_of_Image | [
"Homeomorphisms (Topological Spaces)",
"Set Interiors"
] | [
"Definition:Topological Space",
"Definition:Homeomorphism/Topological Spaces",
"Definition:Interior (Topology)",
"Definition:Interior (Topology)"
] | [
"Definition:Open Set",
"Definition:Homeomorphism/Topological Spaces",
"Definition:Open Mapping",
"Definition:Open Set",
"Definition:Interior (Topology)",
"Definition:Open Set",
"Preimage of Image of Subset under Injection equals Subset",
"Image of Subset under Mapping is Subset of Image",
"Definitio... |
proofwiki-21036 | Homeomorphic Image of Nowhere Dense Set is Nowhere Dense | Let $X$ and $Y$ be topological spaces.
Let $f : X \to Y$ be a homeomorphism.
Let $A$ be a nowhere dense subset of $X$.
Then $f \sqbrk A$ is a nowhere dense subset of $Y$. | Since $A$ is nowhere dense, we have:
:$\paren {A^-}^\circ = \O$
where $A^-$ is the closure of $A$ and $\paren {A^-}^\circ$ is the interior of $A^-$.
From Image of Empty Set is Empty Set, we have:
:$\O = f \sqbrk {\paren {A^-}^\circ}$
So, we have:
{{begin-eqn}}
{{eqn | l = f \sqbrk {\paren {A^-}^\circ}
| r = \paren {f... | Let $X$ and $Y$ be [[Definition:Topological Space|topological spaces]].
Let $f : X \to Y$ be a [[Definition:Homeomorphism (Topological Spaces)|homeomorphism]].
Let $A$ be a [[Definition:Nowhere Dense|nowhere dense]] subset of $X$.
Then $f \sqbrk A$ is a [[Definition:Nowhere Dense|nowhere dense]] subset of $Y$. | Since $A$ is [[Definition:Nowhere Dense|nowhere dense]], we have:
:$\paren {A^-}^\circ = \O$
where $A^-$ is the [[Definition:Topological Closure|closure]] of $A$ and $\paren {A^-}^\circ$ is the [[Definition:Interior (Topology)|interior]] of $A^-$.
From [[Image of Empty Set is Empty Set]], we have:
:$\O = f \sqbrk {\pa... | Homeomorphic Image of Nowhere Dense Set is Nowhere Dense | https://proofwiki.org/wiki/Homeomorphic_Image_of_Nowhere_Dense_Set_is_Nowhere_Dense | https://proofwiki.org/wiki/Homeomorphic_Image_of_Nowhere_Dense_Set_is_Nowhere_Dense | [
"Nowhere Dense",
"Homeomorphisms (Topological Spaces)"
] | [
"Definition:Topological Space",
"Definition:Homeomorphism/Topological Spaces",
"Definition:Nowhere Dense",
"Definition:Nowhere Dense"
] | [
"Definition:Nowhere Dense",
"Definition:Closure (Topology)",
"Definition:Interior (Topology)",
"Image of Empty Set is Empty Set",
"Image of Interior of Set under Homeomorphism is Interior of Image",
"Homeomorphism iff Image of Closure equals Closure of Image",
"Definition:Nowhere Dense",
"Category:Now... |
proofwiki-21037 | Cantor Pairing Function is Primitive Recursive | The Cantor pairing function is primitive recursive. | The Cantor pairing function $\pi : \N^2 \to \N$ is defined as:
:$\map \pi {m, n} = \frac 1 2 \paren {m + n} \paren {m + n + 1} + m$
As Cantor Pairing Function is Well-Defined, the function could also be defined as:
:$\map \pi {m, n} = \map {\operatorname{quot}} {\paren {m + n} \paren {m + n + 1}, 2} + m$
which is primi... | The [[Definition:Cantor Pairing Function|Cantor pairing function]] is [[Definition:Primitive Recursive Function|primitive recursive]]. | The [[Definition:Cantor Pairing Function|Cantor pairing function]] $\pi : \N^2 \to \N$ is defined as:
:$\map \pi {m, n} = \frac 1 2 \paren {m + n} \paren {m + n + 1} + m$
As [[Cantor Pairing Function is Well-Defined]], the function could also be defined as:
:$\map \pi {m, n} = \map {\operatorname{quot}} {\paren {m + n... | Cantor Pairing Function is Primitive Recursive | https://proofwiki.org/wiki/Cantor_Pairing_Function_is_Primitive_Recursive | https://proofwiki.org/wiki/Cantor_Pairing_Function_is_Primitive_Recursive | [
"Cantor Pairing Function"
] | [
"Definition:Cantor Pairing Function",
"Definition:Primitive Recursive/Function"
] | [
"Definition:Cantor Pairing Function",
"Cantor Pairing Function is Well-Defined",
"Definition:Primitive Recursive/Function",
"Definition:Substitution (Mathematical Logic)",
"Addition is Primitive Recursive",
"Quotient is Primitive Recursive",
"Constant Function is Primitive Recursive",
"Category:Cantor... |
proofwiki-21038 | Generalized Sum with Countable Non-zero Summands/Corollary | Let $\set{j_0, j_1, j_2, \ldots}$ and $\set{k_0, k_1, k_2, \ldots}$ be countably infinite enumerations of $J$ and $K$ respectively.
Let $r \in \R_{\mathop > 0}$.
Then:
:the series $\ds \sum_{n \mathop = 1}^\infty v_{j_n}$ converges absolutely to $r$
{{iff}}:
:the series $\ds \sum_{n \mathop = 1}^\infty v_{k_n}$ converg... | From Generalized Sum with Countable Non-zero Summands:
:the series $\ds \sum_{n \mathop = 1}^\infty v_{j_n}$ converges absolutely to $r$
{{iff}}
:the generalized sum $\ds \sum_{i \mathop \in I} v_i$ converges absolutely to $r$
{{iff}}:
:the series $\ds \sum_{n \mathop = 1}^\infty v_{k_n}$ converges absolutely to $r$
{{... | Let $\set{j_0, j_1, j_2, \ldots}$ and $\set{k_0, k_1, k_2, \ldots}$ be [[Definition:Countably Infinite Enumeration|countably infinite enumerations]] of $J$ and $K$ respectively.
Let $r \in \R_{\mathop > 0}$.
Then:
:the [[Definition:Series|series]] $\ds \sum_{n \mathop = 1}^\infty v_{j_n}$ [[Definition:Absolutely Co... | From [[Generalized Sum with Countable Non-zero Summands]]:
:the [[Definition:Series|series]] $\ds \sum_{n \mathop = 1}^\infty v_{j_n}$ [[Definition:Absolutely Convergent Series|converges absolutely]] to $r$
{{iff}}
:the [[Definition:Generalized Sum|generalized sum]] $\ds \sum_{i \mathop \in I} v_i$ [[Definition:Absolut... | Generalized Sum with Countable Non-zero Summands/Corollary | https://proofwiki.org/wiki/Generalized_Sum_with_Countable_Non-zero_Summands/Corollary | https://proofwiki.org/wiki/Generalized_Sum_with_Countable_Non-zero_Summands/Corollary | [
"Generalized Sum with Countable Non-zero Summands"
] | [
"Definition:Enumeration/Countably Infinite",
"Definition:Series",
"Definition:Absolutely Convergent Series",
"Definition:Series",
"Definition:Absolutely Convergent Series"
] | [
"Generalized Sum with Countable Non-zero Summands",
"Definition:Series",
"Definition:Absolutely Convergent Series",
"Definition:Generalized Sum",
"Definition:Generalized Sum/Absolute Net Convergence",
"Definition:Series",
"Definition:Absolutely Convergent Series",
"Category:Generalized Sum with Counta... |
proofwiki-21039 | Length of Basis Representation is Primitive Recursive | Define $\operatorname{basislen} : \N^2 \to \N$ as:
:$\map {\operatorname{basislen}} {b, n} = \begin{cases}m + 1 & : b > 1 \land n > 0 \\n & : b = 1 \\0 & : b = 0 \lor n = 0\end{cases}$
where $\sqbrk {r_m r_{m - 1} \dotsm r_1 r_0}_b$ is the base-$b$ representation of $n$.
Then $\operatorname{basislen}$ is primitive recu... | By:
* Definition by Cases is Primitive Recursive
* Ordering Relations are Primitive Recursive
* Equality Relation is Primitive Recursive
* Constant Function is Primitive Recursive
* Set Operations on Primitive Recursive Relations
all that needs to be shown is that $m + 1$ can be computed from $n$ in the case that $b > ... | Define $\operatorname{basislen} : \N^2 \to \N$ as:
:$\map {\operatorname{basislen}} {b, n} = \begin{cases}m + 1 & : b > 1 \land n > 0 \\n & : b = 1 \\0 & : b = 0 \lor n = 0\end{cases}$
where $\sqbrk {r_m r_{m - 1} \dotsm r_1 r_0}_b$ is the [[Definition:Basis Representation|base-$b$ representation]] of $n$.
Then $\ope... | By:
* [[Definition by Cases is Primitive Recursive]]
* [[Ordering Relations are Primitive Recursive]]
* [[Equality Relation is Primitive Recursive]]
* [[Constant Function is Primitive Recursive]]
* [[Set Operations on Primitive Recursive Relations]]
all that needs to be shown is that $m + 1$ can be computed from $n$ in... | Length of Basis Representation is Primitive Recursive | https://proofwiki.org/wiki/Length_of_Basis_Representation_is_Primitive_Recursive | https://proofwiki.org/wiki/Length_of_Basis_Representation_is_Primitive_Recursive | [
"Basis Representations",
"Primitive Recursive Functions"
] | [
"Definition:Basis Representation",
"Definition:Primitive Recursive/Function"
] | [
"Definition by Cases is Primitive Recursive",
"Ordering Relations are Primitive Recursive",
"Equality Relation is Primitive Recursive",
"Constant Function is Primitive Recursive",
"Set Operations on Primitive Recursive Relations",
"Definition:Basis Representation",
"Exponentiation is Primitive Recursive... |
proofwiki-21040 | Basis Representation is No Longer than Number | Let $b > 1$ and $n > 0$ be natural numbers.
Let the base-$b$ representation of $n$ be:
:$\sqbrk {r_m r_{m - 1} \dotsm r_1 r_0}_b$
Then:
:$n > m$ | {{Tidy}}
;Base case: $(n=1)$
{{begin-eqn}}
{{eqn | l = \sqbrk 1_b
| r = b^0
| c = From Basis Representation Theorem
}}
{{eqn | r = 1
| c = From Zeroth Power of Real Number equals One
}}
{{end-eqn}}
;Inductive case: $(k\implies k+1)$
;; Subcase 1: $\exists i \in \set {0, \dots, m}, j < i \implies r_j =... | Let $b > 1$ and $n > 0$ be [[Definition:Natural Number|natural numbers]].
Let the [[Definition:Basis Representation|base-$b$ representation]] of $n$ be:
:$\sqbrk {r_m r_{m - 1} \dotsm r_1 r_0}_b$
Then:
:$n > m$ | {{Tidy}}
;Base case: $(n=1)$
{{begin-eqn}}
{{eqn | l = \sqbrk 1_b
| r = b^0
| c = From [[Basis Representation Theorem]]
}}
{{eqn | r = 1
| c = From [[Zeroth Power of Real Number equals One]]
}}
{{end-eqn}}
;Inductive case: $(k\implies k+1)$
;; Subcase 1: $\exists i \in \set {0, \dots, m}, j < i \im... | Basis Representation is No Longer than Number | https://proofwiki.org/wiki/Basis_Representation_is_No_Longer_than_Number | https://proofwiki.org/wiki/Basis_Representation_is_No_Longer_than_Number | [
"Number Bases"
] | [
"Definition:Natural Numbers",
"Definition:Basis Representation"
] | [
"Basis Representation Theorem",
"Zeroth Power of Real Number equals One",
"Basis Representation Theorem",
"Telescoping Series/Example 2",
"Zeroth Power of Real Number equals One",
"Basis Representation Theorem",
"Definition:Basis Representation",
"Basis Representation Theorem",
"Telescoping Series/E... |
proofwiki-21041 | Banach-Steinhaus Theorem/Topological Vector Space | Let $\GF \in \set {\R, \C}$.
Let $X$ and $Y$ be topological vector spaces over $\GF$.
Let $\Gamma$ be a set of continuous linear transformations $X \to Y$.
Let $B$ be the set of all $x \in X$ such that:
:$\map \Gamma x = \set {T x : T \in \Gamma}$
is von Neumann-bounded in $Y$.
Suppose that $B$ is not meager in $X$.
Th... | Let $W$ be an open neighborhood of ${\mathbf 0}_Y$.
From Open Neighborhood of Point in Topological Vector Space contains Sum of Open Neighborhoods, there exists an open neighborhood $V$ of ${\mathbf 0}_Y$ such that:
:$V + V \subseteq W$
From Disjoint Compact Set and Closed Set in Topological Vector Space separated by O... | Let $\GF \in \set {\R, \C}$.
Let $X$ and $Y$ be [[Definition:Topological Vector Space|topological vector spaces]] over $\GF$.
Let $\Gamma$ be a [[Definition:Set|set]] of [[Definition:Continuous Mapping|continuous]] [[Definition:Linear Transformation|linear transformations]] $X \to Y$.
Let $B$ be the [[Definition:Set... | Let $W$ be an [[Definition:Open Neighborhood|open neighborhood]] of ${\mathbf 0}_Y$.
From [[Open Neighborhood of Point in Topological Vector Space contains Sum of Open Neighborhoods]], there exists an [[Definition:Open Neighborhood|open neighborhood]] $V$ of ${\mathbf 0}_Y$ such that:
:$V + V \subseteq W$
From [[Disj... | Banach-Steinhaus Theorem/Topological Vector Space | https://proofwiki.org/wiki/Banach-Steinhaus_Theorem/Topological_Vector_Space | https://proofwiki.org/wiki/Banach-Steinhaus_Theorem/Topological_Vector_Space | [
"Banach-Steinhaus Theorem",
"Topological Vector Spaces"
] | [
"Definition:Topological Vector Space",
"Definition:Set",
"Definition:Continuous Mapping",
"Definition:Linear Transformation",
"Definition:Set",
"Definition:Von Neumann-Bounded Subset of Topological Vector Space",
"Definition:Meager Space",
"Definition:Equicontinuous Family of Linear Transformations be... | [
"Definition:Open Neighborhood",
"Open Neighborhood of Point in Topological Vector Space contains Sum of Open Neighborhoods",
"Definition:Open Neighborhood",
"Disjoint Compact Set and Closed Set in Topological Vector Space separated by Open Neighborhood/Corollary",
"Definition:Open Neighborhood",
"Definiti... |
proofwiki-21042 | Countable Union of Meager Sets is Meager | Let $\struct {X, \tau}$ be a topological space.
Let $\family {A_n}_{n \in \N}$ be a countable set of meager subsets of $X$.
Let:
:$\ds A = \bigcup_{n \mathop = 1}^\infty A_n$
Then $A$ is meager in $X$. | For each $n \in \N$, $A_n$ is meager and hence there exists a countable set $\family {A_{n, m} }_{m \in \N}$ of nowhere dense sets in $X$ such that:
:$\ds A_n = \bigcup_{m \mathop = 1}^\infty A_{n, m}$
Hence:
:$\ds A = \bigcup_{n \mathop = 1}^\infty \bigcup_{m \mathop = 1}^\infty A_{n, m} = \bigcup_{\tuple {n, m} \in \... | Let $\struct {X, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $\family {A_n}_{n \in \N}$ be a [[Definition:Countable Set|countable set]] of [[Definition:Meager Space|meager subsets]] of $X$.
Let:
:$\ds A = \bigcup_{n \mathop = 1}^\infty A_n$
Then $A$ is [[Definition:Meager Space|meager]] in ... | For each $n \in \N$, $A_n$ is [[Definition:Meager Space|meager]] and hence there exists a [[Definition:Countable Set|countable set]] $\family {A_{n, m} }_{m \in \N}$ of [[Definition:Nowhere Dense|nowhere dense sets]] in $X$ such that:
:$\ds A_n = \bigcup_{m \mathop = 1}^\infty A_{n, m}$
Hence:
:$\ds A = \bigcup_{n \ma... | Countable Union of Meager Sets is Meager | https://proofwiki.org/wiki/Countable_Union_of_Meager_Sets_is_Meager | https://proofwiki.org/wiki/Countable_Union_of_Meager_Sets_is_Meager | [
"Meager Spaces",
"Set Union"
] | [
"Definition:Topological Space",
"Definition:Countable Set",
"Definition:Meager Space",
"Definition:Meager Space"
] | [
"Definition:Meager Space",
"Definition:Countable Set",
"Definition:Nowhere Dense",
"Cartesian Product of Countable Sets is Countable",
"Definition:Set Union/Countable Union",
"Definition:Nowhere Dense",
"Countable Union of Countable Sets is Countable",
"Definition:Meager Space"
] |
proofwiki-21043 | Join Semilattice Ideal iff Ordered Set Ideal | Let $\struct {S, \vee, \preceq}$ be a join semilattice.
Let $I \subseteq S$ be a non-empty subset of $S$.
Then:
:$I$ is a join semilattice ideal of $\struct {S, \vee, \preceq}$ {{iff}} $I$ is an ordered set ideal of $\struct {S, \preceq}$. | === Necessary Condition ===
Let $I$ be a join semilattice ideal of $\struct {S, \vee, \preceq}$.
To show that $I$ is an ordered set ideal of $\struct {S, \preceq}$ it is sufficient to show:
{{begin-axiom}}
{{axiom | lc= $I$ is a directed subset of $S$:
| q = \forall x, y \in I: \exists z \in I
| m = x \... | Let $\struct {S, \vee, \preceq}$ be a [[Definition:Join Semilattice|join semilattice]].
Let $I \subseteq S$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subset]] of $S$.
Then:
:$I$ is a [[Definition:Join Semilattice Ideal|join semilattice ideal]] of $\struct {S, \vee, \preceq}$ {{iff}} $I$ is an [... | === Necessary Condition ===
Let $I$ be a [[Definition:Join Semilattice Ideal|join semilattice ideal]] of $\struct {S, \vee, \preceq}$.
