nyuuzyou/wikis huggingface dataset:
Characterization for Topological Evaluation Mapping to be Embedding/Necessary Condition

== Theorem ==
Let $X$ be a topological space.
Let $\family {Y_i}_{i \mathop \in I}$ be an indexed family of topological spaces for some indexing set $I$.
Let $\family {f_i : X \to Y_i}_{i \mathop \in I}$ be an indexed family of continuous mappings.
Let $\ds Y = \prod_{i \mathop \in I} Y_i$ be the product space of $\family {Y_i}_{i \mathop \in I}$. 
Let $f : X \to Y$ be the evaluation mapping induced by $\family{f_i}_{i \mathop \in I}$.
Let $f$ be an embedding
Then:
:$(1)\quad$ the topology on $X$ is the initial topology with respect to $\family {f_i}_{i \mathop \in I}$
:$(2)\quad$ the family $\family {f_i}$ separates points
== Proof ==
=== $(1)$ The Topology on $X$ is the Initial Topology ===
Let $f \sqbrk X$ denote the image of $f$.
Let $\tau_{f \sqbrk X}$ be the subspace topology on $f \sqbrk X$.
By definition of embedding:
:$f \restriction_{X \times f \sqbrk X}$ is a homeomorphism between $X$ and $f \sqbrk X$
From Subspace of Product Space has Initial Topology with respect to Restricted Projections:
:$\tau_{f \sqbrk X}$ is the initial topology on $f \sqbrk X$ with respect to the mappings $\family {\pr_i \restriction_{f \sqbrk X} : f \sqbrk X \to Y_i}_{i \mathop \in I}$
Let $\tau$ be the topology on $X$.
From Homeomorphic Topology of Initial Topology is Initial Topology:
:$\tau$ is the initial topology on $X$ with respect to $\family {\pr_i \restriction_{f \sqbrk X} \circ f \restriction_{X \times f \sqbrk X} : X \to Y_i}_{i \mathop \in I}$
We have:
Hence:
:$\tau$ is the initial topology on $X$ with respect to $\family {f_i : X \to Y_i}_{i \mathop \in I}$
=== $(2)$ The Family $\family {f_i}$ Separates Points ===
By definition of embedding:
:$f$ is a homeomorphism between $X$ and $f \sqbrk X$
By definition of homeomorphism:
:$f$ is an injection
From Evaluation Mapping is Injective iff Mappings Separate Points:
:the family $\family {f_i}$ separates points.
Category:Characterization for Topological Evaluation Mapping to be Embedding
Characterization for Topological Evaluation Mapping to be Embedding/Sufficient Condition

== Theorem ==
Let $X$ be a topological space.
Let $\family {Y_i}_{i \mathop \in I}$ be an indexed family of topological spaces for some indexing set $I$.
Let $\family {f_i : X \to Y_i}_{i \mathop \in I}$ be an indexed family of continuous mappings.
Let $\ds Y = \prod_{i \mathop \in I} Y_i$ be the product space of $\family {Y_i}_{i \mathop \in I}$. 
Let $f : X \to Y$ be the evaluation mapping induced by $\family{f_i}_{i \mathop \in I}$.
Let:
:$(1)\quad$ the topology on $X$ be the initial topology with respect to $\family {f_i}_{i \mathop \in I}$
:$(2)\quad$ the family $\family {f_i}$ separates points
Then:
:$f$ is an embedding
== Proof ==
From Evaluation Mapping is Injective iff Mappings Separate Points:
:$f$ is an injection
From Injection to Image is Bijection:
:$f \restriction_{X \times f \sqbrk X} \mathop : X \to f \sqbrk X$ is a bijection
From Topological Evaluation Mapping is Continuous:
:$f$ is continuous
From Continuity of Composite of Inclusion on Mapping:
:$f \restriction_{X \times f \sqbrk X}$ is continuous
Let $\SS = \set{ f_i^{-1} \sqbrk V : i \in I, V \subseteq Y_i \text{ is open}}$.
Let $f_i^{-1} \sqbrk V \in \SS$ for some $i \in I, V \subseteq Y_i$ is open.
Let $\pr_i$ denote the projection from $Y$ to $Y_i$.
We have:
By definition of product topology:
:$\pr_i^{-1} \sqbrk V$ is open in $Y$
By definition of subspace:
:$f \sqbrk {f_i^{-1} \sqbrk V} = \pr_i^{-1} \sqbrk V \cap f \sqbrk X$ is open in $f \sqbrk X$
By definition of restriction:
:$f \restriction_{X \times f \sqbrk X} \sqbrk {f_i^{-1} \sqbrk V} = f \sqbrk {f_i^{-1} \sqbrk V}$
We have shown that:
:$\forall U \in \SS : f \restriction_{X \times f \sqbrk X} \sqbrk U \text{ is open in } Y$
By definition of initial topology:
:$\SS$ is a sub-basis for the topology on $X$
From Injection is Open Mapping iff Image of Sub-Basis Set is Open:
:$f \restriction_{X \times f \sqbrk X}$ is an open mapping
By definition, $f \restriction_{X \times f \sqbrk X}$ is a homeomorphism.
By definition, $f$ is an embedding.
Category:Characterization for Topological Evaluation Mapping to be Embedding
Characterization of Absolute Continuity of Complex Measure

