nyuuzyou/wikis huggingface dataset:
Topological Closure of Subset is Subset of Topological Closure/Proof 2

== Theorem ==
== Proof ==
From the definition of closure:
:$\map \cl H$ is the union of $H$ and its limit points.
Let $x \in \map \cl H$.
If $x \in H$ then $x \in K \implies x \in \map \cl K$.
Otherwise $x$ is a limit point of $H$.
That is, every open set $U$ of $T$ such that $x \in U$ contains $y \in H$ such that $y \ne x$.
But as $y \in H$ it follows that $y \in K$.
So every open set $U$ of $T$ such that $x \in U$ contains $y \in K$ such that $y \ne x$.
This is the definition for a limit point of $K$.
Thus $x \in \map \cl K$.
Category:Topological Closure of Subset is Subset of Topological Closure
Topological Completeness is Topological Property

== Theorem ==
Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be topological spaces which are homeomorphic.
Let $T_1$ be topologically complete.
Then $T_2$ is also topologically complete.
== Proof ==
== Sources ==
* : Part $\text I$: Basic Definitions: Section $5$: Metric Spaces: Complete Metric Spaces
Category:Topologically Complete Spaces
Topological Completeness is Weakly Hereditary

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space which is topologically complete.
Let $V \subseteq S$ be a closed subspace of $T$.
Then $V$ is also topologically complete.
That is, topological completeness is weakly hereditary.
== Proof ==
== Sources ==
* : Part $\text I$: Basic Definitions: Section $5$: Metric Spaces: Complete Metric Spaces
Category:Topologically Complete Spaces
Topological Completeness is not Hereditary

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space which is topologically complete.
Let $H \subseteq S$ be a subset of $S$.
Let $\struct {H, \tau_H}$ be the topological subspace of $T$ induced by $H$.
Then it is not necessarily the case that $\struct {H, \tau_H}$ is also topologically complete.
That is, topological completeness is not hereditary.
== Proof ==
Let $\struct {\R, d}$ denote the real number line under the Euclidean metric.
Let $\struct {\Q, d}$ denote the rational number space under the Euclidean metric.
We have that $\Q \subset \R$ by definition.
Let $T = \struct {\R, \tau_d}$ be the topological space induced on $\R$ by $d$.
By Real Number Line is Complete Metric Space, $\struct {\R, d}$ is a complete metric space.
Thus, by definition, $T$ is topologically complete.
Let $T' = \struct {\Q, \tau_d}$ be the topological space induced on $\Q$ by $d$.
By Rational Number Space is not Complete Metric Space, $\struct {\Q, d}$ is not a complete metric space.
Thus, by definition, $T'$ is not topologically complete.
Hence the result.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $5$: Metric Spaces: Complete Metric Spaces
Category:Topologically Complete Spaces
Topological Dual Space of Hausdorff Locally Convex Space Separates Points

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $\struct {X, \PP}$ be a Hausdorff locally convex space over $\GF$.
Let $X^\ast$ be the topological dual of $X$. 
Let $x \in X$. 
Then $X^\ast$ separates points.
That is, if $x, y \in X$ are such that: 
:$\map f x = \map f y$ for all $f \in X^\ast$
we have that $x = y$.
== Proof ==
Suppose $x \ne y$. 
It suffices to find $f \in X^\ast$ such that $\map f x \ne \map f y$. 
From Finite Topological Space is Compact, $\set {\mathbf 0_X}$ is compact. 
From Compact Subspace of Hausdorff Space is Closed, $\set {\mathbf 0_X}$ is closed.
Since $x \ne y$, we have: 
:$x - y \not \in X \setminus \set {\mathbf 0_X}$
From Existence of Non-Zero Continuous Linear Functional vanishing on Proper Closed Subspace of Hausdorff Locally Convex Space, there exists $f \in X^\ast$ such that: 
:$\map f {x - y} \ne 0$
That is: 
:$\map f x \ne \map f y$
Category:Locally Convex Spaces
Category:Topological Dual Spaces
Topological Equivalence is Equivalence Relation