To show that $I$ is an [[Definition:Ideal (Order Theory)|ordered set ideal]] of $\struct {S, \preceq}$ it is sufficient to show:
{{begin-axiom}}
{{axiom | lc= $I$ is a [[Definition:Di... | Join Semilattice Ideal iff Ordered Set Ideal | https://proofwiki.org/wiki/Join_Semilattice_Ideal_iff_Ordered_Set_Ideal | https://proofwiki.org/wiki/Join_Semilattice_Ideal_iff_Ordered_Set_Ideal | [
"Join Semilattices"
] | [
"Definition:Join Semilattice",
"Definition:Non-Empty Set",
"Definition:Subset",
"Definition:Join Semilattice Ideal",
"Definition:Ideal (Order Theory)"
] | [
"Definition:Join Semilattice Ideal",
"Definition:Ideal (Order Theory)",
"Definition:Directed Subset",
"Definition:Join Semilattice Ideal",
"Definition:Subsemilattice",
"Definition:Join (Order Theory)",
"Definition:Ideal (Order Theory)",
"Definition:Join Semilattice Ideal",
"Definition:Subsemilattice... |
proofwiki-21044 | Equivalence of Definitions of Lattice Ideal | Let $\struct {S, \vee, \wedge, \preceq}$ be a lattice.
Let $I \subseteq S$ be a non-empty subset of $S$.
{{TFAE|def = Lattice Ideal}} | === Definition 1 implies Definition 2 ===
Let $I$ satisify the lattice ideal axioms.
To show that $I$ is a join semilattice ideal it is sufficient to show:
{{begin-axiom}}
{{axiom | lc= $I$ is a lower section of $S$:
| q = \forall x \in F: \forall y \in S
| m = y \preceq x \implies y \in I
}}
{{end-axio... | Let $\struct {S, \vee, \wedge, \preceq}$ be a [[Definition:Lattice (Order Theory)|lattice]].
Let $I \subseteq S$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subset]] of $S$.
{{TFAE|def = Lattice Ideal}} | === Definition 1 implies Definition 2 ===
Let $I$ satisify the [[Axiom:Lattice Ideal Axioms|lattice ideal axioms]].
To show that $I$ is a [[Definition:Join Semilattice Ideal|join semilattice ideal]] it is sufficient to show:
{{begin-axiom}}
{{axiom | lc= $I$ is a [[Definition:Lower Section|lower section]] of $S$:
... | Equivalence of Definitions of Lattice Ideal | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Lattice_Ideal | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Lattice_Ideal | [
"Lattice Ideals"
] | [
"Definition:Lattice (Order Theory)",
"Definition:Non-Empty Set",
"Definition:Subset"
] | [
"Axiom:Lattice Ideal Axioms",
"Definition:Join Semilattice Ideal",
"Definition:Lower Section",
"Axiom:Lattice Ideal Axioms",
"Definition:Sublattice",
"Preceding iff Meet equals Less Operand",
"Definition:Join Semilattice Ideal",
"Definition:Join Semilattice Ideal",
"Definition:Lower Section"
] |
proofwiki-21045 | Finite Subset of Topological Vector Space is von Neumann-Bounded | Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \tau}$ be a topological vector space over $\GF$.
Let $F$ be a finite subset of $X$.
Then $F$ is von Neumann-bounded. | Let $F = \set {v_1, \ldots, v_m}$.
Let $U$ be an open neighborhood of ${\mathbf 0}_X$.
From Multiple of Vector in Topological Vector Space Converges, we have:
:$\ds \frac {v_i} n \to {\mathbf 0}_X$ as $n \to \infty$
for each $1 \le i \le n$.
Hence for each $1 \le i \le n$, there exists $N_i \in \N$ such that:
:$\ds \f... | Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \tau}$ be a [[Definition:Topological Vector Space|topological vector space]] over $\GF$.
Let $F$ be a [[Definition:Finite Set|finite]] [[Definition:Subset|subset]] of $X$.
Then $F$ is [[Definition:Von Neumann-Bounded Subset of Topological Vector Space|von Neumann-bou... | Let $F = \set {v_1, \ldots, v_m}$.
Let $U$ be an [[Definition:Open Neighborhood|open neighborhood]] of ${\mathbf 0}_X$.
From [[Multiple of Vector in Topological Vector Space Converges]], we have:
:$\ds \frac {v_i} n \to {\mathbf 0}_X$ as $n \to \infty$
for each $1 \le i \le n$.
Hence for each $1 \le i \le n$, there... | Finite Subset of Topological Vector Space is von Neumann-Bounded | https://proofwiki.org/wiki/Finite_Subset_of_Topological_Vector_Space_is_von_Neumann-Bounded | https://proofwiki.org/wiki/Finite_Subset_of_Topological_Vector_Space_is_von_Neumann-Bounded | [
"Von Neumann-Bounded Subsets of Topological Vector Spaces"
] | [
"Definition:Topological Vector Space",
"Definition:Finite Set",
"Definition:Subset",
"Definition:Von Neumann-Bounded Subset of Topological Vector Space"
] | [
"Definition:Open Neighborhood",
"Multiple of Vector in Topological Vector Space Converges",
"Definition:Open Neighborhood",
"Definition:Von Neumann-Bounded Subset of Topological Vector Space",
"Category:Von Neumann-Bounded Subsets of Topological Vector Spaces"
] |
proofwiki-21046 | Preimage of Dilation of Set under Linear Transformation is Dilation of Preimage | Let $K$ be a field.
Let $X$ and $Y$ be vector spaces over $K$.
Let $T : X \to Y$ be a linear transformation.
Let $E \subseteq X$ be a non-empty set.
Let $\lambda \in K$.
Then:
:$T^{-1} \sqbrk {\lambda E} = \lambda T^{-1} \sqbrk E$
where $\lambda E$ denotes the dilation of $E$ by $\lambda$. | The result is immediate when $\lambda = 0_K$, since $T^{-1} \sqbrk {\set { {\mathbf 0}_X} } = \set { {\mathbf 0}_X}$.
Now take $\lambda \ne 0_K$.
Let $x \in X$.
We have:
:$x \in T^{-1} \sqbrk {\lambda E}$
{{iff}}:
:$T x \in \lambda E$
From linearity, this is the case {{iff}}:
:$\map T {\lambda^{-1} x} \in E$
This is eq... | Let $K$ be a [[Definition:Field (Abstract Algebra)|field]].
Let $X$ and $Y$ be [[Definition:Vector Space|vector spaces]] over $K$.
Let $T : X \to Y$ be a [[Definition:Linear Transformation|linear transformation]].
Let $E \subseteq X$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Set|set]].
Let $\lambda ... | The result is immediate when $\lambda = 0_K$, since $T^{-1} \sqbrk {\set { {\mathbf 0}_X} } = \set { {\mathbf 0}_X}$.
Now take $\lambda \ne 0_K$.
Let $x \in X$.
We have:
:$x \in T^{-1} \sqbrk {\lambda E}$
{{iff}}:
:$T x \in \lambda E$
From [[Definition:Linear Transformation|linearity]], this is the case {{iff}}:
:$... | Preimage of Dilation of Set under Linear Transformation is Dilation of Preimage | https://proofwiki.org/wiki/Preimage_of_Dilation_of_Set_under_Linear_Transformation_is_Dilation_of_Preimage | https://proofwiki.org/wiki/Preimage_of_Dilation_of_Set_under_Linear_Transformation_is_Dilation_of_Preimage | [
"Dilations of Subsets of Vector Spaces"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Vector Space",
"Definition:Linear Transformation",
"Definition:Non-Empty Set",
"Definition:Set",
"Definition:Linear Combination of Subsets of Vector Space/Dilation"
] | [
"Definition:Linear Transformation",
"Category:Dilations of Subsets of Vector Spaces"
] |
proofwiki-21047 | Homeomorphic Image of Meager Set is Meager | Let $X$ and $Y$ be topological spaces.
Let $f : X \to Y$ be a homeomorphism.
Let $A \subseteq X$ be meager in $X$.
Then $f \sqbrk A$ is meager in $Y$. | Since $A$ is meager in $X$, there exists a countable set $\family {A_n}_{n \in \N}$ of nowhere dense sets in $X$ such that:
:$\ds A = \bigcup_{n \mathop = 1}^\infty A_n$
So, we have:
:$\ds f \sqbrk A = f \sqbrk {\bigcup_{n \mathop = 1}^\infty A_n}$
So, from Image of Union under Mapping:
:$\ds f \sqbrk A = \bigcup_{n \m... | Let $X$ and $Y$ be [[Definition:Topological Space|topological spaces]].
Let $f : X \to Y$ be a [[Definition:Homeomorphism|homeomorphism]].
Let $A \subseteq X$ be [[Definition:Meager Space|meager]] in $X$.
Then $f \sqbrk A$ is [[Definition:Meager Space|meager]] in $Y$. | Since $A$ is [[Definition:Meager Space|meager]] in $X$, there exists a [[Definition:Countable Set|countable set]] $\family {A_n}_{n \in \N}$ of [[Definition:Nowhere Dense|nowhere dense]] sets in $X$ such that:
:$\ds A = \bigcup_{n \mathop = 1}^\infty A_n$
So, we have:
:$\ds f \sqbrk A = f \sqbrk {\bigcup_{n \mathop = ... | Homeomorphic Image of Meager Set is Meager | https://proofwiki.org/wiki/Homeomorphic_Image_of_Meager_Set_is_Meager | https://proofwiki.org/wiki/Homeomorphic_Image_of_Meager_Set_is_Meager | [
"Meager Spaces"
] | [
"Definition:Topological Space",
"Definition:Homeomorphism",
"Definition:Meager Space",
"Definition:Meager Space"
] | [
"Definition:Meager Space",
"Definition:Countable Set",
"Definition:Nowhere Dense",
"Image of Union under Mapping",
"Homeomorphic Image of Nowhere Dense Set is Nowhere Dense",
"Definition:Nowhere Dense",
"Definition:Set Union/Countable Union",
"Definition:Nowhere Dense",
"Definition:Meager Space",
... |
proofwiki-21048 | Banach-Steinhaus Theorem/F-Space | Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \tau_X}$ be an $F$-Space over $\GF$.
Let $\struct {Y, \tau_Y}$ be a topological vector space over $\GF$.
Let $\Gamma$ be a set of continuous linear transformations $X \to Y$ such that for all $x \in X$:
:$\map \Gamma x = \set {T x : T \in \Gamma}$ is von Neumann-bounded in ... | Let $B$ be the set of $x \in X$ such that:
:$\map \Gamma x = \set {T x : T \in \Gamma}$ is von Neumann-bounded in $Y$.
By hypothesis, $B = X$.
Let $d$ be a metric inducing $\tau_X$ such that $\struct {X, d}$ is a complete metric space.
From the Baire Category Theorem, $\struct {X, d}$ is a Baire space.
From Baire Spa... | Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \tau_X}$ be an [[Definition:F-Space|$F$-Space]] over $\GF$.
Let $\struct {Y, \tau_Y}$ be a [[Definition:Topological Vector Space|topological vector space]] over $\GF$.
Let $\Gamma$ be a [[Definition:Set|set]] of [[Definition:Continuous Mapping|continuous]] [[Definition:L... | Let $B$ be the set of $x \in X$ such that:
:$\map \Gamma x = \set {T x : T \in \Gamma}$ is [[Definition:Von Neumann-Bounded Subset of Topological Vector Space|von Neumann-bounded]] in $Y$.
By hypothesis, $B = X$.
Let $d$ be a [[Definition:Metric|metric]] [[Definition:Topology Induced by Metric|inducing]] $\tau_X$ su... | Banach-Steinhaus Theorem/F-Space | https://proofwiki.org/wiki/Banach-Steinhaus_Theorem/F-Space | https://proofwiki.org/wiki/Banach-Steinhaus_Theorem/F-Space | [
"F-Spaces",
"Banach-Steinhaus Theorem"
] | [
"Definition:F-Space",
"Definition:Topological Vector Space",
"Definition:Set",
"Definition:Continuous Mapping",
"Definition:Linear Transformation",
"Definition:Von Neumann-Bounded Subset of Topological Vector Space",
"Definition:Equicontinuous Family of Linear Transformations between Topological Vector ... | [
"Definition:Von Neumann-Bounded Subset of Topological Vector Space",
"Definition:Metric Space/Metric",
"Definition:Topology Induced by Metric",
"Definition:Complete Metric Space",
"Baire Category Theorem",
"Definition:Baire Space",
"Baire Space is Non-Meager",
"Definition:Meager Space",
"Definition:... |
proofwiki-21049 | Norm Topology Induced by Metric Induced by Norm | Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \norm {\, \cdot \,} }$ be a normed vector space over $\GF$.
Let $d$ be the metric induced on $X$ by $\norm {\, \cdot \,}$.
Then the open sets of $\struct {X, d}$ as a metric space are precisely the open sets of $\struct {X, \norm {\, \cdot \,} }$ as a normed vector space. | For $\epsilon > 0$ and $x \in X$, let $\map {B_\epsilon^d} x$ be the open ball in $\struct {X, d}$ with radius $\epsilon$ and center $x$.
For $\epsilon > 0$ and $x \in X$, let $\map {B_\epsilon^{\norm {\, \cdot \,} } } x$ be the open ball in $\struct {X, \norm {\, \cdot \,} }$ with radius $\epsilon$ and center $x$.
{... | Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \norm {\, \cdot \,} }$ be a [[Definition:Normed Vector Space|normed vector space]] over $\GF$.
Let $d$ be the [[Definition:Metric Induced by Norm|metric induced]] on $X$ by $\norm {\, \cdot \,}$.
Then the [[Definition:Open Set of Metric Space|open sets]] of $\struct {X... | For $\epsilon > 0$ and $x \in X$, let $\map {B_\epsilon^d} x$ be the [[Definition:Open Ball|open ball]] in $\struct {X, d}$ with [[Definition:Radius of Open Ball|radius]] $\epsilon$ and [[Definition:Center of Open Ball|center]] $x$.
For $\epsilon > 0$ and $x \in X$, let $\map {B_\epsilon^{\norm {\, \cdot \,} } } x$ b... | Norm Topology Induced by Metric Induced by Norm | https://proofwiki.org/wiki/Norm_Topology_Induced_by_Metric_Induced_by_Norm | https://proofwiki.org/wiki/Norm_Topology_Induced_by_Metric_Induced_by_Norm | [
"Normed Vector Spaces",
"Metric Spaces"
] | [
"Definition:Normed Vector Space",
"Definition:Metric Induced by Norm",
"Definition:Open Set/Metric Space",
"Definition:Metric Space",
"Definition:Open Set/Normed Vector Space",
"Definition:Normed Vector Space"
] | [
"Definition:Open Ball",
"Definition:Open Ball/Radius",
"Definition:Open Ball/Center",
"Definition:Open Ball",
"Definition:Open Ball/Radius",
"Definition:Open Ball/Center",
"Category:Normed Vector Spaces",
"Category:Metric Spaces"
] |
proofwiki-21050 | Banach Space is F-Space | Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \norm {\, \cdot \,} }$ be a Banach space over $\GF$.
From Normed Vector Space is Hausdorff Topological Vector Space, we may consider $\struct {X, \norm {\, \cdot \,} }$ as a topological vector space.
With this identification, $\struct {X, \norm {\, \cdot \,} }$ is an $F$-... | Let $d$ be the metric induced by $\norm {\, \cdot \,}$.
From Norm Topology Induced by Metric Induced by Norm, $d$ induces the topology on $\struct {X, \norm {\, \cdot \,} }$.
Since $\struct {X, \norm {\, \cdot \,} }$ is a Banach space, $\struct {X, d}$ is a complete metric space.
From Metric Induced by Norm is Invaria... | Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \norm {\, \cdot \,} }$ be a [[Definition:Banach Space|Banach space]] over $\GF$.
From [[Normed Vector Space is Hausdorff Topological Vector Space]], we may consider $\struct {X, \norm {\, \cdot \,} }$ as a [[Definition:Topological Vector Space|topological vector space]]... | Let $d$ be the [[Definition:Metric Induced by Norm|metric induced by $\norm {\, \cdot \,}$]].
From [[Norm Topology Induced by Metric Induced by Norm]], $d$ induces the [[Definition:Topology|topology]] on $\struct {X, \norm {\, \cdot \,} }$.
Since $\struct {X, \norm {\, \cdot \,} }$ is a [[Definition:Banach Space|Bana... | Banach Space is F-Space | https://proofwiki.org/wiki/Banach_Space_is_F-Space | https://proofwiki.org/wiki/Banach_Space_is_F-Space | [
"F-Spaces",
"Banach Spaces"
] | [
"Definition:Banach Space",
"Normed Vector Space is Hausdorff Topological Vector Space",
"Definition:Topological Vector Space",
"Definition:F-Space"
] | [
"Definition:Metric Induced by Norm",
"Norm Topology Induced by Metric Induced by Norm",
"Definition:Topology",
"Definition:Banach Space",
"Definition:Complete Metric Space",
"Metric Induced by Norm is Invariant Metric",
"Definition:Invariant Metric on Vector Space",
"Definition:F-Space",
"Category:F... |
proofwiki-21051 | Proper Closed Linear Subspace of Topological Vector Space is Meager | Let $\GF \in \set {\R, \C}$.
Let $X$ be a topological vector space over $\GF$.
Let $D$ be a proper closed linear subspace of $X$.
Then $D$ is meager. | From Set is Closed iff Equals Topological Closure, we have $D^- = D$.
From Proper Linear Subspace of Topological Vector Space has Empty Interior, we then have that $\paren {D^-}^\circ = D^\circ = \O$.
Hence $D$ is nowhere dense, and in particular meager.
{{qed}}
Category:Meager Spaces
Category:Topological Vector Spac... | Let $\GF \in \set {\R, \C}$.
Let $X$ be a [[Definition:Topological Vector Space|topological vector space]] over $\GF$.
Let $D$ be a [[Definition:Proper Subset|proper]] [[Definition:Closed Linear Subspace|closed linear subspace]] of $X$.
Then $D$ is [[Definition:Meager Space|meager]]. | From [[Set is Closed iff Equals Topological Closure]], we have $D^- = D$.
From [[Proper Linear Subspace of Topological Vector Space has Empty Interior]], we then have that $\paren {D^-}^\circ = D^\circ = \O$.