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space.
Let $\mu$ be a measure on $\struct {X, \Sigma}$.
Let $\nu$ be a complex measure on $\struct {X, \Sigma}$.
Then $\nu$ is absolutely continuous with respect to $\mu$ : 
:for all $A \in \Sigma$ with $\map \mu A = 0$, we have $\map \nu A = 0$.
== Proof ==
Let $\tuple {\nu_1, \nu_2, \nu_3, \nu_4}$ be the Jordan decomposition of $\nu$. 
Let $\size \nu$ be the variation of $\nu$.
=== Sufficient Condition ===
Suppose that: 
:for all $A \in \Sigma$ with $\map \mu A = 0$, we have $\map \nu A = 0$.
We aim to show that: 
:for all $A \in \Sigma$ with $\map \mu A = 0$, we have $\map {\size \nu} A = 0$
which will give: 
:$\size \nu$ is absolutely continuous with respect to $\mu$
from which we will obtain: 
:$\nu$ is absolutely continuous with respect to $\mu$.
Suppose that $A \in \Sigma$ has $\map \mu A = 0$. 
From Null Sets Closed under Subset, we have: 
:$\map \mu B = 0$ for each $\Sigma$-measurable $B \subseteq A$. 
Using the assumption on each such $B$, we have: 
:for each $\Sigma$-measurable $B \subseteq A$ we have $\map \nu B = 0$.
From Characterization of Null Sets of Variation of Complex Measure, this implies that: 
:$\map {\size \nu} A = 0$
So:
:for all $A \in \Sigma$ with $\map \mu A = 0$, we have $\map {\size \nu} A = 0$
=== Necessary Condition ===
Suppose that $\nu$ is absolutely continuous with respect to $\mu$.
Then from Absolute Continuity of Complex Measure in terms of Jordan Decomposition, we have: 
:$\nu_1$, $\nu_2$, $\nu_3$ and $\nu_4$ are absolutely continuous with respect to $\mu$.
So:
:for all $A \in \Sigma$ with $\map \mu A = 0$ we have $\map {\nu_1} A = \map {\nu_2} A = \map {\nu_3} A = \map {\nu_4} A = 0$
From the definition of the Jordan decomposition, this implies: 
:$\map \nu A = \map {\nu_1} A - \map {\nu_2} A + i \paren {\map {\nu_3} A - \map {\nu_4} A} = 0$
So:
:for all $A \in \Sigma$ with $\map \mu A = 0$ we have $\map \nu A = 0$.
== Sources ==
* : $4.2$: Absolute Continuity
Category:Complex Measures
Characterization of Absolute Continuity of Signed Measure