== Theorem ==
Let $A$ be a set.
Let $\DD$ be the set of all metrics on $A$.
Let $\sim$ be the relation on $\DD$ defined as:
:$\forall d_1, d_2 \in \DD: d_1 \sim d_2 \iff d_1$ is topologically equivalent to $d_2$
Then $\sim$ is an equivalence relation.
== Proof ==
Let $A$ be a set and let $\DD$ be the set of all metrics on $A$.
In the following, let $d_1, d_2, d_3 \subseteq \DD$ be arbitrary.
Checking in turn each of the criteria for equivalence:
=== Reflexivity ===
Let $d_1$ be a metric on $A$.
Then trivially:
:$U \subseteq A$ is $d_1$-open $\iff$ $U \subseteq A$ is $d_1$-open.
That is, $d_1 \sim d_1$ and so $\sim$ has been shown to be reflexive.
=== Symmetry ===
Let $d_1 \sim d_2$.
That is, let $d_1, d_2$ be topologically equivalent metrics on $A$.
Then by definition:
:$U \subseteq A$ is $d_1$-open $\iff$ $U \subseteq A$ is $d_2$-open.
It follows from Biconditional is Commutative that:
:$U \subseteq A$ is $d_2$-open $\iff$ $U \subseteq A$ is $d_1$-open.
That is, $d_2 \sim d_1$ and so $\sim$ has been shown to be symmetric.
=== Transitivity ===
Let $d_1 \sim d_2$ and $d_2 \sim d_3$.
Then by definition:
:$U \subseteq A$ is $d_1$-open $\iff$ $U \subseteq A$ is $d_2$-open.
:$U \subseteq A$ is $d_2$-open $\iff$ $U \subseteq A$ is $d_3$-open.
Then by Biconditional is Transitive:
:$U \subseteq A$ is $d_1$-open $\iff$ $U \subseteq A$ is $d_3$-open.
That is, $d_1 \sim d_3$ and so $\sim$ has been shown to be transitive.
$\sim$ has been shown to be reflexive, symmetric and transitive.
Hence by definition it is an equivalence relation.
== Sources ==
* : $2$: Continuity generalized: metric spaces: $2.4$: Equivalent metrics: Definition $2.4.1$
Category:Topologically Equivalent Metrics
Category:Examples of Equivalence Relations
Topological Equivalence of Definitions of Spectrum of Locale

== Theorem ==
Let $\struct{L, \vee, \wedge, \preceq}$ be a locale.
=== Spectrum as Completely Prime Filters ===
=== Spectrum as Frame Homomorphisms ===
=== Spectrum as Meet-Irreducible Elements ===
=== Spectrum as Continuous Maps ===
== Proof ==
From 
:* Spectrum of Locale as Completely Prime Filters is Sober Space:
the spectrum of $L$ as completely prime filters is a sober space.
From:
:* Canonical Bijection from Completely Prime Filters to Frame Homomorphisms
:* Characterization of Homeomorphic Topological Spaces
the spectrum of $L$ as frame homomorphisms is a topological space homeomorphic to the spectrum of $L$ as completely prime filters.
From:
:* Canonical Bijection from Completely Prime Filters to Meet Irreducible Elements
:* Characterization of Homeomorphic Topological Spaces
the spectrum of $L$ as meet-irreducible elements is a topological space homeomorphic to the spectrum of $L$ as completely prime filters.
From:
:* Canonical Bijection from Frame Homomorphisms to Continuous Maps
:* Characterization of Homeomorphic Topological Spaces
the spectrum of $L$ as continuous maps is a topological space homeomorphic to the spectrum of $L$ as frame homomorphisms.
The result follows by definition of topological equivalence.
Category:Spectra of Locales
Topological Evaluation Mapping is Continuous