Hence $D$ is [[Definition:Nowhere Dense|nowhere dense]], and in particular [[Definition:Meager Space|meager... | Proper Closed Linear Subspace of Topological Vector Space is Meager | https://proofwiki.org/wiki/Proper_Closed_Linear_Subspace_of_Topological_Vector_Space_is_Meager | https://proofwiki.org/wiki/Proper_Closed_Linear_Subspace_of_Topological_Vector_Space_is_Meager | [
"Meager Spaces",
"Topological Vector Spaces"
] | [
"Definition:Topological Vector Space",
"Definition:Proper Subset",
"Definition:Closed Linear Subspace",
"Definition:Meager Space"
] | [
"Set is Closed iff Equals Topological Closure",
"Proper Linear Subspace of Topological Vector Space has Empty Interior",
"Definition:Nowhere Dense",
"Definition:Meager Space",
"Category:Meager Spaces",
"Category:Topological Vector Spaces"
] |
proofwiki-21052 | Proper Linear Subspace of Topological Vector Space has Empty Interior | Let $\GF \in \set {\R, \C}$.
Let $X$ be a topological vector space over $\GF$.
Let $L$ be a proper linear subspace of $X$.
Then:
:$L^\circ = \O$
where $L^\circ$ denotes the interior of $L$. | {{AimForCont}} that $L^\circ \ne \O$.
Let $x \in L^\circ$.
Then there exists an open neighborhood $U$ of $x$ such that $U \subseteq L$.
From Translation of Open Set in Topological Vector Space is Open, $U - x$ is an open neighborhood of ${\mathbf 0}_X$ with $U - x \subseteq L - x$.
Since $L$ is a linear subspace of $... | Let $\GF \in \set {\R, \C}$.
Let $X$ be a [[Definition:Topological Vector Space|topological vector space]] over $\GF$.
Let $L$ be a [[Definition:Proper Subset|proper]] [[Definition:Linear Subspace|linear subspace]] of $X$.
Then:
:$L^\circ = \O$
where $L^\circ$ denotes the [[Definition:Interior (Topology)|interior]... | {{AimForCont}} that $L^\circ \ne \O$.
Let $x \in L^\circ$.
Then there exists an [[Definition:Open Neighborhood|open neighborhood]] $U$ of $x$ such that $U \subseteq L$.
From [[Translation of Open Set in Topological Vector Space is Open]], $U - x$ is an [[Definition:Open Neighborhood|open neighborhood]] of ${\mathb... | Proper Linear Subspace of Topological Vector Space has Empty Interior | https://proofwiki.org/wiki/Proper_Linear_Subspace_of_Topological_Vector_Space_has_Empty_Interior | https://proofwiki.org/wiki/Proper_Linear_Subspace_of_Topological_Vector_Space_has_Empty_Interior | [
"Topological Vector Spaces"
] | [
"Definition:Topological Vector Space",
"Definition:Proper Subset",
"Definition:Linear Subspace",
"Definition:Interior (Topology)"
] | [
"Definition:Open Neighborhood",
"Translation of Open Set in Topological Vector Space is Open",
"Definition:Open Neighborhood",
"Definition:Linear Subspace",
"Topological Vector Space as Union of Dilations of Open Neighborhood of Origin",
"Definition:Linear Subspace",
"Definition:Proper Subset",
"Defin... |
proofwiki-21053 | Non-Meager Linear Subspace of Topological Vector Space is Everywhere Dense | Let $\GF \in \set {\R, \C}$.
Let $X$ be a topological vector space over $\GF$.
Let $L$ be a proper non-meager linear subspace of $X$.
Then $L$ is everywhere dense. | {{AimForCont}} that $L$ is not everywhere dense.
Then $L^- \ne X$.
From Closure of Linear Subspace of Topological Vector Space is Linear Subspace, $L^-$ is then a proper closed linear subspace of $X$.
From Proper Closed Linear Subspace of Topological Vector Space is Meager, it follows that $L^-$ is meager.
From Subset... | Let $\GF \in \set {\R, \C}$.
Let $X$ be a [[Definition:Topological Vector Space|topological vector space]] over $\GF$.
Let $L$ be a [[Definition:Proper Subset|proper]] [[Definition:Meager Space|non-meager]] [[Definition:Linear Subspace|linear subspace]] of $X$.
Then $L$ is [[Definition:Everywhere Dense|everywhere ... | {{AimForCont}} that $L$ is not [[Definition:Everywhere Dense|everywhere dense]].
Then $L^- \ne X$.
From [[Closure of Linear Subspace of Topological Vector Space is Linear Subspace]], $L^-$ is then a [[Definition:Proper Subset|proper]] [[Definition:Closed Set|closed]] [[Definition:Linear Subspace|linear subspace]] of ... | Non-Meager Linear Subspace of Topological Vector Space is Everywhere Dense | https://proofwiki.org/wiki/Non-Meager_Linear_Subspace_of_Topological_Vector_Space_is_Everywhere_Dense | https://proofwiki.org/wiki/Non-Meager_Linear_Subspace_of_Topological_Vector_Space_is_Everywhere_Dense | [
"Non-Meager Spaces",
"Everywhere Dense",
"Topological Vector Spaces"
] | [
"Definition:Topological Vector Space",
"Definition:Proper Subset",
"Definition:Meager Space",
"Definition:Linear Subspace",
"Definition:Everywhere Dense"
] | [
"Definition:Everywhere Dense",
"Closure of Linear Subspace of Topological Vector Space is Linear Subspace",
"Definition:Proper Subset",
"Definition:Closed Set",
"Definition:Linear Subspace",
"Proper Closed Linear Subspace of Topological Vector Space is Meager",
"Definition:Meager Space",
"Subset of Me... |
proofwiki-21054 | Pointwise Cauchyness of Sequence of Continuous Linear Transformations on Non-Meager Set implies Everywhere Pointwise Cauchyness | Let $\GF \in \set {\R, \C}$.
{{explain|What role does $\GF$ play in the following?}}
Let $X$ and $Y$ be topological vector spaces.
Let $\sequence {T_n}_{n \in \N}$ be a sequence of continuous linear transformations $T_n : X \to Y$ such that:
:the set $C$ of $x \in X$ such that $\sequence {T_n x}_{n \in \N}$ is a Cauchy... | Let $B$ be the set of all $x \in X$ such that:
:$\map \Gamma x = \set {T_n x : n \in \N}$
is von Neumann-bounded in $Y$.
For each $x \in C$, we have that $\sequence {T_n x}_{n \in \N}$ is Cauchy.
From Image of Cauchy Sequence in Topological Vector Space is von Neumann-Bounded, it follows that:
:$\set {T_n x : n \in \N}... | Let $\GF \in \set {\R, \C}$.
{{explain|What role does $\GF$ play in the following?}}
Let $X$ and $Y$ be [[Definition:Topological Vector Space|topological vector spaces]].
Let $\sequence {T_n}_{n \in \N}$ be a [[Definition:Sequence|sequence]] of [[Definition:Continuous Mapping|continuous]] [[Definition:Linear Transfo... | Let $B$ be the [[Definition:Set|set]] of all $x \in X$ such that:
:$\map \Gamma x = \set {T_n x : n \in \N}$
is [[Definition:Von Neumann-Bounded Subset of Topological Vector Space|von Neumann-bounded]] in $Y$.
For each $x \in C$, we have that $\sequence {T_n x}_{n \in \N}$ is [[Definition:Cauchy Sequence|Cauchy]].
Fr... | Pointwise Cauchyness of Sequence of Continuous Linear Transformations on Non-Meager Set implies Everywhere Pointwise Cauchyness | https://proofwiki.org/wiki/Pointwise_Cauchyness_of_Sequence_of_Continuous_Linear_Transformations_on_Non-Meager_Set_implies_Everywhere_Pointwise_Cauchyness | https://proofwiki.org/wiki/Pointwise_Cauchyness_of_Sequence_of_Continuous_Linear_Transformations_on_Non-Meager_Set_implies_Everywhere_Pointwise_Cauchyness | [
"Meager Spaces",
"Cauchy Sequences in Topological Vector Spaces"
] | [
"Definition:Topological Vector Space",
"Definition:Sequence",
"Definition:Continuous Mapping",
"Definition:Linear Transformation",
"Definition:Set",
"Definition:Cauchy Sequence",
"Definition:Meager Space",
"Definition:Cauchy Sequence"
] | [
"Definition:Set",
"Definition:Von Neumann-Bounded Subset of Topological Vector Space",
"Definition:Cauchy Sequence",
"Image of Cauchy Sequence in Topological Vector Space is von Neumann-Bounded",
"Definition:Von Neumann-Bounded Subset of Topological Vector Space",
"Definition:Meager Space",
"Definition:... |
proofwiki-21055 | Scalar Multiple of Cauchy Sequence in Topological Vector Space is Cauchy Sequence | Let $K$ be a topological field.
Let $X$ be a topological vector space over $K$.
Let $\sequence {x_n}_{n \in \N}$ be a Cauchy sequence in $X$.
Let $\lambda \in K$.
Then $\sequence {\lambda x_n}_{n \in \N}$ is a Cauchy sequence in $X$. | First consider the case $\lambda = 0_K$.
Then $\lambda x_n = {\mathbf 0}_X$ for each $n \in \N$.
From Constant Sequence in Topological Space Converges, we have that $\lambda x_n \to {\mathbf 0}_X$ as $n \to \infty$.
From Convergent Sequence in Topological Vector Space is Cauchy, $\sequence {\lambda x_n}_{n \in \N}$ is... | Let $K$ be a [[Definition:Topological Field|topological field]].
Let $X$ be a [[Definition:Topological Vector Space|topological vector space]] over $K$.
Let $\sequence {x_n}_{n \in \N}$ be a [[Definition:Cauchy Sequence in Topological Vector Space|Cauchy sequence]] in $X$.
Let $\lambda \in K$.
Then $\sequence {\l... | First consider the case $\lambda = 0_K$.
Then $\lambda x_n = {\mathbf 0}_X$ for each $n \in \N$.
From [[Constant Sequence in Topological Space Converges]], we have that $\lambda x_n \to {\mathbf 0}_X$ as $n \to \infty$.
From [[Convergent Sequence in Topological Vector Space is Cauchy]], $\sequence {\lambda x_n}_{n ... | Scalar Multiple of Cauchy Sequence in Topological Vector Space is Cauchy Sequence | https://proofwiki.org/wiki/Scalar_Multiple_of_Cauchy_Sequence_in_Topological_Vector_Space_is_Cauchy_Sequence | https://proofwiki.org/wiki/Scalar_Multiple_of_Cauchy_Sequence_in_Topological_Vector_Space_is_Cauchy_Sequence | [
"Cauchy Sequences in Topological Vector Spaces"
] | [
"Definition:Topological Field",
"Definition:Topological Vector Space",
"Definition:Cauchy Sequence/Topological Vector Space",
"Definition:Cauchy Sequence/Topological Vector Space"
] | [
"Constant Sequence in Topological Space Converges",
"Convergent Sequence in Topological Vector Space is Cauchy",
"Definition:Cauchy Sequence/Topological Vector Space",
"Definition:Open Neighborhood",
"Dilation of Open Set in Topological Vector Space is Open",
"Definition:Open Neighborhood",
"Definition:... |
proofwiki-21056 | Sum of Cauchy Sequences in Topological Vector Space is Cauchy Sequence | Let $K$ be a topological field.
Let $X$ be a topological vector space over $K$.
Let $\sequence {x_n}_{n \in \N}$ and $\sequence {y_n}_{n \in \N}$ be Cauchy sequences in $X$.
Then $\sequence {x_n + y_n}_{n \in \N}$ is a Cauchy sequence in $X$. | Let $U$ be an open neighborhood of ${\mathbf 0}_X$.
From Open Neighborhood of Point in Topological Vector Space contains Sum of Open Neighborhoods: Corollary 1, there exists an open neighborhood $V$ of ${\mathbf 0}_X$ such that:
:$V + V \subseteq U$
Since $\sequence {x_n}_{n \in \N}$ is Cauchy, there exists $N_1 \in N$... | Let $K$ be a [[Definition:Topological Field|topological field]].
Let $X$ be a [[Definition:Topological Vector Space|topological vector space]] over $K$.
Let $\sequence {x_n}_{n \in \N}$ and $\sequence {y_n}_{n \in \N}$ be [[Definition:Cauchy Sequence in Topological Vector Space|Cauchy sequences]] in $X$.
Then $\se... | Let $U$ be an [[Definition:Open Neighborhood|open neighborhood]] of ${\mathbf 0}_X$.
From [[Open Neighborhood of Point in Topological Vector Space contains Sum of Open Neighborhoods/Corollary 1|Open Neighborhood of Point in Topological Vector Space contains Sum of Open Neighborhoods: Corollary 1]], there exists an [[D... | Sum of Cauchy Sequences in Topological Vector Space is Cauchy Sequence | https://proofwiki.org/wiki/Sum_of_Cauchy_Sequences_in_Topological_Vector_Space_is_Cauchy_Sequence | https://proofwiki.org/wiki/Sum_of_Cauchy_Sequences_in_Topological_Vector_Space_is_Cauchy_Sequence | [
"Cauchy Sequences in Topological Vector Spaces"
] | [
"Definition:Topological Field",
"Definition:Topological Vector Space",
"Definition:Cauchy Sequence/Topological Vector Space",
"Definition:Cauchy Sequence/Topological Vector Space"
] | [
"Definition:Open Neighborhood",
"Open Neighborhood of Point in Topological Vector Space contains Sum of Open Neighborhoods/Corollary 1",
"Definition:Open Neighborhood",
"Definition:Cauchy Sequence/Topological Vector Space",
"Definition:Cauchy Sequence/Topological Vector Space",
"Definition:Cauchy Sequence... |
proofwiki-21057 | Linear Combination of Cauchy Sequences in Topological Vector Space is Cauchy Sequence | Let $K$ be a topological field.
Let $X$ be a topological vector space over $K$.
Let $\sequence {x_n}_{n \in \N}$ and $\sequence {y_n}_{n \in \N}$ be Cauchy sequences in $X$.
Let $\lambda, \mu \in \GF$.
Then $\sequence {\lambda x_n + \mu y_n}_{n \in \N}$ is a Cauchy sequence in $X$. | From Scalar Multiple of Cauchy Sequence in Topological Vector Space is Cauchy Sequence, $\sequence {\lambda x_n}_{n \in \N}$ and $\sequence {\mu y_n}_{n \in \N}$ are Cauchy in $X$.
From Sum of Cauchy Sequences in Topological Vector Space is Cauchy Sequence, $\sequence {\lambda x_n + \mu y_n}_{n \in \N}$ is Cauchy.
{{qe... | Let $K$ be a [[Definition:Topological FIeld|topological field]].
Let $X$ be a [[Definition:Topological Vector Space|topological vector space]] over $K$.
Let $\sequence {x_n}_{n \in \N}$ and $\sequence {y_n}_{n \in \N}$ be [[Definition:Cauchy Sequence in Topological Vector Space|Cauchy sequences]] in $X$.
Let $\lamb... | From [[Scalar Multiple of Cauchy Sequence in Topological Vector Space is Cauchy Sequence]], $\sequence {\lambda x_n}_{n \in \N}$ and $\sequence {\mu y_n}_{n \in \N}$ are [[Definition:Cauchy Sequence in Topological Vector Space|Cauchy]] in $X$.
From [[Sum of Cauchy Sequences in Topological Vector Space is Cauchy Sequen... | Linear Combination of Cauchy Sequences in Topological Vector Space is Cauchy Sequence | https://proofwiki.org/wiki/Linear_Combination_of_Cauchy_Sequences_in_Topological_Vector_Space_is_Cauchy_Sequence | https://proofwiki.org/wiki/Linear_Combination_of_Cauchy_Sequences_in_Topological_Vector_Space_is_Cauchy_Sequence | [
"Cauchy Sequences in Topological Vector Spaces"
] | [
"Definition:Topological FIeld",
"Definition:Topological Vector Space",
"Definition:Cauchy Sequence/Topological Vector Space",
"Definition:Cauchy Sequence/Topological Vector Space"
] | [
"Scalar Multiple of Cauchy Sequence in Topological Vector Space is Cauchy Sequence",
"Definition:Cauchy Sequence/Topological Vector Space",
"Sum of Cauchy Sequences in Topological Vector Space is Cauchy Sequence",
"Definition:Cauchy Sequence/Topological Vector Space",
"Category:Cauchy Sequences in Topologic... |
proofwiki-21058 | Scalar Multiple of Convergent Sequence in Topological Vector Space is Convergent | Let $K$ be a topological field.
Let $X$ be a topological vector space over $K$.
Let $x \in X$.
Let $\sequence {x_n}_{n \in \N}$ be a convergent sequence in $X$ with:
:$x_n \to x$
Let $\lambda \in K$.
Then $\sequence {\lambda x_n}_{n \in \N}$ is a convergent sequence in $X$ with:
:$\lambda x_n \to \lambda x$. | First consider the case $\lambda = 0_K$.
Then $\lambda x_n = {\mathbf 0}_X$ for each $n \in \N$.
From Constant Sequence in Topological Space Converges, we have that $\lambda x_n \to {\mathbf 0}_X = \lambda x$ as $n \to \infty$.
So $\sequence {\lambda x_n}_{n \in \N}$ converges for $\lambda = 0$.
Now take $\lambda \ne... | Let $K$ be a [[Definition:Topological Field|topological field]].
Let $X$ be a [[Definition:Topological Vector Space|topological vector space]] over $K$.
Let $x \in X$.
Let $\sequence {x_n}_{n \in \N}$ be a [[Definition:Convergent Sequence|convergent sequence]] in $X$ with:
:$x_n \to x$
Let $\lambda \in K$.
Then ... | First consider the case $\lambda = 0_K$.
Then $\lambda x_n = {\mathbf 0}_X$ for each $n \in \N$.
From [[Constant Sequence in Topological Space Converges]], we have that $\lambda x_n \to {\mathbf 0}_X = \lambda x$ as $n \to \infty$.
So $\sequence {\lambda x_n}_{n \in \N}$ [[Definition:Convergent Sequence|converges]]... | Scalar Multiple of Convergent Sequence in Topological Vector Space is Convergent | https://proofwiki.org/wiki/Scalar_Multiple_of_Convergent_Sequence_in_Topological_Vector_Space_is_Convergent | https://proofwiki.org/wiki/Scalar_Multiple_of_Convergent_Sequence_in_Topological_Vector_Space_is_Convergent | [
"Topological Vector Spaces"
] | [
"Definition:Topological Field",
"Definition:Topological Vector Space",
"Definition:Convergent Sequence",
"Definition:Convergent Sequence"
] | [
"Constant Sequence in Topological Space Converges",
"Definition:Convergent Sequence",
"Definition:Open Neighborhood",
"Dilation of Open Set in Topological Vector Space is Open",
"Definition:Open Neighborhood",
"Definition:Convergent Sequence",
"Definition:Open Neighborhood",
"Definition:Convergent Seq... |
proofwiki-21059 | Sum of Convergent Sequences in Topological Vector Space is Convergent | Let $K$ be a topological field.