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space.
Let $\mu$ be a measure on $\struct {X, \Sigma}$.
Let $\nu$ be a signed measure on $\struct {X, \Sigma}$.
Then $\nu$ is absolutely continuous with respect to $\mu$ : 
:for all $A \in \Sigma$ with $\map \mu A = 0$, we have $\map \nu A = 0$.
== Proof ==
Let $\tuple {\nu^+, \nu^-}$ be the Jordan decomposition of $\nu$. 
Let $\size \nu$ be the variation of $\nu$.
=== Sufficient Condition ===
Suppose that: 
:for all $A \in \Sigma$ with $\map \mu A = 0$, we have $\map \nu A = 0$.
We aim to show that: 
:for all $A \in \Sigma$ with $\map \mu A = 0$, we have $\map {\size \nu} A = 0$
which will give: 
:$\size \nu$ is absolutely continuous with respect to $\mu$
from which we will obtain: 
:$\nu$ is absolutely continuous with respect to $\mu$.
Suppose that $A \in \Sigma$ has $\map \mu A = 0$. 
From Null Sets Closed under Subset, we have: 
:$\map \mu B = 0$ for each $\Sigma$-measurable $B \subseteq A$. 
Using the assumption on each such $B$, we have: 
:for each $\Sigma$-measurable $B \subseteq A$ we have $\map \nu B = 0$.
From Characterization of Null Sets of Variation of Signed Measure, this implies that: 
:$\map {\size \nu} A = 0$
So:
:for all $A \in \Sigma$ with $\map \mu A = 0$, we have $\map {\size \nu} A = 0$
=== Necessary Condition ===
Suppose that $\nu$ is absolutely continuous with respect to $\mu$.
Then from Absolute Continuity of Signed Measure in terms of Jordan Decomposition, we have: 
:$\nu^+$ and $\nu^-$ are absolutely continuous with respect to $\mu$.
So:
:for all $A \in \Sigma$ with $\map \mu A = 0$ we have $\map {\nu^+} A = \map {\nu^-} A = 0$
From the definition of the Jordan decomposition, this implies: 
:$\map \nu A = \map {\nu^+} A - \map {\nu^-} A = 0$
So:
:for all $A \in \Sigma$ with $\map \mu A = 0$ we have $\map \nu A = 0$.
== Sources ==
* : $4.2$: Absolute Continuity
Category:Absolutely Continuous Signed Measures
Category:Absolutely Continuous Measures
Category:Signed Measures
Characterization of Affine Transformations

== Theorem ==
Let $\EE$ and $\FF$ be affine spaces over a field $K$.
Let $\LL: \EE \to \FF$ be a mapping.
Then $\LL$ is an affine transformation :
:$\forall p, q \in \EE: \forall \lambda \in K: \map \LL {\lambda p + \paren {1 - \lambda} q} = \lambda \map \LL p + \paren {1 - \lambda} \map \LL q$
where $\lambda p + \paren {1 - \lambda} q$ and $\lambda \map \LL p + \paren {1 - \lambda} \map \LL q$ denote barycenters.
== Proof ==
=== Sufficient Condition ===
Let $\LL$ be an affine transformation.
Let $L$ be the tangent map.
Let $r \in \EE$ be any point.
Then by definition we have:
:$\lambda p + \paren {1 - \lambda} q = r + \lambda \vec{r p} + \paren {1 - \lambda} \vec{r q}$
Thus we find:
=== Necessary Condition ===
Suppose that for all points $p, q \in \EE$ and all $\lambda \in \R$:
:$\map \LL {\lambda p + \paren {1 - \lambda} q} = \lambda \map \LL p + \paren {1 - \lambda} \map \LL q$
Let $E$ be the difference space of $\EE$.
Fix a point $p \in \EE$, and define for all $u \in E$:
:$\map L u = \map \LL {p + u} - \map \LL p$
Let $q = p + u$.
Then:
:$\map \LL q = \map \LL p + \map L u$
So to show that $\LL$ is affine, we are required to prove that $L$ is linear.
That is, we want to show that for all $\lambda \in k$ and all $u, v \in E$:
:$\map L {\lambda u} = \lambda \map L u$
and:
:$\map L {u + v} = \map L u + \map L v$
First of all:
Now it is to be shown that
:$\map L {u + v} = \map L u + \map L v$
First:
:$p + u + v = \dfrac 1 2 \paren {p + 2 u} + \dfrac 1 2 \paren {p + 2 v}$
Now:
From the above calculation:
:$\map L {u + v} = \map \LL {p + u + v} - \map \LL p = \map L u + \map L v$
This shows that $L$ is linear, and therefore concludes the proof.
Category:Affine Geometry
Characterization of Almost Everywhere Zero