== 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 $\family{f_i}_{i \mathop \in I}$.
Then:
:$f$ is continuous
== Proof ==
For each $i \in I$, let:
:$\pr_i : Y \to Y_i$ denote the $i$th projection on $Y$
From Composite of Evaluation Mapping and Projection:
:$\forall i \in I : \pr_i \circ f = f_i$
By assumption:
:$\forall i \in I : \pr_i \circ f$ is continuous
From Continuous Mapping to Product Space:
:$f$ is continuous
Category:Evaluation Mappings (Topological Spaces)
Topological Group is Hausdorff iff Identity is Closed

== Theorem ==
Let $G$ be a topological group.
Let $e$ be its identity element.
Then $G$ is Hausdorff $\left\{ {e}\right\}$ is closed in $G$.
== Proof ==
=== Necessary Condition ===
Suppose $G$ is Hausdorff.
By T2 Space is T1 Space, $\left\{ {e}\right\}$ is closed.
=== Sufficient Condition ===
Suppose $\left\{ {e}\right\}$ is closed.
Let $f: G \times G \to G$ be defined as:
:$f \left({g h}\right) = g h^{-1}$
Because $f$ is continuous, $f^{-1} \left({e}\right)$ is closed.
We have that $f^{-1} \left({e}\right)$ is the diagonal set on $G$.
By Hausdorff Space iff Diagonal Set on Product is Closed, $G$ is Hausdorff.
== Also see ==
* Topological Group is T1 iff T2
* Group is Hausdorff iff Discrete Subgroups are Closed
Category:Topological Groups
Category:Hausdorff Spaces
Topological Group is T1 iff T2

== Theorem ==
Let $G$ be a topological group.
Then $G$ is a $T_1$ space $G$ is Hausdorff.
== Proof ==
=== Necessary Condition ===
Follows directly from $T_2$ Space is $T_1$ Space.
=== Sufficient Condition ===
Let $G$ be a $T_1$ space.
Then $\left\{ {e}\right\}$ is closed.
By Topological Group is Hausdorff iff Identity is Closed, $G$ is Hausdorff.
Category:Topological Groups
Category:T1 Spaces
Category:Hausdorff Spaces
Topological Manifold is Locally Compact

== Theorem ==
Let $M$ be a topological manifold.
Then $M$ is a locally compact space.
== Proof ==
By definition of manifold:
:$M$ is a locally Euclidean space.
The result follows from Locally Euclidean Space is Locally Compact.
Category:Topological Manifolds
Category:Locally Compact Spaces
Topological Manifold is Locally Path-Connected

== Theorem ==
Let $M$ be a topological manifold.
Then $M$ is a locally path-connected space.
== Proof ==
By definition of manifold:
:$M$ is a locally Euclidean space
The result follows from Locally Euclidean Space is Locally Path-Connected
== Sources ==
* : Chapter $4.$ Connectedness and Compactness, Proposition $4.23$
Category:Manifolds
Topological Product of Compact Spaces/Finite Product

== Theorem ==
Let $T_1, T_2, \ldots, T_n$ be topological spaces.
Let $\ds \prod_{i \mathop = 1}^n T_i$ be the product space of $T_1, T_2, \ldots, T_n$.
Then $\ds \prod_{i \mathop = 1}^n T_i$ is compact all of $T_1, T_2, \ldots, T_n$ are compact.
== Proof ==
Proof by induction:
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:$\ds \prod_{i \mathop = 1}^n T_i$ is compact all of $T_1, T_2, \ldots, T_n$ are compact
=== Basis for the Induction ===
$\map P 1$ is the case:
:$T_1$ is compact $T_1$ is compact
which is trivially true.
Thus $\map P 1$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
:$\ds \prod_{i \mathop = 1}^k T_i$ is compact all of $T_1, T_2, \ldots, T_k$ are compact
from which it is to be shown that:
:$\ds \prod_{i \mathop = 1}^{k + 1} T_i$ is compact all of $T_1, T_2, \ldots, T_{k + 1}$ are compact
=== Induction Step ===
This is the induction step:
We have:
:$\ds \prod_{i \mathop = 1}^{k + 1} T_i = \paren {\prod_{i \mathop = 1}^k T_i} \times T_{k + 1}$
Hence:
So $\map P k \implies \map P {k + 1}$ and thus it follows by the Principle of Mathematical Induction that for any $n \ge 1$:
:$\ds \prod_{i \mathop = 1}^n T_i$ is compact all of $T_1, T_2, \ldots, T_n$ are compact
== Also see ==
* Tychonoff's Theorem, where this result is extended to the topological product of any infinite number of topological spaces.
== Sources ==
* : $5$: Compact spaces: $5.6$: Compactness and Constructions: Remark $5.6.3$
Category:Topology
Topological Properties of Non-Archimedean Division Rings