Let $X$ be a topological vector space over $K$.
Let $x, y \in X$.
Let $\sequence {x_n}_{n \in \N}$ and $\sequence {y_n}_{n \in \N}$ be convergent sequences with:
:$x_n \to x$
and:
:$y_n \to y$
Then $\sequence {x_n + y_n}_{n \in \N}$ converges with:
:$x_n + y_n \to x + y$ | Let $W$ be an open neighborhood of $x + y$.
From Open Neighborhood of Point in Topological Vector Space contains Sum of Open Neighborhoods there exists an open neighborhood $U$ of $x$ and an open neighborhood $V$ of $y$ such that:
:$U + V \subseteq W$
Since $\sequence {x_n}_{n \in \N}$ converges to $x$, there exists $N... | Let $K$ be a [[Definition:Topological Field|topological field]].
Let $X$ be a [[Definition:Topological Vector Space|topological vector space]] over $K$.
Let $x, y \in X$.
Let $\sequence {x_n}_{n \in \N}$ and $\sequence {y_n}_{n \in \N}$ be [[Definition:Convergent Sequence|convergent sequences]] with:
:$x_n \to x$
a... | Let $W$ be an [[Definition:Open Neighborhood|open neighborhood]] of $x + y$.
From [[Open Neighborhood of Point in Topological Vector Space contains Sum of Open Neighborhoods]] there exists an [[Definition:Open Neighborhood|open neighborhood]] $U$ of $x$ and an [[Definition:Open Neighborhood|open neighborhood]] $V$ of ... | Sum of Convergent Sequences in Topological Vector Space is Convergent | https://proofwiki.org/wiki/Sum_of_Convergent_Sequences_in_Topological_Vector_Space_is_Convergent | https://proofwiki.org/wiki/Sum_of_Convergent_Sequences_in_Topological_Vector_Space_is_Convergent | [
"Topological Vector Spaces"
] | [
"Definition:Topological Field",
"Definition:Topological Vector Space",
"Definition:Convergent Sequence",
"Definition:Convergent Sequence"
] | [
"Definition:Open Neighborhood",
"Open Neighborhood of Point in Topological Vector Space contains Sum of Open Neighborhoods",
"Definition:Open Neighborhood",
"Definition:Open Neighborhood",
"Definition:Convergent Sequence",
"Definition:Convergent Sequence",
"Definition:Open Neighborhood",
"Definition:C... |
proofwiki-21060 | Linear Combination of Convergent Sequences in Topological Vector Space is Convergent | Let $K$ be a topological field.
Let $X$ be a topological vector space over $K$.
Let $x, y \in X$ and $\lambda, \mu \in K$.
Let $\sequence {x_n}_{n \in \N}$ and $\sequence {y_n}_{n \in \N}$ be convergent sequences with:
:$x_n \to x$
and:
:$y_n \to y$
Then $\sequence {\lambda x_n + \mu y_n}_{n \in \N}$ converges with:
... | From Scalar Multiple of Convergent Sequence in Topological Vector Space is Convergent, we have that:
:$\sequence {\lambda x_n}_{n \in \N}$ converges to $\lambda x$
and:
:$\sequence {\mu y_n}_{n \in \N}$ converges to $\mu y$.
From Sum of Convergent Sequences in Topological Vector Space is Convergent, $\sequence {\lambda... | Let $K$ be a [[Definition:Topological Field|topological field]].
Let $X$ be a [[Definition:Topological Vector Space|topological vector space]] over $K$.
Let $x, y \in X$ and $\lambda, \mu \in K$.
Let $\sequence {x_n}_{n \in \N}$ and $\sequence {y_n}_{n \in \N}$ be [[Definition:Convergent Sequence|convergent sequen... | From [[Scalar Multiple of Convergent Sequence in Topological Vector Space is Convergent]], we have that:
:$\sequence {\lambda x_n}_{n \in \N}$ [[Definition:Convergent Sequence|converges]] to $\lambda x$
and:
:$\sequence {\mu y_n}_{n \in \N}$ [[Definition:Convergent Sequence|converges]] to $\mu y$.
From [[Sum of Conver... | Linear Combination of Convergent Sequences in Topological Vector Space is Convergent | https://proofwiki.org/wiki/Linear_Combination_of_Convergent_Sequences_in_Topological_Vector_Space_is_Convergent | https://proofwiki.org/wiki/Linear_Combination_of_Convergent_Sequences_in_Topological_Vector_Space_is_Convergent | [
"Topological Vector Spaces"
] | [
"Definition:Topological Field",
"Definition:Topological Vector Space",
"Definition:Convergent Sequence",
"Definition:Convergent Sequence"
] | [
"Scalar Multiple of Convergent Sequence in Topological Vector Space is Convergent",
"Definition:Convergent Sequence",
"Definition:Convergent Sequence",
"Sum of Convergent Sequences in Topological Vector Space is Convergent",
"Definition:Convergent Sequence",
"Category:Topological Vector Spaces"
] |
proofwiki-21061 | Constant Sequence in Topological Space Converges | Let $X$ be a topological space.
Let $x \in X$.
Define a sequence $\sequence {x_n}_{n \in \N}$ by:
:$x_n = x$ for each $n \in \N$.
Then $\sequence {x_n}_{n \in \N}$ converges to $x$. | Let $U$ be an open neighborhood of $x$.
Then we have $x_n = x \in U$ for all $n \in \N$.
Hence $\sequence {x_n}_{n \in \N}$ converges to $x$.
{{qed}}
Category:Convergence
mww034dbrt84w84623o1xuwp9ay3x6p | Let $X$ be a [[Definition:Topological Space|topological space]].
Let $x \in X$.
Define a [[Definition:Sequence|sequence]] $\sequence {x_n}_{n \in \N}$ by:
:$x_n = x$ for each $n \in \N$.
Then $\sequence {x_n}_{n \in \N}$ [[Definition:Convergent Sequence|converges]] to $x$. | Let $U$ be an [[Definition:Open Neighborhood|open neighborhood]] of $x$.
Then we have $x_n = x \in U$ for all $n \in \N$.
Hence $\sequence {x_n}_{n \in \N}$ [[Definition:Convergent Sequence|converges]] to $x$.
{{qed}}
[[Category:Convergence]]
mww034dbrt84w84623o1xuwp9ay3x6p | Constant Sequence in Topological Space Converges | https://proofwiki.org/wiki/Constant_Sequence_in_Topological_Space_Converges | https://proofwiki.org/wiki/Constant_Sequence_in_Topological_Space_Converges | [
"Convergence"
] | [
"Definition:Topological Space",
"Definition:Sequence",
"Definition:Convergent Sequence"
] | [
"Definition:Open Neighborhood",
"Definition:Convergent Sequence",
"Category:Convergence"
] |
proofwiki-21062 | Sum of Convergent Nets in Topological Vector Space is Convergent | Let $K$ be a topological field.
Let $X$ be a topological vector space over $K$.
Let $\struct {\Lambda, \preceq}$ be a directed set.
Let $x, y \in X$.
Let $\family {x_\lambda}_{\lambda \in \Lambda}$ and $\family {y_\lambda}_{\lambda \in \Lambda}$ be nets converging to $x$ and $y$ respectively.
Then the net $\family {x_... | For ease of reading, let $\succeq$ be the inverse relation of $\preceq$.
Let $W$ be an open neighborhood of $x + y$.
From Open Neighborhood of Point in Topological Vector Space contains Sum of Open Neighborhoods there exists an open neighborhood $U$ of $x$ and an open neighborhood $V$ of $y$ such that:
:$U + V \subsete... | Let $K$ be a [[Definition:Topological Field|topological field]].
Let $X$ be a [[Definition:Topological Vector Space|topological vector space]] over $K$.
Let $\struct {\Lambda, \preceq}$ be a [[Definition:Directed Set|directed set]].
Let $x, y \in X$.
Let $\family {x_\lambda}_{\lambda \in \Lambda}$ and $\family {y_... | For ease of reading, let $\succeq$ be the [[Definition:Inverse Relation|inverse relation]] of $\preceq$.
Let $W$ be an [[Definition:Open Neighborhood|open neighborhood]] of $x + y$.
From [[Open Neighborhood of Point in Topological Vector Space contains Sum of Open Neighborhoods]] there exists an [[Definition:Open Nei... | Sum of Convergent Nets in Topological Vector Space is Convergent | https://proofwiki.org/wiki/Sum_of_Convergent_Nets_in_Topological_Vector_Space_is_Convergent | https://proofwiki.org/wiki/Sum_of_Convergent_Nets_in_Topological_Vector_Space_is_Convergent | [
"Moore-Smith Sequences",
"Nets (Set Theory)",
"Nets (Set Theory)",
"Topological Vector Spaces"
] | [
"Definition:Topological Field",
"Definition:Topological Vector Space",
"Definition:Directed Preordering",
"Definition:Net (Set Theory)",
"Definition:Convergent Net",
"Definition:Net (Set Theory)",
"Definition:Convergent Net"
] | [
"Definition:Inverse Relation",
"Definition:Open Neighborhood",
"Open Neighborhood of Point in Topological Vector Space contains Sum of Open Neighborhoods",
"Definition:Open Neighborhood",
"Definition:Open Neighborhood",
"Definition:Convergent Net",
"Definition:Convergent Net",
"Definition:Directed Pre... |
proofwiki-21063 | Scalar Multiple of Convergent Net in Topological Vector Space is Convergent | Let $K$ be a topological field.
Let $X$ be a topological vector space over $K$.
Let $\struct {\Lambda, \preceq}$ be a directed set.
Let $x \in X$ and $\mu \in K$.
Let $\family {x_\lambda}_{\lambda \in \Lambda}$ be a net converging to $x$.
Then $\family {\mu x_\lambda}_{\lambda \in \Lambda}$ converges to $\mu x$. | For ease of reading, let $\succeq$ be the inverse relation of $\preceq$.
First consider the case $\mu = 0_K$.
Then $\mu x_n = {\mathbf 0}_X$ for each $n \in \N$.
From Constant Net is Convergent, we have that $\mu x_\lambda$ converges to ${\mathbf 0}_X$.
So $\sequence {\lambda x_n}_{n \in \N}$ converges to $\mu x$ fo... | Let $K$ be a [[Definition:Topological Field|topological field]].
Let $X$ be a [[Definition:Topological Vector Space|topological vector space]] over $K$.
Let $\struct {\Lambda, \preceq}$ be a [[Definition:Directed Set|directed set]].
Let $x \in X$ and $\mu \in K$.
Let $\family {x_\lambda}_{\lambda \in \Lambda}$ be ... | For ease of reading, let $\succeq$ be the [[Definition:Inverse Relation|inverse relation]] of $\preceq$.
First consider the case $\mu = 0_K$.
Then $\mu x_n = {\mathbf 0}_X$ for each $n \in \N$.
From [[Constant Net is Convergent]], we have that $\mu x_\lambda$ [[Definition:Convergent Net|converges]] to ${\mathbf 0}_... | Scalar Multiple of Convergent Net in Topological Vector Space is Convergent | https://proofwiki.org/wiki/Scalar_Multiple_of_Convergent_Net_in_Topological_Vector_Space_is_Convergent | https://proofwiki.org/wiki/Scalar_Multiple_of_Convergent_Net_in_Topological_Vector_Space_is_Convergent | [
"Nets (Set Theory)",
"Topological Vector Spaces"
] | [
"Definition:Topological Field",
"Definition:Topological Vector Space",
"Definition:Directed Preordering",
"Definition:Net (Set Theory)",
"Definition:Convergent Net",
"Definition:Convergent Net"
] | [
"Definition:Inverse Relation",
"Constant Net is Convergent",
"Definition:Convergent Net",
"Definition:Convergent Net",
"Definition:Open Neighborhood",
"Dilation of Open Set in Topological Vector Space is Open",
"Definition:Open Neighborhood",
"Definition:Convergent Sequence",
"Definition:Open Neighb... |
proofwiki-21064 | Characterization of Convex Absorbing Set in Vector Space | Let $\GF \in \set {\R, \C}$.
Let $X$ be a vector space over $\GF$.
Let $C$ be a convex set.
Then $C$ is absorbing {{iff}}:
:for each $x \in X$ there exists $t \in \R_{> 0}$ such that $x \in t C$. | === Necessary Condition ===
Suppose that $C$ is absorbing.
Then for each $x \in X$ there exists $s \in \R_{> 0}$ such that $x \in t C$ for $s \in \C$ with $\cmod s \ge t$.
In particular, $x \in t C$.
{{qed|lemma}} | Let $\GF \in \set {\R, \C}$.
Let $X$ be a [[Definition:Vector Space|vector space]] over $\GF$.
Let $C$ be a [[Definition:Convex Set (Vector Space)|convex set]].
Then $C$ is [[Definition:Absorbing Set|absorbing]] {{iff}}:
:for each $x \in X$ there exists $t \in \R_{> 0}$ such that $x \in t C$. | === Necessary Condition ===
Suppose that $C$ is [[Definition:Absorbing Set|absorbing]].
Then for each $x \in X$ there exists $s \in \R_{> 0}$ such that $x \in t C$ for $s \in \C$ with $\cmod s \ge t$.
In particular, $x \in t C$.
{{qed|lemma}} | Characterization of Convex Absorbing Set in Vector Space | https://proofwiki.org/wiki/Characterization_of_Convex_Absorbing_Set_in_Vector_Space | https://proofwiki.org/wiki/Characterization_of_Convex_Absorbing_Set_in_Vector_Space | [
"Absorbing Sets",
"Convex Sets (Vector Spaces)"
] | [
"Definition:Vector Space",
"Definition:Convex Set (Vector Space)",
"Definition:Absorbing Set"
] | [
"Definition:Absorbing Set",
"Definition:Absorbing Set",
"Definition:Absorbing Set"
] |
proofwiki-21065 | Dilation of Convex Set containing Zero Vector by Real Number between 0 and 1 | Let $\GF \in \set {\R, \C}$.
Let $X$ be a vector space over $\GF$.
Let $C$ be a convex set with ${\mathbf 0}_X \in C$.
Let $t \in \closedint 0 1$.
Then:
:$t C \subseteq C$ | Let $x \in t C$.
Then we have $x \in t C + \paren {1 - t} {\mathbf 0}_X$.
Since ${\mathbf 0}_X \in C$, we have $x \in t C + \paren {1 - t} C$.
By definition 2 of a convex set, we have $t C + \paren {1 - t} C \subseteq C$.
So we have $x \in C$.
So $x \in t C$ implies that $x \in C$.
So $t C \subseteq C$.
{{qed}}
Categor... | Let $\GF \in \set {\R, \C}$.
Let $X$ be a [[Definition:Vector Space|vector space]] over $\GF$.
Let $C$ be a [[Definition:Convex Set (Vector Space)|convex set]] with ${\mathbf 0}_X \in C$.
Let $t \in \closedint 0 1$.
Then:
:$t C \subseteq C$ | Let $x \in t C$.
Then we have $x \in t C + \paren {1 - t} {\mathbf 0}_X$.
Since ${\mathbf 0}_X \in C$, we have $x \in t C + \paren {1 - t} C$.
By [[Definition:Convex Set (Vector Space)/Definition 2|definition 2 of a convex set]], we have $t C + \paren {1 - t} C \subseteq C$.
So we have $x \in C$.
So $x \in t C$ im... | Dilation of Convex Set containing Zero Vector by Real Number between 0 and 1 | https://proofwiki.org/wiki/Dilation_of_Convex_Set_containing_Zero_Vector_by_Real_Number_between_0_and_1 | https://proofwiki.org/wiki/Dilation_of_Convex_Set_containing_Zero_Vector_by_Real_Number_between_0_and_1 | [
"Convex Sets (Vector Spaces)"
] | [
"Definition:Vector Space",
"Definition:Convex Set (Vector Space)"
] | [
"Definition:Convex Set (Vector Space)/Definition 2",
"Category:Convex Sets (Vector Spaces)"
] |
proofwiki-21066 | Superset of Absorbing Set is Absorbing | Let $\GF \in \set {\R, \C}$.
Let $X$ be a vector space over $\GF$.
Let $A$ be an absorbing set.
Let $B \supseteq A$.
Then $B$ is absorbing. | Let $x \in X$.
Since $A$ is absorbing, there exists $t \in \R_{> 0}$ such that:
:$x \in t A$ for $\cmod \alpha \ge t$.
Since $A \subseteq B$, we obtain:
:$x \in t B$ for $\cmod \alpha \ge t$.
{{qed}}
Category:Absorbing Sets
42aktvnfybdn2rnwsciepjkibi8qpw5 | Let $\GF \in \set {\R, \C}$.
Let $X$ be a [[Definition:Vector Space|vector space]] over $\GF$.
Let $A$ be an [[Definition:Absorbing Set|absorbing set]].
Let $B \supseteq A$.
Then $B$ is [[Definition:Absorbing Set|absorbing]]. | Let $x \in X$.
Since $A$ is [[Definition:Absorbing Set|absorbing]], there exists $t \in \R_{> 0}$ such that:
:$x \in t A$ for $\cmod \alpha \ge t$.
Since $A \subseteq B$, we obtain:
:$x \in t B$ for $\cmod \alpha \ge t$.
{{qed}}
[[Category:Absorbing Sets]]
42aktvnfybdn2rnwsciepjkibi8qpw5 | Superset of Absorbing Set is Absorbing | https://proofwiki.org/wiki/Superset_of_Absorbing_Set_is_Absorbing | https://proofwiki.org/wiki/Superset_of_Absorbing_Set_is_Absorbing | [
"Absorbing Sets"
] | [
"Definition:Vector Space",
"Definition:Absorbing Set",
"Definition:Absorbing Set"
] | [
"Definition:Absorbing Set",
"Category:Absorbing Sets"
] |
proofwiki-21067 | Image of Absorbing Set under Surjective Linear Transformation is Absorbing | Let $\GF \in \set {\R, \C}$.
Let $X$ and $Y$ be vector spaces over $\GF$.
Let $A \subseteq X$ be an absorbing set.