== Theorem ==
<onlyinclude>
Let $\struct {X, \Sigma}$ be a measurable space.
Let $\mu$ be a measure on $\struct {X, \Sigma}$.
Let $f : X \to \overline \R$ be a measurable function.
Then:
:$f = 0$ $\mu$-almost everywhere
:
:$\ds \forall A \in \Sigma : \int \paren {\chi_A \cdot f} \rd \mu = 0$
where:
:$\chi_A$ is the characteristic function of $A$
== Proof ==
=== Necessary condition ===
First, by Measurable Function Zero A.E. iff Absolute Value has Zero Integral:
:$\ds \int \size f \rd \mu = 0$
Let $A\in\Sigma$.
Let $\paren {\chi_A \cdot f}^+$, $\paren {\chi_A \cdot f}^-$ be the positive and negative parts of $\chi_A \cdot f$, respectively.
Observe:
:$\paren {\chi_A \cdot f}^+ \le \size f$
and:
:$\paren {\chi_A \cdot f}^- \le \size f$
Therefore, by Integral of Positive Measurable Function is Monotone:
:$\ds \int \paren {\chi_A \cdot f}^+ \rd \mu \le \int \size f \rd \mu = 0$
and:
:$\ds \int \paren {\chi_A \cdot f}^- \rd \mu \le \int \size f \rd \mu = 0$
This means by definition of integral:
:$\ds \int \paren { \chi_A \cdot f } \rd \mu = 0$.
=== Sufficient condition ===
Similarly:
:$\map \mu {\set {-f > 1 / n} } = 0$
Altogether:
:$\map \mu {\set {\size f > 1 / n} } = 0$
Therefore:
Category:Measure Theory
Characterization of Analytic Basis by Local Bases

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $P$ be a set of subsets of $S$ such that
:$P \subseteq \tau$
and
:for all $p \in S$: there exists local basis $B$ at $p: B \subseteq P$
Then $P$ is basis of $T$.
== Proof ==
By assumption:
:$P \subseteq \tau$
Let $U$ be an open subset of $S$.
Define:
:$X := \set {V \in P: V \subseteq U}$
By definition of subset:
:$X \subseteq P$
We will prove that:
:$\forall u \in S: u \in U \iff \exists Z \in X: u \in Z$
Let $u \in S$.
We will prove that:
:$u \in U \implies \exists Z \in X: u \in Z$
Assume that:
:$u \in U$
By assumption:
:there exists local basis $B$ at $u: B \subseteq P$.
By definition of local basis:
:$\exists V \in B: V \subseteq U$
Thus by definitions of subset and $X$:
:$V \in X$
Thus by definition of local basis:
:$u \in V$
Assume that:
:$\exists Z \in X: u \in Z$
By definition of $X$:
:$Z \subseteq U$
Thus by definition of subset:
:$u \in U$
Thus by definition of union:
:$U = \bigcup X$
Hence $P$ is basis of $L$.
== Sources ==
* 
Category:Topology
Characterization of Boundary by Basis

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $\BB \subseteq \tau$ be a basis.
Let $A$ be a subset of $T$.
Let $x$ be a point of $T$.
Then $x \in \partial A$ :
:for every $U \in \BB$:
::if $x \in U$
::then $A \cap U \ne \O$ and $\relcomp S A \cap U \ne \O$
where:
:$\relcomp S A = S \setminus A$ denotes the complement of $A$ in $S$
:$\partial A$ denotes the boundary of $A$ in $T$.
== Proof ==
=== Sufficient Condition ===
Let $x \in \partial A$.
Let $U \in \BB$.
By definition of basis, $U$ is an open set of $T$.
Thus from Characterization of Boundary by Open Sets:
:if $x \in U$
::then $A \cap U \ne \O$ and $\relcomp S A \cap U \ne \O$.
=== Necessary Condition ===
Let $x$ be such that for every $U \in \BB$:
:if $x \in U$
:then $A \cap U \ne \O$ and $\relcomp S A \cap U \ne \O$.
By Characterization of Boundary by Open Sets, to prove that $x \in \partial A$ it is enough to prove
that:
:for every open set $U$ of $T$:
::if $x \in U$ then $A \cap U \ne \O$ and $\relcomp S A \cap U \ne \O$.
Let $U$ be an open set of $T$.
Let $x \in U$.
By definition of (analytic) basis, there exists $V \in \BB$ such that:
:$x \in V \subseteq U$
By assumption:
:$A \cap V \ne \O$
and:
:$\relcomp S A \cap V \ne \O$
From the corollary to Set Intersection Preserves Subsets:
:$A \cap V \subseteq A \cap U$
and:
:$\relcomp S A \cap V \subseteq \relcomp S A \cap U$
So:
:$A \cap U \ne \O$ and $\relcomp S A \cap U \ne \O$
and hence the result.
== Sources ==
* 
* 
Category:Set Boundaries
Characterization of Boundary by Open Sets