== Theorem ==
Let $\struct {R, \norm {\,\cdot\,} }$ be a normed division ring with non-Archimedean norm $\norm {\,\cdot\,}$.
For $a \in R$ and $\epsilon \in \R_{>0}$ let:
:$\map {B_\epsilon} a$ denote the open $\epsilon$-ball of $a$ in $\struct {R, \norm{\,\cdot\,} }$
:$\map { {B_\epsilon}^-} a$ denote the closed $\epsilon$-ball of $a$ in $\struct {R, \norm{\,\cdot\,}}$
Let $x, y \in R$.
Let $r, s \in \R_{>0}$.
The following results hold:
=== Centers of Open Balls ===
=== Centers of Closed Balls ===
=== Open Balls are Clopen ===
=== Closed Balls are Clopen ===
=== Spheres are Clopen ===
=== Intersection of Open Balls ===
=== Intersection of Closed Balls ===
== Sources ==
* : $\S 2.3$: Topology: Proposition $2.3.6$
Category:Normed Division Rings
Category:Non-Archimedean Norms
Category:Topological Properties of Non-Archimedean Division Rings
Topological Properties of Non-Archimedean Division Rings/Centers of Closed Balls

== Theorem ==
Let $\struct {R, \norm {\,\cdot\,} }$ be a normed division ring with non-Archimedean norm $\norm {\,\cdot\,}$,
For $a \in R$ and $\epsilon \in \R_{>0}$ let:
:$\map { {B_\epsilon}^-} a$ denote the closed $\epsilon$-ball of $a$ in $\struct {R, \norm {\,\cdot\,} }$
Let $x, y \in R$.
Let $r \in \R_{\gt 0}$.
Then:
:$y \in \map { {B_r}^-} x \implies \map { {B_r}^-} y = \map { {B_r}^-} x$
== Proof ==
Let $y \in \map { {B_r}^-} x$.
Let $a \in \map { {B_r}^-} y$.
By the definition of an closed ball, then:
:$\norm {a - y} \le r$
:$\norm {y - x} \le r$
Hence:
By the definition of a closed ball, then: 
:$a \in \map { {B_r}^-} x$.
Hence:
:$\map { {B_r}^-} y \subseteq \map { {B_r}^-} x$
By Norm of Negative then:
:$\norm {x - y} \le r$
By the definition of a closed ball, then:
:$x \in \map { {B_r}^-} y$
Similarly it follows that:
:$\map { {B_r}^-} x \subseteq \map { {B_r}^-} y$
By set equality:
:$\map { {B_r}^-} x = \map { {B_r}^-} y$
== Sources ==
* : $\S 2.3$: Topology, Proposition $2.3.6 \ \text {(ii)}$
Category:Topological Properties of Non-Archimedean Division Rings
Topological Properties of Non-Archimedean Division Rings/Centers of Open Balls