Let $T : X \to Y$ be a surjective linear transformation.
Then $T \sqbrk A$ is an absorbing set. | Let $y \in Y$.
Since $T$ is surjective, there exists $x \in X$ such that $y = T x$.
Since $A$ is absorbing, there exists $t \in \R_{> 0}$ such that:
:$x \in \alpha A$ for $\alpha \in \C$ with $\cmod \alpha \ge t$.
Then:
:$y = T x \in T \sqbrk {\alpha A}$ for $\alpha \in \C$ with $\cmod \alpha \ge t$.
From Image of Dil... | Let $\GF \in \set {\R, \C}$.
Let $X$ and $Y$ be [[Definition:Vector Space|vector spaces]] over $\GF$.
Let $A \subseteq X$ be an [[Definition:Absorbing Set|absorbing set]].
Let $T : X \to Y$ be a [[Definition:Surjection|surjective]] [[Definition:Linear Transformation|linear transformation]].
Then $T \sqbrk A$ is ... | Let $y \in Y$.
Since $T$ is [[Definition:Surjection|surjective]], there exists $x \in X$ such that $y = T x$.
Since $A$ is [[Definition:Absorbing Set|absorbing]], there exists $t \in \R_{> 0}$ such that:
:$x \in \alpha A$ for $\alpha \in \C$ with $\cmod \alpha \ge t$.
Then:
:$y = T x \in T \sqbrk {\alpha A}$ for $\a... | Image of Absorbing Set under Surjective Linear Transformation is Absorbing | https://proofwiki.org/wiki/Image_of_Absorbing_Set_under_Surjective_Linear_Transformation_is_Absorbing | https://proofwiki.org/wiki/Image_of_Absorbing_Set_under_Surjective_Linear_Transformation_is_Absorbing | [
"Absorbing Sets"
] | [
"Definition:Vector Space",
"Definition:Absorbing Set",
"Definition:Surjection",
"Definition:Linear Transformation",
"Definition:Absorbing Set"
] | [
"Definition:Surjection",
"Definition:Absorbing Set",
"Image of Dilation of Set under Linear Transformation is Dilation of Image",
"Definition:Absorbing Set",
"Category:Absorbing Sets"
] |
proofwiki-21068 | Closure of Absorbing Set is Absorbing | Let $\GF \in \set {\R, \C}$.
Let $X$ be a vector space over $\GF$.
Let $A \subseteq X$ be an absorbing set.
Then $A^-$ is absorbing. | From the definition of closure, we have $A \subseteq A^-$.
From Superset of Absorbing Set is Absorbing, $A^-$ is absorbing.
{{qed}}
Category:Absorbing Sets
leiv0e5m97pokef7xdw42qjv3lfqims | Let $\GF \in \set {\R, \C}$.
Let $X$ be a [[Definition:Vector Space|vector space]] over $\GF$.
Let $A \subseteq X$ be an [[Definition:Absorbing Set|absorbing set]].
Then $A^-$ is [[Definition:Absorbing Set|absorbing]]. | From the definition of [[Definition:Topological Closure|closure]], we have $A \subseteq A^-$.
From [[Superset of Absorbing Set is Absorbing]], $A^-$ is [[Definition:Absorbing Set|absorbing]].
{{qed}}
[[Category:Absorbing Sets]]
leiv0e5m97pokef7xdw42qjv3lfqims | Closure of Absorbing Set is Absorbing | https://proofwiki.org/wiki/Closure_of_Absorbing_Set_is_Absorbing | https://proofwiki.org/wiki/Closure_of_Absorbing_Set_is_Absorbing | [
"Absorbing Sets"
] | [
"Definition:Vector Space",
"Definition:Absorbing Set",
"Definition:Absorbing Set"
] | [
"Definition:Closure (Topology)",
"Superset of Absorbing Set is Absorbing",
"Definition:Absorbing Set",
"Category:Absorbing Sets"
] |
proofwiki-21069 | Finite Intersection of Absorbing Sets is Absorbing | Let $\GF \in \set {\R, \C}$.
Let $X$ be a vector space over $\GF$.
Let $A_1, \ldots, A_n \subseteq X$ be absorbing sets.
Then:
:$\ds \bigcap_{j \mathop = 1}^n A_j$ is absorbing. | Let $x \in X$ and $j \in \set {1, \ldots, n}$.
Since $A_j$ is absorbing, there exists $t_j \in \R_{> 0}$ such that:
:$x \in \alpha A_j$ for $\alpha \in \C$ with $\cmod \alpha \ge t_j$.
Let $t = \max \set {t_1, \ldots, t_n}$.
Then, we have:
:$\ds x \in \bigcap_{j \mathop = 1}^n \paren {\alpha A_j}$ for $\alpha \in \C$ ... | Let $\GF \in \set {\R, \C}$.
Let $X$ be a [[Definition:Vector Space|vector space]] over $\GF$.
Let $A_1, \ldots, A_n \subseteq X$ be [[Definition:Absorbing Set|absorbing sets]].
Then:
:$\ds \bigcap_{j \mathop = 1}^n A_j$ is [[Definition:Absorbing Set|absorbing]]. | Let $x \in X$ and $j \in \set {1, \ldots, n}$.
Since $A_j$ is [[Definition:Absorbing Set|absorbing]], there exists $t_j \in \R_{> 0}$ such that:
:$x \in \alpha A_j$ for $\alpha \in \C$ with $\cmod \alpha \ge t_j$.
Let $t = \max \set {t_1, \ldots, t_n}$.
Then, we have:
:$\ds x \in \bigcap_{j \mathop = 1}^n \paren {\... | Finite Intersection of Absorbing Sets is Absorbing | https://proofwiki.org/wiki/Finite_Intersection_of_Absorbing_Sets_is_Absorbing | https://proofwiki.org/wiki/Finite_Intersection_of_Absorbing_Sets_is_Absorbing | [
"Absorbing Sets"
] | [
"Definition:Vector Space",
"Definition:Absorbing Set",
"Definition:Absorbing Set"
] | [
"Definition:Absorbing Set",
"Dilation of Intersection of Subsets of Vector Space",
"Definition:Absorbing Set",
"Category:Absorbing Sets"
] |
proofwiki-21070 | Non-Zero Dilation of Absorbing Set is Absorbing | Let $\GF \in \set {\R, \C}$.
Let $X$ be a vector space over $\GF$.
Let $A \subseteq X$ be an absorbing set.
Let $\lambda \in \GF \setminus \set 0$.
Then $\lambda A$ is absorbing. | Let $x \in X$.
Then there exists $t \in \R_{> 0}$ such that:
:$x \in \alpha A$ for $\cmod \alpha \ge t$.
That is:
:$\ds x \in \frac \alpha \lambda \paren {\lambda A}$ for $\cmod \alpha \ge t$.
Since the map $\alpha \mapsto \alpha/\lambda$ is a bijection from $\cmod \alpha \ge t$ to $\cmod \alpha \ge t/\cmod \lambda$,... | Let $\GF \in \set {\R, \C}$.
Let $X$ be a [[Definition:Vector Space|vector space]] over $\GF$.
Let $A \subseteq X$ be an [[Definition:Absorbing Set|absorbing set]].
Let $\lambda \in \GF \setminus \set 0$.
Then $\lambda A$ is [[Definition:Absorbing Set|absorbing]]. | Let $x \in X$.
Then there exists $t \in \R_{> 0}$ such that:
:$x \in \alpha A$ for $\cmod \alpha \ge t$.
That is:
:$\ds x \in \frac \alpha \lambda \paren {\lambda A}$ for $\cmod \alpha \ge t$.
Since the [[Definition:Mapping|map]] $\alpha \mapsto \alpha/\lambda$ is a [[Definition:Bijection|bijection]] from $\cmod \... | Non-Zero Dilation of Absorbing Set is Absorbing | https://proofwiki.org/wiki/Non-Zero_Dilation_of_Absorbing_Set_is_Absorbing | https://proofwiki.org/wiki/Non-Zero_Dilation_of_Absorbing_Set_is_Absorbing | [
"Dilations of Subsets of Vector Spaces",
"Absorbing Sets"
] | [
"Definition:Vector Space",
"Definition:Absorbing Set",
"Definition:Absorbing Set"
] | [
"Definition:Mapping",
"Definition:Bijection",
"Definition:Absorbing Set",
"Category:Dilations of Subsets of Vector Spaces",
"Category:Absorbing Sets"
] |
proofwiki-21071 | Sum of Absorbing Sets is Absorbing | Let $\GF \in \set {\R, \C}$.
Let $X$ be a vector space over $\GF$.
Let $A, B \subseteq X$ be absorbing sets.
Then $A + B$ is absorbing. | Let $u \in A + B$.
Then there exists $a \in A$, $b \in B$ such that $u = a + b$.
Now, there exists $t_1 \in \R_{> 0}$ such that:
:$a \in \alpha A$ for $\alpha \in \C$ with $\cmod \alpha \ge t_1$
There also exists $t_2 \in \R_{> 0}$ such that:
:$b \in \alpha B$ for $\alpha \in \C$ with $\cmod \alpha \ge t_2$
Let $t = ... | Let $\GF \in \set {\R, \C}$.
Let $X$ be a [[Definition:Vector Space|vector space]] over $\GF$.
Let $A, B \subseteq X$ be [[Definition:Absorbing Set|absorbing sets]].
Then $A + B$ is [[Definition:Absorbing Set|absorbing]]. | Let $u \in A + B$.
Then there exists $a \in A$, $b \in B$ such that $u = a + b$.
Now, there exists $t_1 \in \R_{> 0}$ such that:
:$a \in \alpha A$ for $\alpha \in \C$ with $\cmod \alpha \ge t_1$
There also exists $t_2 \in \R_{> 0}$ such that:
:$b \in \alpha B$ for $\alpha \in \C$ with $\cmod \alpha \ge t_2$
Let $... | Sum of Absorbing Sets is Absorbing | https://proofwiki.org/wiki/Sum_of_Absorbing_Sets_is_Absorbing | https://proofwiki.org/wiki/Sum_of_Absorbing_Sets_is_Absorbing | [
"Absorbing Sets"
] | [
"Definition:Vector Space",
"Definition:Absorbing Set",
"Definition:Absorbing Set"
] | [
"Dilation of Subset of Vector Space Distributes over Sum",
"Definition:Absorbing Set",
"Category:Absorbing Sets"
] |
proofwiki-21072 | Sum of Compact Subsets of Topological Vector Space is Compact | Let $K$ be a topological field.
Let $X$ be a topological vector space over $K$.
Let $A$ and $B$ be compact (topological) subspaces of $X$.
Then $A + B$ is compact. | Since $X$ is a topological vector space, the map $+ : X \times X \to X$ defined by:
:$\map + {x, y} = x + y$
for each $x, y \in X$ is continuous.
We have:
:$A + B = + \sqbrk {A \times B}$
From Tychonoff's Theorem, $A \times B$ is compact.
From Continuous Image of Compact Space is Compact, $+ \sqbrk {A \times B} = A + B... | Let $K$ be a [[Definition:Topological Field|topological field]].
Let $X$ be a [[Definition:Topological Vector Space|topological vector space]] over $K$.
Let $A$ and $B$ be [[Definition:Compact Subspace|compact (topological) subspaces]] of $X$.
Then $A + B$ is [[Definition:Compact Topological Subspace|compact]]. | Since $X$ is a [[Definition:Topological Vector Space|topological vector space]], the map $+ : X \times X \to X$ defined by:
:$\map + {x, y} = x + y$
for each $x, y \in X$ is [[Definition:Continuous Mapping|continuous]].
We have:
:$A + B = + \sqbrk {A \times B}$
From [[Tychonoff's Theorem]], $A \times B$ is [[Definiti... | Sum of Compact Subsets of Topological Vector Space is Compact | https://proofwiki.org/wiki/Sum_of_Compact_Subsets_of_Topological_Vector_Space_is_Compact | https://proofwiki.org/wiki/Sum_of_Compact_Subsets_of_Topological_Vector_Space_is_Compact | [
"Compact Topological Spaces",
"Topological Vector Spaces"
] | [
"Definition:Topological Field",
"Definition:Topological Vector Space",
"Definition:Compact Topological Space/Subspace",
"Definition:Compact Topological Space/Subspace"
] | [
"Definition:Topological Vector Space",
"Definition:Continuous Mapping",
"Tychonoff's Theorem",
"Definition:Compact Topological Space/Subspace",
"Continuous Image of Compact Space is Compact",
"Definition:Compact Topological Space/Subspace",
"Category:Compact Topological Spaces",
"Category:Topological ... |
proofwiki-21073 | Quotient Mapping on Quotient Topological Vector Space is Open Mapping | Let $K$ be a topological field.
Let $\struct {X, \tau}$ be a topological vector space over $K$.
Let $N$ be a closed linear subspace of $X$.
Let $\struct {X/N, \tau_N}$ be the quotient topological vector space of $X$ modulo $N$.
Let $\pi : \struct {X, \tau} \to \struct {X/N, \tau_N}$ be the quotient mapping.
Then $\pi$ ... | Let $V \in \tau$.
From Quotient Mapping is Linear Transformation, $\pi$ is a linear transformation.
From Preimage of Image of Linear Transformation, we have:
:$\pi^{-1} \sqbrk {\pi \sqbrk V} = \ker \pi + V$
From Kernel of Quotient Mapping, we have $\ker \pi = N$ and so:
:$\pi^{-1} \sqbrk {\pi \sqbrk V} = N + V$
From Su... | Let $K$ be a [[Definition:Topological Field|topological field]].
Let $\struct {X, \tau}$ be a [[Definition:Topological Vector Space|topological vector space]] over $K$.
Let $N$ be a [[Definition:Closed Linear Subspace|closed linear subspace]] of $X$.
Let $\struct {X/N, \tau_N}$ be the [[Definition:Quotient Topologic... | Let $V \in \tau$.
From [[Quotient Mapping is Linear Transformation]], $\pi$ is a [[Definition:Linear Transformation|linear transformation]].
From [[Preimage of Image of Linear Transformation]], we have:
:$\pi^{-1} \sqbrk {\pi \sqbrk V} = \ker \pi + V$
From [[Kernel of Quotient Mapping]], we have $\ker \pi = N$ and s... | Quotient Mapping on Quotient Topological Vector Space is Open Mapping | https://proofwiki.org/wiki/Quotient_Mapping_on_Quotient_Topological_Vector_Space_is_Open_Mapping | https://proofwiki.org/wiki/Quotient_Mapping_on_Quotient_Topological_Vector_Space_is_Open_Mapping | [
"Quotient Topological Vector Spaces",
"Open Mappings"
] | [
"Definition:Topological Field",
"Definition:Topological Vector Space",
"Definition:Closed Linear Subspace",
"Definition:Quotient Topological Vector Space",
"Definition:Quotient Mapping",
"Definition:Open Mapping"
] | [
"Quotient Mapping is Linear Transformation",
"Definition:Linear Transformation",
"Preimage of Image of Linear Transformation",
"Kernel of Quotient Mapping",
"Sum of Set and Open Set in Topological Vector Space is Open",
"Definition:Open Set",
"Definition:Quotient Topology",
"Definition:Open Set"
] |
proofwiki-21074 | Open Neighborhood of Dilation of Point in Topological Vector Space contains Pointwise Scalar Multiplication of Open Neighborhood of Scalar with Open Neighborhood of Vector | Let $K$ be a topological field.
Let $X$ be a topological vector space over $K$.
Let $\lambda \in K$ and $x \in X$.
Let $U$ be an open neighborhood of $\lambda x$.
Then there exists an open neighborhood $D$ of $\lambda$ in $K$ and an open neighborhood $V$ in $X$ such that:
:$D V \subseteq U$
where:
:$D V = \set {\mu y ... | Equip $K \times X$ with the product topology.
From Box Topology on Finite Product Space is Product Topology, this is precisely the box topology.
Define $m : K \times X \to X$ by:
:$\map m {\mu, y} = \mu y$
for each $\tuple {\mu, y} \in K \times X$.
From the definition of a topological vector space, $m$ is continuous.
... | Let $K$ be a [[Definition:Topological Field|topological field]].
Let $X$ be a [[Definition:Topological Vector Space|topological vector space]] over $K$.
Let $\lambda \in K$ and $x \in X$.
Let $U$ be an [[Definition:Open Neighborhood|open neighborhood]] of $\lambda x$.
Then there exists an [[Definition:Open Neighb... | Equip $K \times X$ with the [[Definition:Product Topology|product topology]].
From [[Box Topology on Finite Product Space is Product Topology]], this is precisely the [[Definition:Box Topology|box topology]].
Define $m : K \times X \to X$ by:
:$\map m {\mu, y} = \mu y$
for each $\tuple {\mu, y} \in K \times X$.
Fr... | Open Neighborhood of Dilation of Point in Topological Vector Space contains Pointwise Scalar Multiplication of Open Neighborhood of Scalar with Open Neighborhood of Vector | https://proofwiki.org/wiki/Open_Neighborhood_of_Dilation_of_Point_in_Topological_Vector_Space_contains_Pointwise_Scalar_Multiplication_of_Open_Neighborhood_of_Scalar_with_Open_Neighborhood_of_Vector | https://proofwiki.org/wiki/Open_Neighborhood_of_Dilation_of_Point_in_Topological_Vector_Space_contains_Pointwise_Scalar_Multiplication_of_Open_Neighborhood_of_Scalar_with_Open_Neighborhood_of_Vector | [
"Topological Vector Spaces"
] | [
"Definition:Topological Field",
"Definition:Topological Vector Space",
"Definition:Open Neighborhood",
"Definition:Open Neighborhood",
"Definition:Open Neighborhood"
] | [
"Definition:Product Topology",
"Box Topology on Finite Product Space is Product Topology",
"Definition:Box Topology",
"Definition:Topological Vector Space",
"Definition:Continuous Mapping (Topology)",
"Definition:Continuous Mapping (Topology)/Point",
"Definition:Open Neighborhood",
"Basis for Box Topo... |
proofwiki-21075 | Image of Pointwise Scalar Multiplication of Subset of Scalars with Subset of Vectors under Linear Transformation | Let $K$ be a field.
Let $X$ and $Y$ be vector spaces over $K$.
Let $T : X \to Y$ be a linear transformation.
Let $S \subseteq K$ and $D \subseteq X$ be non-empty sets.