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $A$ be a subset of $T$.
Let $x$ be a point of $T$.
Then $x \in \partial A$ :
:for every open set $U$ of $T$:
::if $x \in U$
::then $A \cap U \ne \O$ and $\relcomp S A \cap U \ne \O$
where:
:$\relcomp S A = S \setminus A$ denotes the complement of $A$ in $S$
:$\partial A$ denotes the boundary of $A$.
== Proof ==
=== Sufficient Condition ===
Let $x \in \partial A$.
Then by Boundary is Intersection of Closure with Closure of Complement:
:$x \in \paren {\relcomp S A}^-$ and $x \in A^-$
where $A^-$ denotes the closure of $A$.
Hence by Condition for Point being in Closure, for every open set $U$ of $T$:
:$x \in U \implies A \cap U \ne \O$
and:
:$x \in U \implies \relcomp S A \cap U \ne \O$
=== Necessary Condition ===
Let $x$ be such that for every open set $U$ of $T$:
:if $x \in U$
:then $A \cap U \ne \O$ and $\relcomp S A \cap U \ne \O$.
Then by Condition for Point being in Closure:
:$x \in \paren {\relcomp S A}^-$ and $x \in A^-$.
Hence by Boundary is Intersection of Closure with Closure of Complement:
:$x \in \partial A$
== Sources ==
* 
* 
Category:Set Boundaries
Characterization of Cauchy Sequence in Topological Vector Space in terms of Local Basis

== Theorem ==
Let $\struct {X, \tau}$ be a topological vector space.
Let $\BB$ be a local basis for $\mathbf 0_X$ in $\struct {X, \tau}$.
Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence. 
Then $\sequence {x_n}_{n \mathop \in \N}$ is Cauchy :
:for each $V \in \BB$ there exists $N \in \N$ such that $x_n - x_m \in V$ for $n, m \ge N$. 
== Proof ==
=== Necessary Condition ===
Suppose that $\sequence {x_n}_{n \mathop \in \N}$ is Cauchy.
Then: 
:for each open neighborhood $V$ of $\mathbf 0_X$ there exists $N \in \N$ such that: 
::$x_n - x_m \in V$ for each $n, m \ge N$. 
Since $\BB$ consists of open neighbourhoods of $\mathbf 0_X$, we in particular have:
:for each $V \in \BB$ there exists $N \in \N$ such that $x_n - x_m \in V$ for $n, m \ge N$. 
=== Sufficient Condition ===
Suppose that: 
:for each $V \in \mathcal B$ there exists $N \in \N$ such that $x_n - x_m \in V$ for $n, m \ge N$. 
Let $U$ be an open neighborhood of $\mathbf 0_X$.
Then there exists $V \in \mathcal B$ such that $V \subseteq U$. 
Then there exists $N \in \N$ such that $x_n - x_m \in V \subseteq U$ for $n, m \ge N$. 
So $\sequence {x_n}_{n \mathop \in \N}$ is Cauchy.
Category:Local Bases
Category:Cauchy Sequences in Topological Vector Spaces
Characterization of Character on Banach Algebra