== Theorem ==
Let $\struct {R, \norm {\,\cdot\,} }$ be a normed division ring with non-Archimedean norm $\norm {\,\cdot\,}$,
For $a \in R$ and $\epsilon \in \R_{>0}$ let:
:$\map {B_\epsilon} a$ denote the open $\epsilon$-ball of $a$ in $\struct {R, \norm {\,\cdot\,} }$
Let $x, y \in R$.
Let $r \in \R_{>0}$.
Then:
:$y \in \map {B_r} x \implies \map {B_r} y = \map {B_r} x$
== Proof ==
Let $y \in \map {B_r} x$.
Let $a \in \map {B_r} y$.
By the definition of an open ball, then:
:$\norm {a - y} < r$
:$\norm {y - x} < r$
Hence:
By the definition of an open ball: 
:$a \in \map {B_r} x$
Hence:
:$\map {B_r} y \subseteq \map {B_r} x$
By Norm of Negative:
:$\norm {x - y} < r$
By the definition of an open ball:
:$x \in \map {B_r} y$
Similarly it follows that:
:$\map {B_r} x \subseteq \map {B_r} y$
By set equality:
:$\map {B_r} y = \map {B_r} x$
== Sources ==
* : $\S 2.3$: Topology, Proposition $2.3.6 \ \text {(i)}$
* : $\S 1.2$ Normed Fields: Remark $1.17$; $\S 2.1$ Elementary Topological Properties: Proposition $2.4$ 
Category:Topological Properties of Non-Archimedean Division Rings
Topological Properties of Non-Archimedean Division Rings/Closed Balls are Clopen

== Theorem ==
Let $\struct {R, \norm {\,\cdot\,}}$ be a normed division ring with non-Archimedean norm $\norm {\,\cdot\,}$,
Let $x \in R$.
Let $r \in \R_{\gt 0}$.
Let $\map { {B_r}^-} x$ be the closed $r$-ball of $x$ in $\struct {R,\norm {\,\cdot\,} }$
Then:
:The closed $r$-ball of $x$, $\map { {B_r}^-} x$, is both open and closed in the metric induced by $\norm {\,\cdot\,}$.
== Proof ==
Let $d$ be the metric induced by the norm $\norm {\,\cdot\,}$.
By the definition of a closed ball in $\norm {\,\cdot\,}$ then:
:$\map { {B_r}^-} x$ is a closed ball in the metric space $\struct {R, d}$.
By Closed Ball is Closed in Metric Space then $\map { {B_r}^-} c$ is closed in $d$.
So it remains to show that $\map { {B_r}^-} x$ is open in $d$.
Let $y \in \map { {B_r}^-} x$.
By Centers of Closed Balls then:
:$\map { {B_r}^-} y = \map { {B_r}^-} x$
By the definition of an open ball then:
:$y \in \map {B_r} y \subseteq \map { {B_r}^-} y = \map { {B_r}^-} x$
By the definition of an open set in a metric space, $\map { {B_r}^-} x$ is open.
== Sources ==
* : $\S 2.3$: Topology, Proposition $2.3.6 \, \text {(iv)}$
Category:Topological Properties of Non-Archimedean Division Rings
Topological Properties of Non-Archimedean Division Rings/Intersection of Closed Balls

== Theorem ==
Let $\struct {R, \norm {\,\cdot\,} }$ be a normed division ring with non-Archimedean norm $\norm {\,\cdot\,}$,
For $a \in R$ and $\epsilon \in \R_{>0}$ let:
:$\map { {B_\epsilon}^-} a$ denote the closed $\epsilon$-ball of $a$ in $\struct {R, \norm {\,\cdot\,} }$
Let $x, y \in R$.
Let $r, s \in \R_{>0}$.
Then:
:$\map { {B_r}^-} x \cap \map { {B_s}^-} y \ne \O \iff \map { {B_r}^-} x \subseteq \map { {B_s}^-} y$ or $\map { {B_s}^-} y \subseteq \map { {B_r}^-} x$
== Proof ==
=== Necessary Condition ===
Let $z \in \map { {B_r}^-} x \cap \map { {B_s}^-} y$.
If $r \le s$ then:
Similarly, if $s \le r$ then:
=== Sufficient Condition ===
Let:
:$\map { {B_r}^-} x \subseteq \map { {B_s}^-} y$ 
or:
:$\map { {B_s}^-} y \subseteq \map { {B_r}^-} x$
By the definition of an open ball then:
:$x \in \map { {B_r}^-} x \ne \O$
:$y \in \map { {B_s}^-} y \ne \O$
The result follows.
== Sources ==
* : $\S 2.3$: Topology, Proposition $2.3.6 \, \text {(vi)}$
Category:Topological Properties of Non-Archimedean Division Rings
Topological Properties of Non-Archimedean Division Rings/Intersection of Open Balls