Then:
:$T \sqbrk {S D} = S T \sqbrk D$
where:
:$S D = \set {\lambda x : \lambda \in S, \, x \in D}$ | We have:
{{begin-eqn}}
{{eqn | l = T \sqbrk {S D}
| r = T \sqbrk {\bigcup_{s \mathop \in S} s D}
}}
{{eqn | r = \bigcup_{s \mathop \in S} T \sqbrk {s D}
| c = Image of Union under Mapping
}}
{{eqn | r = \bigcup_{s \mathop \in S} s T \sqbrk D
| c = Image of Dilation of Set under Linear Transformation is Dilation ... | Let $K$ be a [[Definition:Field (Abstract Algebra)|field]].
Let $X$ and $Y$ be [[Definition:Vector Space|vector spaces]] over $K$.
Let $T : X \to Y$ be a [[Definition:Linear Transformation|linear transformation]].
Let $S \subseteq K$ and $D \subseteq X$ be [[Definition:Non-Empty Set|non-empty sets]].
Then:
:$T \s... | We have:
{{begin-eqn}}
{{eqn | l = T \sqbrk {S D}
| r = T \sqbrk {\bigcup_{s \mathop \in S} s D}
}}
{{eqn | r = \bigcup_{s \mathop \in S} T \sqbrk {s D}
| c = [[Image of Union under Mapping]]
}}
{{eqn | r = \bigcup_{s \mathop \in S} s T \sqbrk D
| c = [[Image of Dilation of Set under Linear Transformation is Dil... | Image of Pointwise Scalar Multiplication of Subset of Scalars with Subset of Vectors under Linear Transformation | https://proofwiki.org/wiki/Image_of_Pointwise_Scalar_Multiplication_of_Subset_of_Scalars_with_Subset_of_Vectors_under_Linear_Transformation | https://proofwiki.org/wiki/Image_of_Pointwise_Scalar_Multiplication_of_Subset_of_Scalars_with_Subset_of_Vectors_under_Linear_Transformation | [
"Vector Spaces"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Vector Space",
"Definition:Linear Transformation",
"Definition:Non-Empty Set"
] | [
"Image of Union under Mapping",
"Image of Dilation of Set under Linear Transformation is Dilation of Image",
"Category:Vector Spaces"
] |
proofwiki-21076 | Existence of Banach Limits | Let $\struct {\map {\ell^\infty} \R, \norm \cdot_\infty}$ be the normed vector space of bounded sequences on $\R$.
Let $\struct {\paren {\map {\ell^\infty} \R}^\ast, \norm \cdot_{\paren {\ell^\infty}^\ast} }$ be the normed dual space of $\struct {\map {\ell^\infty} \R, \norm \cdot_\infty}$.
Then there exists a Banach ... | Let $\map c \R$ be the set of the convergent real sequences.
Then:
:$\map c \R \subseteq \map {\ell^\infty} \R$
is a linear subspace in view of:
:Convergent Real Sequence is Bounded
:Linear Combination of Convergent Sequences in Topological Vector Space is Convergent
Define a mapping $f_0 : \map c \R \to \R$ by:
:$\ds... | Let $\struct {\map {\ell^\infty} \R, \norm \cdot_\infty}$ be the [[Definition:Normed Vector Space of Bounded Sequences|normed vector space of bounded sequences on $\R$]].
Let $\struct {\paren {\map {\ell^\infty} \R}^\ast, \norm \cdot_{\paren {\ell^\infty}^\ast} }$ be the [[Definition:Normed Dual Space|normed dual spa... | Let $\map c \R$ be the [[Definition:Set|set]] of the [[Definition:Convergent Real Sequence|convergent real sequences]].
Then:
:$\map c \R \subseteq \map {\ell^\infty} \R$
is a [[Definition:Vector Subspace|linear subspace]] in view of:
:[[Convergent Real Sequence is Bounded]]
:[[Linear Combination of Convergent Sequenc... | Existence of Banach Limits | https://proofwiki.org/wiki/Existence_of_Banach_Limits | https://proofwiki.org/wiki/Existence_of_Banach_Limits | [
"Banach Limits"
] | [
"Definition:Space of Bounded Sequences/Normed Vector Space",
"Definition:Normed Dual Space",
"Definition:Banach Limit"
] | [
"Definition:Set",
"Definition:Convergent Sequence/Real Numbers",
"Definition:Vector Subspace",
"Convergent Real Sequence is Bounded",
"Linear Combination of Convergent Sequences in Topological Vector Space is Convergent",
"Definition:Mapping",
"Definition:Mapping",
"Definition:Mapping"
] |
proofwiki-21077 | Properties of Product of Identity plus Operator Raised to Powers of 2 | Let $X$ be a Banach space.
Let $\map \LL X$ be the set of all linear transformations.
Let $\map {CL} X$ be a continuous linear transformation sapce.
Let $\norm {\, \cdot \,}$ be the supremum operator norm.
Let $A \in \map {CL} X$ be such that $\norm A < 1$.
Let $I$ be the identity mapping.
Let $\circ$ be the compositio... | === $\paren {I - A} \circ P_n = I - A^{2^{n + 1} }$ ===
This will be a proof by induction. | Let $X$ be a [[Definition:Banach Space|Banach space]].
Let $\map \LL X$ be the [[Definition:Set of All Linear Transformations/Vector Space|set of all linear transformations]].
Let $\map {CL} X$ be a [[Definition:Continuous Linear Transformation Space|continuous linear transformation sapce]].
Let $\norm {\, \cdot \,}... | === $\paren {I - A} \circ P_n = I - A^{2^{n + 1} }$ ===
This will be a [[Principle of Mathematical Induction|proof by induction]]. | Properties of Product of Identity plus Operator Raised to Powers of 2 | https://proofwiki.org/wiki/Properties_of_Product_of_Identity_plus_Operator_Raised_to_Powers_of_2 | https://proofwiki.org/wiki/Properties_of_Product_of_Identity_plus_Operator_Raised_to_Powers_of_2 | [
"Continuous Linear Transformations",
"Banach Spaces",
"Proofs by Induction",
"Convergent Sequences",
"Inverse Mappings"
] | [
"Definition:Banach Space",
"Definition:Set of All Linear Transformations/Vector Space",
"Definition:Continuous Linear Transformation Space",
"Definition:Supremum Operator Norm",
"Definition:Identity Mapping",
"Definition:Composition of Mappings",
"Definition:Sequence",
"Definition:Convergent Sequence"... | [
"Principle of Mathematical Induction"
] |
proofwiki-21078 | Locally Convex Space is Topological Vector Space/Corollary | Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \PP}$ be a Hausdorff locally convex space over $\GF$ equipped with the standard topology $\tau$.
Then $\struct {X, \tau}$ is a Hausdorff topological vector space. | From Locally Convex Space is Topological Vector Space, $\struct {X, \tau}$ is a topological vector space.
From Locally Convex Space is Hausdorff iff induces Hausdorff Topology, $\struct {X, \tau}$ is a Hausdorff space.
So $\struct {X, \tau}$ is a Hausdorff topological vector space.
{{qed}}
Category:Hausdorff Topologica... | Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \PP}$ be a [[Definition:Locally Convex Space/Hausdorff|Hausdorff locally convex space]] over $\GF$ equipped with the [[Definition:Locally Convex Space/Standard Topology|standard topology]] $\tau$.
Then $\struct {X, \tau}$ is a [[Definition:Hausdorff Topological Vector ... | From [[Locally Convex Space is Topological Vector Space]], $\struct {X, \tau}$ is a [[Definition:Topological Vector Space|topological vector space]].
From [[Locally Convex Space is Hausdorff iff induces Hausdorff Topology]], $\struct {X, \tau}$ is a [[Definition:Hausdorff Space|Hausdorff space]].
So $\struct {X, \tau... | Locally Convex Space is Topological Vector Space/Corollary | https://proofwiki.org/wiki/Locally_Convex_Space_is_Topological_Vector_Space/Corollary | https://proofwiki.org/wiki/Locally_Convex_Space_is_Topological_Vector_Space/Corollary | [
"Hausdorff Topological Vector Spaces",
"Locally Convex Space is Topological Vector Space"
] | [
"Definition:Locally Convex Space/Hausdorff",
"Definition:Locally Convex Space/Standard Topology",
"Definition:Hausdorff Topological Vector Space"
] | [
"Locally Convex Space is Topological Vector Space",
"Definition:Topological Vector Space",
"Locally Convex Space is Hausdorff iff induces Hausdorff Topology",
"Definition:T2 Space",
"Definition:Hausdorff Topological Vector Space",
"Category:Hausdorff Topological Vector Spaces",
"Category:Locally Convex ... |
proofwiki-21079 | Topological Vector Space is Hausdorff iff T1 | Let $K$ be a topological field.
Let $\struct {X, \tau}$ be a topological vector space over $K$.
{{TFAE}}
:$(1): \quad \struct {X, \tau}$ is Hausdorff
:$(2): \quad \struct {X, \tau}$ is a $T_1$ space
:$(3): \quad$ for each $x \in X \setminus \set { {\mathbf 0}_X}$, there exists an open neighborhood $U_x$ of ${\mathbf 0}... | === $(1)$ implies $(2)$ ===
This is precisely $T_2$ Space is $T_1$ Space.
{{qed|lemma}} | Let $K$ be a [[Definition:Topological Field|topological field]].
Let $\struct {X, \tau}$ be a [[Definition:Topological Vector Space|topological vector space]] over $K$.
{{TFAE}}
:$(1): \quad \struct {X, \tau}$ is [[Definition:Hausdorff Space|Hausdorff]]
:$(2): \quad \struct {X, \tau}$ is a [[Definition:T1 Space|$T_1... | === $(1)$ implies $(2)$ ===
This is precisely [[T2 Space is T1 Space|$T_2$ Space is $T_1$ Space]].
{{qed|lemma}} | Topological Vector Space is Hausdorff iff T1 | https://proofwiki.org/wiki/Topological_Vector_Space_is_Hausdorff_iff_T1 | https://proofwiki.org/wiki/Topological_Vector_Space_is_Hausdorff_iff_T1 | [
"Hausdorff Spaces",
"T1 Spaces",
"Topological Vector Spaces"
] | [
"Definition:Topological Field",
"Definition:Topological Vector Space",
"Definition:T2 Space",
"Definition:T1 Space",
"Definition:Open Neighborhood"
] | [
"T2 Space is T1"
] |
proofwiki-21080 | Equivalence of Definitions of Hausdorff Topological Vector Space | Let $K$ be a topological field.
Let $\struct {X, \tau}$ be a topological vector space over $K$.
{{TFAE|def = Hausdorff Topological Vector Space}}
=== Definition 1 ===
{{:Definition:Hausdorff Topological Vector Space/Definition 1}}
=== Definition 2 ===
{{:Definition:Hausdorff Topological Vector Space/Definition 2}} | From Topological Vector Space is Hausdorff iff T1, $\struct {X, \tau}$ is Hausdorff {{iff}} it is $T_1$ space.
From definition 3 of a $T_1$ space, it follows that $\struct {X, \tau}$ is Hausdorff {{iff}}:
:for each $x \in X$, $\set x$ is closed in $\struct {X, \tau}$.
{{qed}}
Category:Hausdorff Topological Vector Space... | Let $K$ be a [[Definition:Topological Field|topological field]].
Let $\struct {X, \tau}$ be a [[Definition:Topological Vector Space|topological vector space]] over $K$.
{{TFAE|def = Hausdorff Topological Vector Space}}
=== [[Definition:Hausdorff Topological Vector Space/Definition 1|Definition 1]] ===
{{:Definitio... | From [[Topological Vector Space is Hausdorff iff T1]], $\struct {X, \tau}$ is [[Definition:Hausdorff Space|Hausdorff]] {{iff}} it is [[Definition:T1 Space|$T_1$ space]].
From [[Definition:T1 Space/Definition 3|definition 3 of a $T_1$ space]], it follows that $\struct {X, \tau}$ is [[Definition:Hausdorff Space|Hausdorf... | Equivalence of Definitions of Hausdorff Topological Vector Space | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Hausdorff_Topological_Vector_Space | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Hausdorff_Topological_Vector_Space | [
"Hausdorff Topological Vector Spaces"
] | [
"Definition:Topological Field",
"Definition:Topological Vector Space",
"Definition:Hausdorff Topological Vector Space/Definition 1",
"Definition:Hausdorff Topological Vector Space/Definition 2"
] | [
"Topological Vector Space is Hausdorff iff T1",
"Definition:T2 Space",
"Definition:T1 Space",
"Definition:T1 Space/Definition 3",
"Definition:T2 Space",
"Definition:Closed Set",
"Category:Hausdorff Topological Vector Spaces"
] |
proofwiki-21081 | Characterization of Hausdorff Topological Vector Space | Let $K$ be a topological field.
Let $\struct {X, \tau}$ be a topological vector space.
{{TFAE}}
{{begin-itemize}}
{{item|(1):|$\struct {X, \tau}$ is a Hausdorff topological vector space}}
{{item|(2):|the intersection of all open neighborhoods of ${\mathbf 0}_X$ in $\struct {X, \tau}$ is $\set { {\mathbf 0}_X}$}}
{{item... | === $(1)$ implies $(2)$ ===
Suppose that $\struct {X, \tau}$ is a Hausdorff topological vector space.
Let $\FF$ be the set of open neighborhoods of ${\mathbf 0}_X$.
From Topological Vector Space is Hausdorff iff $T_1$, $\struct {X, \tau}$ is $T_1$.
Clearly:
:$\set { {\mathbf 0}_X} \subseteq \bigcap \FF$
Now let $x \in ... | Let $K$ be a [[Definition:Topological Field|topological field]].
Let $\struct {X, \tau}$ be a [[Definition:Topological Vector Space|topological vector space]].
{{TFAE}}
{{begin-itemize}}
{{item|(1):|$\struct {X, \tau}$ is a [[Definition:Hausdorff Topological Vector Space|Hausdorff topological vector space]]}}
{{item... | === $(1)$ implies $(2)$ ===
Suppose that $\struct {X, \tau}$ is a [[Definition:Hausdorff Topological Vector Space|Hausdorff topological vector space]].
Let $\FF$ be the [[Definition:Set|set]] of [[Definition:Open Neighborhood|open neighborhoods]] of ${\mathbf 0}_X$.
From [[Topological Vector Space is Hausdorff iff T... | Characterization of Hausdorff Topological Vector Space | https://proofwiki.org/wiki/Characterization_of_Hausdorff_Topological_Vector_Space | https://proofwiki.org/wiki/Characterization_of_Hausdorff_Topological_Vector_Space | [
"Hausdorff Topological Vector Spaces"
] | [
"Definition:Topological Field",
"Definition:Topological Vector Space",
"Definition:Hausdorff Topological Vector Space",
"Definition:Set Intersection",
"Definition:Open Neighborhood",
"Definition:Closed Set"
] | [
"Definition:Hausdorff Topological Vector Space",
"Definition:Set",
"Definition:Open Neighborhood",
"Topological Vector Space is Hausdorff iff T1",
"Definition:T1 Space",
"Definition:Open Neighborhood",
"Definition:Set",
"Definition:Open Neighborhood",
"Definition:Open Neighborhood",
"Definition:Op... |
proofwiki-21082 | Dilation of Complement of Set in Vector Space | Let $K$ be a field.
Let $X$ be a vector space over $K$.
Let $F \subseteq X$ be a non-empty set.
Let $\lambda \in K$.
Then:
:$X \setminus \paren {\lambda F} = \lambda \paren {X \setminus F}$ | It is immediate from the definition of a dilation that if $x \in F$ we have $\lambda x \in \lambda F$.
Conversely, if $\lambda x \in \lambda F$, we have $\lambda x = \lambda y$ for some $y \in F$.
That is, $\lambda \paren {x - y} = 0$.
Since $\lambda \ne 0$, it follows that $x = y$, and so $x \in F$.
So we have $x \i... | Let $K$ be a [[Definition:Field (Abstract Algebra)|field]].
Let $X$ be a [[Definition:Vector Space|vector space]] over $K$.
Let $F \subseteq X$ be a [[Definition:Non-Empty Set|non-empty set]].
Let $\lambda \in K$.
Then:
:$X \setminus \paren {\lambda F} = \lambda \paren {X \setminus F}$ | It is immediate from the definition of a [[Definition:Dilation of Subset of Vector Space|dilation]] that if $x \in F$ we have $\lambda x \in \lambda F$.
Conversely, if $\lambda x \in \lambda F$, we have $\lambda x = \lambda y$ for some $y \in F$.
That is, $\lambda \paren {x - y} = 0$.
Since $\lambda \ne 0$, it fol... | Dilation of Complement of Set in Vector Space | https://proofwiki.org/wiki/Dilation_of_Complement_of_Set_in_Vector_Space | https://proofwiki.org/wiki/Dilation_of_Complement_of_Set_in_Vector_Space | [
"Dilations of Subsets of Vector Spaces"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Vector Space",
"Definition:Non-Empty Set"
] | [
"Definition:Linear Combination of Subsets of Vector Space/Dilation",
"Category:Dilations of Subsets of Vector Spaces"
] |
proofwiki-21083 | Translation of Complement of Set in Vector Space | Let $K$ be a field.
Let $X$ be a vector space over $K$.
Let $F \subseteq X$ be a non-empty set.
Let $x \in X$.
Then:
:$X \setminus \paren {F + x} = \paren {X \setminus F} + x$ | It is immediate that if:
:$y \in F$
we have:
:$y + x \in F + x$.
Conversely, if:
:$y + x \in F + x$
then:
:$y + x = u + x$
for some $u \in F$.
That is:
:$y = u$
and so:
:$y \in F$.
Hence we have:
:$y \in F$ {{iff}} $y + x \in F + x$
Hence for $y \in X$ we have:
:$y + x \not \in F + x$ {{iff}} $y \not \in F$
That is:
:$... | Let $K$ be a [[Definition:Field (Abstract Algebra)|field]].
Let $X$ be a [[Definition:Vector Space|vector space]] over $K$.
Let $F \subseteq X$ be a [[Definition:Non-Empty Set|non-empty set]].
Let $x \in X$.
Then:
:$X \setminus \paren {F + x} = \paren {X \setminus F} + x$ | It is immediate that if:
:$y \in F$
we have:
:$y + x \in F + x$.
Conversely, if:
:$y + x \in F + x$
then:
:$y + x = u + x$
for some $u \in F$.
That is:
:$y = u$
and so:
:$y \in F$.