== Theorem ==
Let $\struct {A, \norm {\, \cdot \,} }$ be a Banach algebra over $\C$.
Let $\struct {A_+, \norm {\, \cdot \,}_{A_+} }$ be the unitization of $\struct {A, \norm {\, \cdot \,} }$.
Let $\phi : A_+ \to \C$ be a character.
Then either:
:$(1) \quad \map \phi {x, \lambda} = \lambda$ for each $\tuple {x, \lambda} \in A_+$
:$(2) \quad $ there exists a character $\widetilde \phi$ on $A$ such that $\map \phi {x, \lambda} = \map {\widetilde \phi} x + \lambda$ for each $\tuple {x, \lambda} \in A_+$.
Conversely, every map of the form given in $(2)$ is a character.
== Proof ==
Note that since $\phi$ is linear, we have:
:$\map \phi {x, \lambda} = \map \phi {x, 0} + \lambda \map \phi {0, 1}$
Then from Character on Unital Banach Algebra is Unital Algebra Homomorphism, we have:
:$\map \phi {0, 1} = 1$
so that:
:$\map \phi {x, \lambda} = \map \phi {x, 0} + \lambda$
We first show that:
:$\map \phi {x, \lambda} = \lambda$
defines a character.
Let $\tuple {x, \lambda}, \tuple {y, \mu} \in A_+$ and $t \in \C$. 
Then we have:
hence $\phi$ is linear.
We also have:
hence $\phi$ is a character.
In this case $\map \phi {x, 0} = 0$ for each $x \in A$. 
Now suppose that $\map \phi {x, 0} \ne 0$ for each $x \in A$. 
We show that:
:$\map {\widetilde \phi} x = \map \phi {x, 0}$
is a character on $A$.
For $\tuple {x, \lambda}, \tuple {y, \mu} \in A_+$ and $t \in \C$, we have:
and:
So $\widetilde \phi$ is a character on $A$ and $\phi$ has the form:
:$\map \phi {x, \lambda} = \map {\widetilde \phi} x + \lambda$
where $\widetilde \phi$ is a character on $A$.
So every character on $A_+$ has the form in $(2)$. 
Conversely, let $\widetilde \phi$ be a character on $A$.
Define:
:$\map \phi {x, \lambda} = \map {\widetilde \phi} x + \lambda$
for each $\tuple {x, \lambda} \in A_+$.
Let $\tuple {x, \lambda}, \tuple {y, \mu} \in A_+$ and $t \in \C$.
Then we have:
and:
So $\phi$ is a character.
Category:Characters (Banach Algebras)
Category:Unital Banach Algebras
Characterization of Class Membership

== Theorem ==
Let $A$ and $B$ be classes.
Then:
:$\forall A, B: \paren {A \in B \iff \exists x: \paren {A = x \land x \in B} }$
where $x$ is specifically a set.
== Proof ==
Let $V$ denote the universal class.
By Class is Subclass of Universal Class, $A \subseteq V$ and $B \subseteq V$.
By definition of universal class, every element of $V$ is a set.
Hence every element of $B$ is a set.
So if $A \in B$, then it follows that $A$ is itself a set.
Hence the result.
== Also see ==
* Definition:Universal Class
* Definition:Class (Zermelo-Fraenkel), where class membership is taken to be a definitional abbreviation
== Sources ==
* : $\S 6.3$
Category:Class Theory
Characterization of Closable Densely-Defined Linear Operators in terms of Closure of Graph

== Theorem ==
Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space over $\C$.
Consider $\HH \times \HH$ with the direct product norm $\norm {\, \cdot \,}_{\HH \times \HH}$.
Let $\tuple {\map D T, T}$ be a densely-defined linear operator.
Let $\map \GG T$ be the graph of $T$.
Then $T$ is closable :
:whenever $\tuple { {\mathbf 0}_X, y} \in \map \cl {\map \GG T}$, we have $y = {\mathbf 0}_Y$
where $\map \cl {\map \GG T}$ is the closure of $\map \GG T$ taken in $\struct {\HH \times \HH, \norm {\, \cdot \,}_{\HH \times \HH} }$.
== Proof ==
=== Necessary Condition ===
Suppose that $T$ is closed.
Then $\map \cl {\map \GG T} = \map \GG T$ from Set is Closed iff Equals Topological Closure.
Then if $\tuple { {\mathbf 0}_X, y} \in \map \GG T$, we have $y = \map T { {\mathbf 0}_X}$ from the definition of the graph. 
So $y = {\mathbf 0}_Y$ since $T$ is linear.
=== Sufficient Condition ===
Suppose that:
:whenever $\tuple { {\mathbf 0}_X, y} \in \map \cl {\map \GG T}$, we have $y = {\mathbf 0}_Y$.
Define:
:$\map D S = \set {x \in \HH : \tuple {x, y} \in \map \cl {\map \GG T} \text { for some } y \in \HH}$
From Graph of Linear Transformation is Vector Subspace, $\map \GG T$ is a vector subspace of $\HH \times \HH$. 
From Closure of Subspace of Normed Vector Space is Subspace, $\map \cl {\map \GG T}$ is a vector subspace of $\HH \times \HH$.
From Linear Transformation defined from Graph, we can define $S : \map D S \to Y$ by:
:$S x$ is the unique $y \in \HH$ such that $\tuple {x, y} \in \map \cl {\map \GG T}$
with $\map \GG S = \map \cl {\map \GG T}$.
Hence $\map \GG S$ is closed, so $S$ is a closed densely-defined operator.
For $x \in \map D T$, we have $\tuple {x, T x}, \tuple {x, S x} \in \map \cl {\map \GG T}$, hence we have $T x = S x$ by uniqueness.
So $S$ extends $T$ and is our desired map.
Category:Closed Linear Transformations
Category:Closable Densely-Defined Linear Operators
Characterization of Closed Ball in P-adic Numbers