== Theorem ==
Let $\struct {R, \norm {\,\cdot\,} }$ be a normed division ring with non-Archimedean norm $\norm {\,\cdot\,}$,
For $a \in R$ and $\epsilon \in \R_{>0}$ let:
:$\map {B_\epsilon} a$ denote the open $\epsilon$-ball of $a$ in $\struct {R, \norm {\,\cdot\,} }$
Let $x, y \in R$.
Let $r, s \in \R_{\gt 0}$.
Then:
:$\map {B_r} x \cap \map {B_s} y \ne \O \iff \map {B_r} x \subseteq \map {B_s} y$ or $\map {B_s} y \subseteq \map {B_r} x$
== Proof ==
=== Necessary Condition ===
Let $z \in \map {B_r} x \cap \map {B_s} y$.
If $r \le s$ then:
Similarly, if $s \le r$ then:
=== Sufficient Condition ===
Let 
:$\map {B_r} x \subseteq \map {B_s} y$ 
or 
:$\map {B_s} y \subseteq \map {B_r} x$
By the definition of an open ball then:
:$x \in \map {B_r} x \ne \O$
:$y \in \map {B_s} y \ne \O$
The result follows.
== Sources ==
* : $\S 2.3$: Topology, Proposition $2.3.6 \, (v)$
* : $\S 2.1$ Elementary Topological Properties, Proposition $2.5$
Category:Topological Properties of Non-Archimedean Division Rings
Topological Properties of Non-Archimedean Division Rings/Open Balls are Clopen

== Theorem ==
Let $\struct {R, \norm {\,\cdot\,}}$ be a normed division ring with non-Archimedean norm $\norm {\,\cdot\,}$,
Let $x \in R$.
Let $r \in \R_{>0}$.
Let $\map {B_r} x$ denote the open $r$-ball of $x$ in $\struct {R, \norm {\,\cdot\,} }$
Then:
:The open $r$-ball of $x$, $\map {B_r} x$, is both open and closed in the metric induced by $\norm {\,\cdot\,}$.
== Proof ==
Let $d$ be the metric induced by the norm $\norm {\,\cdot\,}$.
By the definition of an open ball in $\norm {\,\cdot\,}$:
:$\map {B_r} x$ is an open ball in the metric space $\struct {R, d}$.
By Open Ball of Metric Space is Open Set then $\map {B_r} x$ is open in $\struct {R, d}$.
So it remains to show that $\map {B_r} x$ is closed in $\struct {R, d}$.
Let $\map \cl {\map {B_r} x}$ denote the closure of $\map {B_r} x$.
Let $y \in \map \cl {\map {B_r} x}$.
By the definition of the closure of $\map {B_r} x$ then:
:$\forall s > 0: \map {B_s} y \cap \map {B_r} x \ne \O$
In particular:
:$\map {B_r} y \cap \map {B_r} x \ne \O$
Let $z \in \map {B_r} y \cap \map {B_r} x$.
By Centers of Open Balls:
:$\map {B_r} y = B_r \paren{z} = \map {B_r} x$
By the definition of an open ball:
:$y \in \map {B_r} y = \map {B_r} x$.
Hence:
:$\map \cl {\map {B_r} x} \subseteq \map {B_r} x$
By Subset of Metric Space is Subset of its Closure then:
:$\map {B_r} x \subseteq \map \cl {\map {B_r} x}$
So by definition of set equality:
:$\map \cl {\map {B_r} x} = \map {B_r} x$
By Set is Closed iff Equals Topological Closure then $\map {B_r} x$ is closed.
== Sources ==
* : $\S 2.3$: Topology, Proposition $2.3.6 \ \text {(iii)}$
* : $\S 2.1$ Elementary Topological Properties, Proposition $2.3$ 
Category:Topological Properties of Non-Archimedean Division Rings
Topological Properties of Non-Archimedean Division Rings/Spheres are Clopen