Hence we have:
:$y \in F$ {{iff}} $y + x \in F + x$
Hence for $y \in X$ we have:
:$y + x \not \in F + x$ {{iff}} $y \not \in F$
That ... | Translation of Complement of Set in Vector Space | https://proofwiki.org/wiki/Translation_of_Complement_of_Set_in_Vector_Space | https://proofwiki.org/wiki/Translation_of_Complement_of_Set_in_Vector_Space | [
"Translation of Subsets of Vector Spaces"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Vector Space",
"Definition:Non-Empty Set"
] | [
"Category:Translation of Subsets of Vector Spaces"
] |
proofwiki-21084 | Translation of Closed Set in Topological Vector Space is Closed Set | Let $K$ be a topological field.
Let $X$ be a topological vector space over $K$.
Let $F$ be a closed set in $X$.
Let $x \in X$.
Then $F + x$ is a closed set in $X$. | We aim to show that $X \setminus \paren {F + x}$ is open.
We are given $F$ is closed in $X$.
So, $X \setminus F$ is open.
It follows from Translation of Open Set in Topological Vector Space is Open that $\paren {X \setminus F} + x$ is open.
From Translation of Complement of Set in Vector Space, we have:
:$X \setminus \... | Let $K$ be a [[Definition:Topological Field|topological field]].
Let $X$ be a [[Definition:Topological Vector Space|topological vector space]] over $K$.
Let $F$ be a [[Definition:Closed Set (Topology)|closed set]] in $X$.
Let $x \in X$.
Then $F + x$ is a [[Definition:Closed Set (Topology)|closed set]] in $X$. | We aim to show that $X \setminus \paren {F + x}$ is [[Definition:Open Set|open]].
We are [[Definition:Given|given]] $F$ is [[Definition:Closed Set|closed]] in $X$.
So, $X \setminus F$ is [[Definition:Open Set|open]].
It follows from [[Translation of Open Set in Topological Vector Space is Open]] that $\paren {X \set... | Translation of Closed Set in Topological Vector Space is Closed Set/Proof 1 | https://proofwiki.org/wiki/Translation_of_Closed_Set_in_Topological_Vector_Space_is_Closed_Set | https://proofwiki.org/wiki/Translation_of_Closed_Set_in_Topological_Vector_Space_is_Closed_Set/Proof_1 | [
"Translation of Subsets of Vector Spaces",
"Topological Vector Spaces",
"Translation of Closed Set in Topological Vector Space is Closed Set"
] | [
"Definition:Topological Field",
"Definition:Topological Vector Space",
"Definition:Closed Set/Topology",
"Definition:Closed Set/Topology"
] | [
"Definition:Open Set",
"Definition:Given",
"Definition:Closed Set",
"Definition:Open Set",
"Translation of Open Set in Topological Vector Space is Open",
"Definition:Open Set",
"Translation of Complement of Set in Vector Space",
"Definition:Open Set",
"Definition:Open Set",
"Definition:Closed Set"... |
proofwiki-21085 | Translation of Closed Set in Topological Vector Space is Closed Set | Let $K$ be a topological field.
Let $X$ be a topological vector space over $K$.
Let $F$ be a closed set in $X$.
Let $x \in X$.
Then $F + x$ is a closed set in $X$. | Define a mapping $T_{-x} : X \to X$ by:
:$\map {c_\lambda} y = y + x$
for each $y \in X$.
From Translation Mapping on Topological Vector Space is Homeomorphism, $T_{-x}$ is a homeomorphism.
From Definition 4 of a homeomorphism, $T_{-x}$ is therefore a closed mapping.
Hence $T_{-x} \sqbrk F = F + x$ is closed.
{{qed}} | Let $K$ be a [[Definition:Topological Field|topological field]].
Let $X$ be a [[Definition:Topological Vector Space|topological vector space]] over $K$.
Let $F$ be a [[Definition:Closed Set (Topology)|closed set]] in $X$.
Let $x \in X$.
Then $F + x$ is a [[Definition:Closed Set (Topology)|closed set]] in $X$. | Define a [[Definition:Mapping|mapping]] $T_{-x} : X \to X$ by:
:$\map {c_\lambda} y = y + x$
for each $y \in X$.
From [[Translation Mapping on Topological Vector Space is Homeomorphism]], $T_{-x}$ is a [[Definition:Homeomorphism|homeomorphism]].
From [[Definition:Homeomorphism/Topological Spaces/Definition 4|Definiti... | Translation of Closed Set in Topological Vector Space is Closed Set/Proof 2 | https://proofwiki.org/wiki/Translation_of_Closed_Set_in_Topological_Vector_Space_is_Closed_Set | https://proofwiki.org/wiki/Translation_of_Closed_Set_in_Topological_Vector_Space_is_Closed_Set/Proof_2 | [
"Translation of Subsets of Vector Spaces",
"Topological Vector Spaces",
"Translation of Closed Set in Topological Vector Space is Closed Set"
] | [
"Definition:Topological Field",
"Definition:Topological Vector Space",
"Definition:Closed Set/Topology",
"Definition:Closed Set/Topology"
] | [
"Definition:Mapping",
"Translation Mapping on Topological Vector Space is Homeomorphism",
"Definition:Homeomorphism",
"Definition:Homeomorphism/Topological Spaces/Definition 4",
"Definition:Closed Mapping",
"Definition:Closed Set"
] |
proofwiki-21086 | Image of Linear Combination of Subsets of Vector Space under Linear Transformation | Let $K$ be a field.
Let $X$ and $Y$ be vector spaces over $K$.
Let $T : X \to Y$ be a linear transformation.
Let $\lambda, \mu \in K$.
Let $A, B \subseteq X$.
Then:
:$T \sqbrk {\lambda A + \mu B} = \lambda T \sqbrk A + \mu T \sqbrk B$ | We have:
{{begin-eqn}}
{{eqn | l = T \sqbrk {\bigcup_{x \in A} \paren {\lambda x + \mu B} }
| r = \bigcup_{x \in A} T \sqbrk {\lambda x + \mu B}
| c = Image of Union under Mapping
}}
{{eqn | r = \bigcup_{x \in A} \paren {\lambda T x + T \sqbrk {\mu B} }
| c = Image of Translation of Set under Linear Transformati... | Let $K$ be a [[Definition:Field (Abstract Algebra)|field]].
Let $X$ and $Y$ be [[Definition:Vector Space|vector spaces]] over $K$.
Let $T : X \to Y$ be a [[Definition:Linear Transformation|linear transformation]].
Let $\lambda, \mu \in K$.
Let $A, B \subseteq X$.
Then:
:$T \sqbrk {\lambda A + \mu B} = \lambda T \... | We have:
{{begin-eqn}}
{{eqn | l = T \sqbrk {\bigcup_{x \in A} \paren {\lambda x + \mu B} }
| r = \bigcup_{x \in A} T \sqbrk {\lambda x + \mu B}
| c = [[Image of Union under Mapping]]
}}
{{eqn | r = \bigcup_{x \in A} \paren {\lambda T x + T \sqbrk {\mu B} }
| c = [[Image of Translation of Set under Linear Transf... | Image of Linear Combination of Subsets of Vector Space under Linear Transformation | https://proofwiki.org/wiki/Image_of_Linear_Combination_of_Subsets_of_Vector_Space_under_Linear_Transformation | https://proofwiki.org/wiki/Image_of_Linear_Combination_of_Subsets_of_Vector_Space_under_Linear_Transformation | [
"Linear Combination of Subsets of Vector Spaces"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Vector Space",
"Definition:Linear Transformation"
] | [
"Image of Union under Mapping",
"Image of Translation of Set under Linear Transformation is Translation of Image",
"Translation of Union of Subsets of Vector Space",
"Image of Dilation of Set under Linear Transformation is Dilation of Image",
"Image of Union under Mapping",
"Image of Dilation of Set under... |
proofwiki-21087 | Quotient Topological Vector Space is Hausdorff iff Linear Subspace is Closed | Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \tau}$ be a topological vector space over $\GF$.
Let $N$ be a linear subspace of $X$.
Let $X/N$ be the quotient vector space of $X$ modulo $N$.
Let $\tau_N$ be the quotient topology on $X/N$.
Then $\struct {X/N, \tau_N}$ is Hausdorff {{iff}}:
:$N$ is a closed linear subspac... | Let $\pi : X \to X/N$ be the quotient mapping.
=== Necessary Condition ===
Suppose that $\struct {X/N, \tau_N}$ is Hausdorff.
From Characterization of Hausdorff Topological Vector Space, $\set { {\mathbf 0}_{X/N} }$ is closed in $\struct {X/N, \tau_N}$.
From the definition of the quotient topology, $\pi$ is continuous.... | Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \tau}$ be a [[Definition:Topological Vector Space|topological vector space]] over $\GF$.
Let $N$ be a [[Definition:Linear Subspace|linear subspace]] of $X$.
Let $X/N$ be the [[Definition:Quotient Vector Space|quotient vector space of $X$ modulo $N$]].
Let $\tau_N$ be th... | Let $\pi : X \to X/N$ be the [[Definition:Quotient Mapping|quotient mapping]].
=== Necessary Condition ===
Suppose that $\struct {X/N, \tau_N}$ is [[Definition:Hausdorff Space|Hausdorff]].
From [[Characterization of Hausdorff Topological Vector Space]], $\set { {\mathbf 0}_{X/N} }$ is [[Definition:Closed Set|closed]... | Quotient Topological Vector Space is Hausdorff iff Linear Subspace is Closed/Proof 1 | https://proofwiki.org/wiki/Quotient_Topological_Vector_Space_is_Hausdorff_iff_Linear_Subspace_is_Closed | https://proofwiki.org/wiki/Quotient_Topological_Vector_Space_is_Hausdorff_iff_Linear_Subspace_is_Closed/Proof_1 | [
"Quotient Topological Vector Space is Hausdorff iff Linear Subspace is Closed",
"Quotient Topological Vector Spaces",
"Hausdorff Topological Vector Spaces",
"Quotient Topological Vector Space is Hausdorff iff Linear Subspace is Closed"
] | [
"Definition:Topological Vector Space",
"Definition:Linear Subspace",
"Definition:Quotient Vector Space",
"Definition:Quotient Topology",
"Definition:T2 Space",
"Definition:Closed Linear Subspace"
] | [
"Definition:Quotient Mapping",
"Definition:T2 Space",
"Characterization of Hausdorff Topological Vector Space",
"Definition:Closed Set",
"Definition:Quotient Topology",
"Definition:Continuous Mapping",
"Definition:Closed Set",
"Kernel of Quotient Mapping",
"Definition:Closed Set",
"Definition:Clos... |
proofwiki-21088 | Preimage of Image of Linear Transformation | Let $K$ be a field.
Let $X$ and $Y$ be vector spaces over $K$.
Let $T : X \to Y$ be a linear transformation.
Let $A \subseteq X$ be a non-empty set.
Then $T^{-1} \sqbrk {T \sqbrk A} = \ker T + A$ | Let $x \in T^{-1} \sqbrk {T \sqbrk A}$.
Then $T x \in T \sqbrk A$.
Then there exists $y \in A$ such that $T x = T y$.
Hence we have $\map T {x - y} = 0$.
Hence $x - y \in \ker T$.
So $x \in y + \ker T$.
We have $y + \ker T \subseteq A + \ker T$, so we obtain:
:if $x \in \in T^{-1} \sqbrk {T \sqbrk A}$ then $x \in \ker... | Let $K$ be a [[Definition:Field (Abstract Algebra)|field]].
Let $X$ and $Y$ be [[Definition:Vector Space|vector spaces]] over $K$.
Let $T : X \to Y$ be a [[Definition:Linear Transformation|linear transformation]].
Let $A \subseteq X$ be a [[Definition:Non-Empty Set|non-empty set]].
Then $T^{-1} \sqbrk {T \sqbrk A}... | Let $x \in T^{-1} \sqbrk {T \sqbrk A}$.
Then $T x \in T \sqbrk A$.
Then there exists $y \in A$ such that $T x = T y$.
Hence we have $\map T {x - y} = 0$.
Hence $x - y \in \ker T$.
So $x \in y + \ker T$.
We have $y + \ker T \subseteq A + \ker T$, so we obtain:
:if $x \in \in T^{-1} \sqbrk {T \sqbrk A}$ then $x \i... | Preimage of Image of Linear Transformation | https://proofwiki.org/wiki/Preimage_of_Image_of_Linear_Transformation | https://proofwiki.org/wiki/Preimage_of_Image_of_Linear_Transformation | [
"Linear Transformations"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Vector Space",
"Definition:Linear Transformation",
"Definition:Non-Empty Set"
] | [
"Definition:Linear Transformation",
"Category:Linear Transformations"
] |
proofwiki-21089 | Sum of Closed Linear Subspace and Finite-Dimensional Subspace of Hausdorff Topological Vector Space | Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \tau}$ be a Hausdorff topological vector space over $\GF$.
Let $N$ be a closed linear subspace of $X$.
Let $F$ be a finite dimensional linear subspace of $X$.
Then $N + F$ is closed in $\struct {X, \tau}$. | Let $\struct {X/N, \tau_N}$ be the quotient topological vector space of $X$ modulo $N$.
From Characterization of Hausdorff Topological Vector Space, $X/N$ is Hausdorff.
Let $\pi : X \to X/N$ be the quotient mapping.
From Image of Linear Transformation is Submodule and Dimension of Image of Vector Space under Linear Tra... | Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \tau}$ be a [[Definition:Hausdorff Topological Vector Space|Hausdorff topological vector space]] over $\GF$.
Let $N$ be a [[Definition:Closed Linear Subspace|closed linear subspace]] of $X$.
Let $F$ be a [[Definition:Finite Dimensional Vector Space|finite dimensional lin... | Let $\struct {X/N, \tau_N}$ be the [[Definition:Quotient Topological Vector Space|quotient topological vector space of $X$ modulo $N$]].
From [[Characterization of Hausdorff Topological Vector Space]], $X/N$ is [[Definition:Hausdorff Space|Hausdorff]].
Let $\pi : X \to X/N$ be the [[Definition:Quotient Mapping|quotie... | Sum of Closed Linear Subspace and Finite-Dimensional Subspace of Hausdorff Topological Vector Space | https://proofwiki.org/wiki/Sum_of_Closed_Linear_Subspace_and_Finite-Dimensional_Subspace_of_Hausdorff_Topological_Vector_Space | https://proofwiki.org/wiki/Sum_of_Closed_Linear_Subspace_and_Finite-Dimensional_Subspace_of_Hausdorff_Topological_Vector_Space | [
"Quotient Topological Vector Spaces",
"Hausdorff Topological Vector Spaces"
] | [
"Definition:Hausdorff Topological Vector Space",
"Definition:Closed Linear Subspace",
"Definition:Dimension of Vector Space/Finite",
"Definition:Closed Set"
] | [
"Definition:Quotient Topological Vector Space",
"Characterization of Hausdorff Topological Vector Space",
"Definition:T2 Space",
"Definition:Quotient Mapping",
"Image of Linear Transformation is Submodule",
"Dimension of Image of Vector Space under Linear Transformation is Bounded Above by Dimension of Ve... |
proofwiki-21090 | Consistency Principle for Binary Mess | Let $S$ be a set.
Let $M$ be a binary mess on $S$.
Then there exists a mapping $f : S \to \Bbb B$ such that:
:$f$ is consistent with $M$ | {{explain|There's a lot going on here. I'm sure that at least some of it is more general than what is needed to prove the result, so should be pulled out and turned into separate results.}}
{{Proofread}}
Let $I$ be the set of all finite subsets of $S$.
Recall that by definition of binary mess:
:$\ds M \subseteq \bigcup... | Let $S$ be a [[Definition:Set|set]].
Let $M$ be a [[Definition:Binary Mess|binary mess]] on $S$.
Then there exists a [[Definition:Mapping|mapping]] $f : S \to \Bbb B$ such that:
:$f$ is [[Definition:Consistent Mapping|consistent]] with $M$ | {{explain|There's a lot going on here. I'm sure that at least some of it is more general than what is needed to prove the result, so should be pulled out and turned into separate results.}}
{{Proofread}}
Let $I$ be the [[Definition:Set|set]] of all [[Definition:Finite Subset|finite subsets]] of $S$.
Recall that by d... | Consistency Principle for Binary Mess | https://proofwiki.org/wiki/Consistency_Principle_for_Binary_Mess | https://proofwiki.org/wiki/Consistency_Principle_for_Binary_Mess | [
"Binary Messes"
] | [
"Definition:Set",
"Definition:Binary Mess",
"Definition:Mapping",
"Definition:Binary Mess/Consistent Mapping"
] | [
"Definition:Set",
"Definition:Finite Subset",
"Definition:Binary Mess",
"Definition:Set",
"Definition:Mapping",
"Definition:Finite Subset",
"Definition:Boolean Domain",
"Axiom:Binary Mess Axioms",
"Definition:Empty Set",
"Definition:Set of All Mappings",
"Definition:Binary Mess",
"Definition:F... |
proofwiki-21091 | Direct Product of Topological Vector Spaces is Hausdorff iff Hausdorff Factor Spaces | Let $K$ be a topological field.
Let $I$ be a set.
Let $\family {X_i}_{i \in I}$ be an $I$-indexed family of topological vector spaces over $K$.
Let:
:$\ds X = \prod_{i \mathop \in I} X_i$
be the direct product of $\family {X_i}_{i \in I}$.
Equip $X$ with the product topology.
Then $X$ is Hausdorff {{iff}} $X_i$ is Hau... | For each $i \in I$, let $\FF_i$ be the set of open neighborhoods of ${\mathbf 0}_{X_i}$, where ${\mathbf 0}_{X_i}$ is the zero vector of $X_i$.
Let $\FF$ be the set of open neighborhoods of ${\mathbf 0}_X$. | Let $K$ be a [[Definition:Topological Field|topological field]].
Let $I$ be a [[Definition:Set|set]].
Let $\family {X_i}_{i \in I}$ be an [[Definition:Indexed Family of Sets|$I$-indexed family]] of [[Definition:Topological Vector Space|topological vector spaces]] over $K$.
Let:
:$\ds X = \prod_{i \mathop \in I} X_i... | For each $i \in I$, let $\FF_i$ be the [[Definition:Set|set]] of [[Definition:Open Neighborhood|open neighborhoods]] of ${\mathbf 0}_{X_i}$, where ${\mathbf 0}_{X_i}$ is the [[Definition:Zero Vector|zero vector]] of $X_i$.