== Theorem ==
Let $p$ be a prime number.
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.
Let $\Z_p$ be the $p$-adic integers.
For any $\epsilon \in \R_{>0}$ and $a \in \Q_p$ let $\map {{B_\epsilon}^-} a$ denote the closed ball of center $a$ of radius $\epsilon$. 
Let $x, y \in \Q_p$.
Let $n \in \Z$.
::$(1)\quad x \in \map {B^{\,-}_{p^{-n}}} y$
::$(2)\quad \norm{x -y}_p \le p^{-n}$
::$(3)\quad \map {B^{\,-}_{p^{-n}}} x = \map {B^{\,-}_{p^{-n}}} y$
::$(4)\quad x - y \in p^n \Z_p$
::$(5)\quad x + p^n \Z_p = y + p^n \Z_p$
== Proof ==
From P-adic Numbers form Non-Archimedean Valued Field:
:$\norm {\,\cdot\,}_p$ is a non-Archimedean norm.
=== Condition $(1)$ iff Condition $(2)$ ===
This follows directly from the definition of a closed ball in the $p$-adic numbers.
=== Condition $(1)$ iff Condition $(3)$ ===
By definition, $\map {B^{\,-}_{p^{-n}}} y$ is a closed ball in a non-Archimedean norm $\norm {\,\cdot\,}_p$.
From Centers of Closed Balls in Non-Archimedean Division Rings:
:$x \in \map {B^{\,-}_{p^{-n}}} y \leadsto \map {B^{\,-}_{p^{-n}}} x = \map {B^{\,-}_{p^{-n}}} y$
From Center is Element of Closed Ball in P-adic Numbers:
:$\map {B^{\,-}_{p^{-n}}} x = \map {B^{\,-}_{p^{-n}}} y \leadsto x \in \map {B^{\,-}_{p^{-n}}} x = \map {B^{\,-}_{p^{-n}}} y$
=== Condition $(2)$ iff Condition $(4)$ ===
=== Condition $(3)$ iff Condition $(5)$ ===
From Closed Balls of P-adic Number,
:$\map {B^{\,-}_{p^{-n}}} x = x + p^n \Z_p$
and 
:$\map {B^{\,-}_{p^{-n}}} y = y + p^n \Z_p$
Hence:
:$\map {B^{\,-}_{p^{-n}}} x = \map {B^{\,-}_{p^{-n}}} y$ $x + p^n \Z_p = y + p^n \Z_p$
== Sources ==
* : $\S 3.3$ Exploring $\Q_p$: Lemma $3.3.5$
Category:Topology of P-adic Numbers
Characterization of Closed Set by Open Cover

== Theorem ==
Let $T = \struct{S, \tau}$ be a topological space.
Let $\UU$ be an open cover of $T$.
For each $U \in \UU$, let $\tau_U$ denote the subspace topology on $U$.
Let $F \subseteq S$.
Then $F$ is closed in $T$ :
:$\forall U \in \UU: F \cap U$ is closed in $\struct{U, \tau_U}$
== Proof ==
=== Necessary Condition ===
This follows immediately from Closed Set in Topological Subspace.
=== Sufficient Condition ===
Let:
:$F \cap U$ be closed in $\struct{U, \tau_U}$ for each $U \in \UU$
By definition of closed set:
:$U \setminus \paren{F \cap U}$ is open in $\struct{U, \tau_U}$ for each $U \in \UU$
We have:
Hence:
:$\forall U \in \UU : U \cap \paren {S \setminus F}$ is open in $\struct{U, \tau_U}$ for each $U \in \UU$
From Characterization of Open Set by Open Cover:
:$S \setminus F$ is open in $T$
By definition of closed set:
:$F$ is closed in $T$
Category:Closed Sets
Category:Covers