== Theorem ==
Let $\struct {R, \norm {\,\cdot\,} }$ be a normed division ring with non-Archimedean norm $\norm{\,\cdot\,}$,
Let $x \in R$.
Let $r \in \R_{>0}$.
Let $\map {S_r} x$ denote the $r$-sphere of $x$ in $\struct {R, \norm {\,\cdot\,}}$
Then:
:The $r$-sphere of $x$, $\map {S_r} x$, is both open and closed in the metric induced by $\norm {\,\cdot\,}$.
== Proof ==
We have:
Let $d$ be the metric induced by the norm $\norm {\,\cdot\,}$.
By Open Balls Are Clopen then $\map {B_r} x$ is both open and closed in $d$.
By Metric Induces Topology then $R \setminus \map {B_r} x$ is is both open and closed in $d$.
By Closed Balls Are Clopen then $\map { {B_r}^-} x$ is both open and closed in $d$.
By Metric Induces Topology then the intersection of a finite number of open sets is open. 
Hence $\map {S_r} x$ is open in the metric space $\struct {R, d}$.
By Intersection of Closed Sets is Closed in Topological Space then $\map {S_r} x$ is closed in $\struct {R, d}$.
The result follows.
== Sources ==
* : $\S 2.3$: Topology, Problem $51$
* : $\S 2.1$ Elementary topological properties, Proposition $2.6$
Category:Topological Properties of Non-Archimedean Division Rings
Topological Space Induced by Neighborhood Space is Topological Space

== Theorem ==
Let $\struct {S, \NN}$ be a neighborhood space.
Let $\struct {S, \tau}$ be the topological space induced by $\struct {S, \NN}$.
Then $\struct {S, \tau}$ is a topological space.
== Proof ==
From Neighborhood Space is Topological Space, $\struct {S, \NN}$ is a topological space.
Consequently, the open sets of $\struct {S, \tau}$ are exactly the open sets of $\struct {S, \NN}$.
Category:Neighborhood Spaces
Topological Space Separated by Mappings is Hausdorff

== Theorem ==
Let $X$ be a topological space.
Let $\family {Y_i}_{i \mathop \in I}$ be an indexed family of Hausdorff 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.
Suppose $\family {f_i : X \to Y_i}_{i \mathop \in I}$ separates the points of $X$.
Then, $X$ is Hausdorff.
== Proof ==
Let $x \ne y$ be elements of $X$.
By definition of separating points, there exists some $i \in I$ such that:
:$\map {f_i} x \ne \map {f_i} y$
As $Y_i$ is Hausdorff, there exist open sets $U, V \subseteq Y_i$ such that:
:$\map {f_i} x \in U$
:$\map {f_i} y \in V$
:$U \cap V = \O$
By definition of continuous mapping, we have that:
:$f_i^{-1} \sqbrk U$ and $f_i^{-1} \sqbrk V$ are open sets of $X$
By the definition of preimage, it is immediate that:
:$x \in f_i^{-1} \sqbrk U$
:$y \in f_i^{-1} \sqbrk V$
It remains to show that $f_i^{-1} \sqbrk U$ and $f_i^{-1} \sqbrk V$ are disjoint.
 there is some $z \in X$ such that:
:$z \in f_i^{-1} \sqbrk U$
:$z \in f_i^{-1} \sqbrk V$
Then, by definition of preimage:
:$\map {f_i} z \in U$
:$\map {f_i} z \in V$
contradicting the fact that $U$ and $V$ are themselves disjoint.
By Proof by Contradiction, it follows that $f_i^{-1} \sqbrk U$ and $f_i^{-1} \sqbrk V$ are disjoint.
As $x \ne y$ were arbitrary, it follows that $X$ is Hausdorff by definition.
Category:Hausdorff Spaces