Let $\FF$ be the [[Definition:Set|set]] of [[Definition:Open Neighborhood|open neighborhoods]] o... | Direct Product of Topological Vector Spaces is Hausdorff iff Hausdorff Factor Spaces | https://proofwiki.org/wiki/Direct_Product_of_Topological_Vector_Spaces_is_Hausdorff_iff_Hausdorff_Factor_Spaces | https://proofwiki.org/wiki/Direct_Product_of_Topological_Vector_Spaces_is_Hausdorff_iff_Hausdorff_Factor_Spaces | [
"Direct Product of Vector Spaces",
"Topological Vector Spaces",
"Hausdorff Topological Vector Spaces",
"Direct Product of Topological Vector Spaces is Hausdorff iff Hausdorff Factor Spaces"
] | [
"Definition:Topological Field",
"Definition:Set",
"Definition:Indexing Set/Family of Sets",
"Definition:Topological Vector Space",
"Definition:Module Direct Product",
"Definition:Product Topology",
"Definition:T2 Space",
"Definition:T2 Space"
] | [
"Definition:Set",
"Definition:Open Neighborhood",
"Definition:Zero Vector",
"Definition:Set",
"Definition:Open Neighborhood"
] |
proofwiki-21092 | Radius of Curvature in Whewell Form | The '''radius of curvature''' $\kappa$ of $C$ at a point $P$ can be expressed in the form of a Whewell equation as:
:$\rho = \size {\dfrac {\d s} {\d \psi} }$
where:
:$s$ is the arc length of $C$
:$\psi$ is the turning angle of $C$
:$\size {\, \cdot \,}$ denotes the absolute value function. | By definition, the radius of curvature $\rho$ is given by:
:$\rho = \dfrac 1 {\size \kappa}$
where $\kappa$ is the curvature, given in Whewell form as:
:$\kappa = \dfrac {\d \psi} {\d s}$
Hence the result.
{{qed}} | The '''[[Definition:Radius of Curvature|radius of curvature]]''' $\kappa$ of $C$ at a [[Definition:Point|point]] $P$ can be expressed in the form of a [[Definition:Whewell Equation|Whewell equation]] as:
:$\rho = \size {\dfrac {\d s} {\d \psi} }$
where:
:$s$ is the [[Definition:Arc Length|arc length]] of $C$
:$\psi$ i... | By definition, the [[Definition:Radius of Curvature|radius of curvature]] $\rho$ is given by:
:$\rho = \dfrac 1 {\size \kappa}$
where $\kappa$ is the [[Definition:Curvature|curvature]], given in [[Definition:Curvature/Whewell Form|Whewell form]] as:
:$\kappa = \dfrac {\d \psi} {\d s}$
Hence the result.
{{qed}} | Radius of Curvature in Whewell Form | https://proofwiki.org/wiki/Radius_of_Curvature_in_Whewell_Form | https://proofwiki.org/wiki/Radius_of_Curvature_in_Whewell_Form | [
"Radius of Curvature"
] | [
"Definition:Radius of Curvature",
"Definition:Point",
"Definition:Intrinsic Equation/Whewell Equation",
"Definition:Arc Length",
"Definition:Turning Angle",
"Definition:Absolute Value"
] | [
"Definition:Radius of Curvature",
"Definition:Curvature",
"Definition:Curvature/Whewell Form"
] |
proofwiki-21093 | Taylor's Theorem/Two Variables | Let $f: \R^2 \to \R$ be a real-valued function which is appropriately differentiable in a neighborhood of the point $\tuple {a, b}$.
Then for $\sqrt {h^2 + k^2} < R$ for some $R \in \R$:
{{begin-eqn}}
{{eqn | l = \map f {a + h, b + k}
| r = \frac 1 {0!} \map f {a, b}
| c =
}}
{{eqn | o =
| ro= +
... | {{ProofWanted}}
{{Namedfor|Brook Taylor}} | Let $f: \R^2 \to \R$ be a [[Definition:Real-Valued Function|real-valued function]] which is appropriately [[Definition:Differentiability Class|differentiable]] in a [[Definition:Neighborhood|neighborhood]] of the [[Definition:Point|point]] $\tuple {a, b}$.
Then for $\sqrt {h^2 + k^2} < R$ for some $R \in \R$:
{{begin... | {{ProofWanted}}
{{Namedfor|Brook Taylor}} | Taylor's Theorem/Two Variables | https://proofwiki.org/wiki/Taylor's_Theorem/Two_Variables | https://proofwiki.org/wiki/Taylor's_Theorem/Two_Variables | [
"Taylor's Theorem"
] | [
"Definition:Real-Valued Function",
"Definition:Differentiability Class",
"Definition:Neighborhood",
"Definition:Point"
] | [] |
proofwiki-21094 | Chain Rule for Partial Derivatives | Let $F: \R^2 \to \R$ be a real-valued function of $2$ variables.
Let $X: \R^2 \to \R$ and $Y: \R^2 \to \R$ also be real-valued functions of $2$ variables.
Let $F = \map f {x, y}$ be such that:
{{begin-eqn}}
{{eqn | l = x
| r = \map X {u, v}
}}
{{eqn | l = y
| r = \map Y {u, v}
}}
{{end-eqn}}
Then:
:$F = \ma... | {{ProofWanted|Discussed here: https://www.ma.imperial.ac.uk/~jdg/AECHAIN.PDF which is appropriate considering this result I'm reading from a reference book written by staff at Imperial.<br/>Use Cauchy-Riemann Equations (a special case) as a basis for this, and make that page dependent upon this.}} | Let $F: \R^2 \to \R$ be a [[Definition:Real-Valued Function|real-valued function]] of $2$ [[Definition:Independent Variable|variables]].
Let $X: \R^2 \to \R$ and $Y: \R^2 \to \R$ also be [[Definition:Real-Valued Function|real-valued functions]] of $2$ [[Definition:Independent Variable|variables]].
Let $F = \map f {x... | {{ProofWanted|Discussed here: https://www.ma.imperial.ac.uk/~jdg/AECHAIN.PDF which is appropriate considering this result I'm reading from a reference book written by staff at Imperial.<br/>Use [[Cauchy-Riemann Equations]] (a special case) as a basis for this, and make that page dependent upon this.}} | Chain Rule for Partial Derivatives | https://proofwiki.org/wiki/Chain_Rule_for_Partial_Derivatives | https://proofwiki.org/wiki/Chain_Rule_for_Partial_Derivatives | [
"Chain Rule for Partial Derivatives",
"Partial Differentiation"
] | [
"Definition:Real-Valued Function",
"Definition:Independent Variable",
"Definition:Real-Valued Function",
"Definition:Independent Variable"
] | [
"Cauchy-Riemann Equations"
] |
proofwiki-21095 | Almost Convergent Sequence/Examples/Sequence of alternating zeros and ones converges almost to one half | Let $\sequence {x_n}_{n \in \N}$ be the sequence defined by:
:$x_n = \begin{cases} 0 & : n \equiv 0 \pmod 2 \\ 1 & : n \equiv 1 \pmod 2 \end{cases}$
where $\bmod$ denotes the congruence modulo.
Then $\sequence {x_n}_{n \in \N}$ almost converges to $1/2$. | Let $\phi$ be a Banach limit.
Let $S$ be the left shift operator on $\map {\ell^\infty} \R$.
Let $\mathbf 1 := \sequence {1, 1, 1, \ldots}$.
Then:
{{begin-eqn}}
{{eqn | l = \mathbf 1
| r = \sequence {1, 0, 1, 0, \ldots} + \sequence {0, 1, 0, 1, \ldots}
}}
{{eqn | r = \map S {\sequence {x_n} } + \sequence {x_n}
... | Let $\sequence {x_n}_{n \in \N}$ be the [[Definition:Sequence|sequence]] defined by:
:$x_n = \begin{cases} 0 & : n \equiv 0 \pmod 2 \\ 1 & : n \equiv 1 \pmod 2 \end{cases}$
where $\bmod$ denotes the [[Definition:Congruence Modulo Integer|congruence modulo]].
Then $\sequence {x_n}_{n \in \N}$ [[Definition:Almost Conve... | Let $\phi$ be a [[Definition:Banach Limit|Banach limit]].
Let $S$ be the [[Definition:Left Shift Operator|left shift operator]] on $\map {\ell^\infty} \R$.
Let $\mathbf 1 := \sequence {1, 1, 1, \ldots}$.
Then:
{{begin-eqn}}
{{eqn | l = \mathbf 1
| r = \sequence {1, 0, 1, 0, \ldots} + \sequence {0, 1, 0, 1, \ld... | Almost Convergent Sequence/Examples/Sequence of alternating zeros and ones converges almost to one half | https://proofwiki.org/wiki/Almost_Convergent_Sequence/Examples/Sequence_of_alternating_zeros_and_ones_converges_almost_to_one_half | https://proofwiki.org/wiki/Almost_Convergent_Sequence/Examples/Sequence_of_alternating_zeros_and_ones_converges_almost_to_one_half | [
"Definitions/Banach Limits"
] | [
"Definition:Sequence",
"Definition:Congruence (Number Theory)/Integers",
"Definition:Almost Convergent Sequence"
] | [
"Definition:Banach Limit",
"Definition:Left Shift Operator",
"Definition:Banach Limit",
"Definition:Banach Limit",
"Category:Definitions/Banach Limits"
] |
proofwiki-21096 | Meet Semilattice Filter iff Ordered Set Filter | Let $\struct {S, \wedge, \preceq}$ be a meet semilattice.
Let $F \subseteq S$ be a non-empty subset of $S$.
Then:
:$F$ is a meet semilattice filter of $\struct {S, \wedge, \preceq}$ {{iff}} $F$ is an ordered set filter of $\struct {S, \preceq}$. | === Necessary Condition ===
Let $F$ be a meet semilattice filter of $\struct {S, \wedge, \preceq}$.
To show that $F$ is an ordered set filter of $\struct {S, \preceq}$ it is sufficient to show:
{{begin-axiom}}
{{axiom | q = \forall x, y \in F: \exists z \in F
| m = z \preceq x \text{ and } z \preceq y
}}
{{end-... | Let $\struct {S, \wedge, \preceq}$ be a [[Definition:Meet Semilattice|meet semilattice]].
Let $F \subseteq S$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subset]] of $S$.
Then:
:$F$ is a [[Definition:Meet Semilattice Filter|meet semilattice filter]] of $\struct {S, \wedge, \preceq}$ {{iff}} $F$ i... | === Necessary Condition ===
Let $F$ be a [[Definition:Meet Semilattice Filter|meet semilattice filter]] of $\struct {S, \wedge, \preceq}$.
To show that $F$ is an [[Definition:Filter|ordered set filter]] of $\struct {S, \preceq}$ it is sufficient to show:
{{begin-axiom}}
{{axiom | q = \forall x, y \in F: \exists z \in... | Meet Semilattice Filter iff Ordered Set Filter | https://proofwiki.org/wiki/Meet_Semilattice_Filter_iff_Ordered_Set_Filter | https://proofwiki.org/wiki/Meet_Semilattice_Filter_iff_Ordered_Set_Filter | [
"Meet Semilattices"
] | [
"Definition:Meet Semilattice",
"Definition:Non-Empty Set",
"Definition:Subset",
"Definition:Meet Semilattice Filter",
"Definition:Filter"
] | [
"Definition:Meet Semilattice Filter",
"Definition:Filter",
"Definition:Meet Semilattice Filter",
"Definition:Subsemilattice",
"Definition:Meet (Order Theory)",
"Definition:Filter",
"Definition:Meet Semilattice Filter",
"Definition:Subsemilattice",
"Definition:Filter",
"Definition:Meet (Order Theor... |
proofwiki-21097 | Equivalence of Definitions of Lattice Filter | Let $\struct {S, \vee, \wedge, \preceq}$ be a lattice.
Let $F \subseteq S$ be a non-empty subset of $S$.
{{TFAE|def = Lattice Filter}} | === Definition 1 implies Definition 2 ===
Let $F$ satisify the lattice filter axioms.
To show that $F$ is a meet semilattice filter it is sufficient to show:
{{begin-axiom}}
{{axiom | lc= $F$ is a upper section of $S$:
| q = \forall x \in F: \forall y \in S
| m = x \preceq y \implies y \in F
}}
{{end-ax... | Let $\struct {S, \vee, \wedge, \preceq}$ be a [[Definition:Lattice (Order Theory)|lattice]].
Let $F \subseteq S$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subset]] of $S$.
{{TFAE|def = Lattice Filter}} | === Definition 1 implies Definition 2 ===
Let $F$ satisify the [[Axiom:Lattice Filter Axioms|lattice filter axioms]].
To show that $F$ is a [[Definition:Meet Semilattice Filter|meet semilattice filter]] it is sufficient to show:
{{begin-axiom}}
{{axiom | lc= $F$ is a [[Definition:Upper Section|upper section]] of $S$:... | Equivalence of Definitions of Lattice Filter | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Lattice_Filter | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Lattice_Filter | [
"Lattice Filters"
] | [
"Definition:Lattice (Order Theory)",
"Definition:Non-Empty Set",
"Definition:Subset"
] | [
"Axiom:Lattice Filter Axioms",
"Definition:Meet Semilattice Filter",
"Definition:Upper Section",
"Axiom:Lattice Filter Axioms",
"Definition:Sublattice",
"Preceding iff Join equals Larger Operand",
"Definition:Meet Semilattice Filter",
"Definition:Meet Semilattice Filter",
"Definition:Upper Section"
... |
proofwiki-21098 | Root is Commutative | Let $x \in \R_{> 0}$ be a (strictly) positive real number.
Let $a$ and $b$ be nonzero integers.
Then:
:$\sqrt [a] {\sqrt [b] x} = \sqrt [b] {\sqrt [a] x}$ | Let $y = \sqrt [a] {\sqrt [b] x}$.
Then
{{begin-eqn}}
{{eqn | l = \sqrt [a] {\sqrt [b] x}
| r = y
| c = {{hypothesis}}
}}
{{eqn | ll= \leadsto
| l = \sqrt [b] x
| r = y^a
| c = {{Defof|Root of Number}}
}}
{{eqn | ll= \leadsto
| l = x
| r = \paren {y^a}^b
| c = {{Defof|Roo... | Let $x \in \R_{> 0}$ be a [[Definition:Strictly Positive Real Number|(strictly) positive real number]].
Let $a$ and $b$ be nonzero [[Definition:Integer|integers]].
Then:
:$\sqrt [a] {\sqrt [b] x} = \sqrt [b] {\sqrt [a] x}$ | Let $y = \sqrt [a] {\sqrt [b] x}$.
Then
{{begin-eqn}}
{{eqn | l = \sqrt [a] {\sqrt [b] x}
| r = y
| c = {{hypothesis}}
}}
{{eqn | ll= \leadsto
| l = \sqrt [b] x
| r = y^a
| c = {{Defof|Root of Number}}
}}
{{eqn | ll= \leadsto
| l = x
| r = \paren {y^a}^b
| c = {{Defof|R... | Root is Commutative | https://proofwiki.org/wiki/Root_is_Commutative | https://proofwiki.org/wiki/Root_is_Commutative | [
"Roots of Numbers"
] | [
"Definition:Strictly Positive/Real Number",
"Definition:Integer"
] | [
"Product of Indices of Real Number",
"Integer Multiplication is Commutative",
"Product of Indices of Real Number",
"Existence and Uniqueness of Positive Root of Positive Real Number",
"Existence and Uniqueness of Positive Root of Positive Real Number",
"Category:Roots of Numbers"
] |
proofwiki-21099 | Force between Infinite Parallel Straight Conductors carrying Steady Current | Let $s_1$ and $s_2$ be wires in a vacuum carrying steady currents $I_1$ and $I_2$.
Let $s_1$ and $s_2$ be parallel and (effectively) infinitely long.
Then the magnitude of the force per unit distance between $s_1$ and $s_2$ is given by:
:$\dfrac {\d F} {\d l} = \dfrac {\mu_0 I_1 I_2} {2 \pi r}$
where:
:$r$ denotes the ... | Let $s_1$ lie along the $z$ axis.
Define $I_1$ and $I_2$ to be positive when flowing in the positive $z$ direction and negative otherwise.
Let $s_2$ pass through the point $\tuple {r, 0, 0}$.
From Magnetic Field of Infinite Straight Conductor carrying Steady Current, we have:
:$\map {\mathbf B} {\mathbf r} = \dfrac {\m... | Let $s_1$ and $s_2$ be [[Definition:Wire (Electricity)|wires]] in a [[Definition:Vacuum|vacuum]] carrying [[Definition:Steady Current|steady currents]] $I_1$ and $I_2$.
Let $s_1$ and $s_2$ be [[Definition:Parallel Lines|parallel]] and (effectively) [[Definition:Infinite|infinitely]] [[Definition:Length (Linear Measure... | Let $s_1$ lie along the [[Definition:Z-Axis|$z$ axis]].
Define $I_1$ and $I_2$ to be [[Definition:Positive|positive]] when flowing in the [[Definition:Positive Direction|positive $z$ direction]] and [[Definition:Negative|negative]] otherwise.
Let $s_2$ pass through the [[Definition:Point|point]] $\tuple {r, 0, 0}$.
... | Force between Infinite Parallel Straight Conductors carrying Steady Current | https://proofwiki.org/wiki/Force_between_Infinite_Parallel_Straight_Conductors_carrying_Steady_Current | https://proofwiki.org/wiki/Force_between_Infinite_Parallel_Straight_Conductors_carrying_Steady_Current | [
"Magnetic Field of Infinite Straight Conductor carrying Steady Current",
"Electric Current",
"Magnetic Forces",
"Electromagnetism"
] | [
"Definition:Wire (Electricity)",
"Definition:Vacuum",
"Definition:Steady Current",
"Definition:Parallel (Geometry)/Lines",
"Definition:Infinite",
"Definition:Linear Measure/Length",
"Definition:Magnitude",
"Definition:Force",
"Definition:Linear Measure/Length",
"Definition:Distance between Paralle... | [
"Definition:Axis/Z-Axis",
"Definition:Positive",
"Definition:Axis/Positive Direction",
"Definition:Negative",
"Definition:Point",
"Magnetic Field of Infinite Straight Conductor carrying Steady Current",
"Definition:Magnetic Field",
"Magnetic Force on Conductor carrying Steady Current",
"Newton's Law... |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.