nyuuzyou/wikis huggingface dataset:
Commutators are Identity iff Group is Abelian

== Theorem ==
Let $\struct {G, \circ}$ be a group whose identity is $e$.
For $g, h \in G$, let $\sqbrk {g, h}$ denote the commutator of $g$ and $h$.
Then $\struct {G, \circ}$ is abelian :
:$\forall g, h \in G: \sqbrk {g, h} = e$
== Proof ==
=== Necessary Condition ===
Let $\struct {G, \circ}$ be such that:
:$\forall g, h \in G: \sqbrk {g, h} = e$
From Commutator is Identity iff Elements Commute:
:$\forall g, h \in G: g \circ h = h \circ g$
Hence $\struct {G, \circ}$ is abelian by definition.
=== Sufficient Condition ===
Let $\struct {G, \circ}$ be an abelian group.
Then by definition:
:$\forall g, h \in G: g \circ h = h \circ g$
From Commutator is Identity iff Elements Commute:
:$\forall g, h \in G: \sqbrk {g, h} = e$
Hence the result.
== Sources ==
* 
Category:Group Commutators
Category:Abelian Groups
Compact Closure is Directed

== Theorem ==
Let $L = \left({S, \vee, \preceq}\right)$ be a bounded below join semilattice.
Let $x \in S$.
Then $x^{\mathrm{compact} }$ is directed
where $x^{\mathrm{compact} }$ denotes the compact closure of $x$.
== Proof ==
By Bottom in Compact Closure:
:$\bot \in x^{\mathrm{compact} }$
where $\bot$ denotes the smallest element in $L$.
Thus by definition:
:$x^{\mathrm{compact} }$ is non-empty.
Let $y, z \in x^{\mathrm{compact} }$
By definition of compact closure:
:$y$ and $z$ are compact elements and $y \preceq x$ and $z \preceq x$
By definitions of supremum and upper bound:
:$y \vee z \preceq x$
By definition of compact element:
:$y \ll y$ and $z \ll z$
where $\ll$ denotes the way below relation.
By Join Succeeds Operands:
:$y \preceq y \vee z$ and $z \preceq y \vee z$
By Preceding and Way Below implies Way Below and definition of reflexivity:
:$y \ll y \vee z$ and $z \ll y \vee z$
By Join is Way Below if Operands are Way Below:
:$y \vee z \ll y \vee z$
By definition:
:$y \vee z$ is compact element.
By definition of compact closure:
:$y \vee z \in x^{\mathrm{compact} }$
Thus
:$\exists v \in x^{\mathrm{compact} }: y \preceq v \land z \preceq v$
== Sources ==
* 
Category:Way Below Relation
Compact Closure is Increasing

== Theorem ==
Let $\left({S, \preceq}\right)$ be an ordered set.
Let $x, y \in S$ such that
:$x \preceq y$
Then $x^{\mathrm{compact} } \subseteq y^{\mathrm{compact} }$
where $x^{\mathrm{compact} }$ denotes the compact closure of $x$.
== Proof ==
Let $z \in x^{\mathrm{compact} }$
By definition of compact closure:
:$z$ is a compact element and $z \preceq x$
By definition of transitivity:
:$z \preceq y$
Thus by definition of compact closure:
:$z \in y^{\mathrm{compact} }$
== Sources ==
* 
Category:Way Below Relation
Compact Closure is Intersection of Lower Closure and Compact Subset

== Theorem ==
Let $L = \struct {S, \vee, \preceq}$ be a bounded below join semilattice.
Let $x \in S$.
Then $x^{\mathrm {compact} } = x^\preceq \cap \map K L$
where
:$x^{\mathrm {compact} }$ denotes the compact closure of $x$,
:$x^\preceq$ denotes the lower closure of $x$,
:$\map K L$ denotes the compact subset of $L$.
== Proof ==
:$y \in x^{\mathrm {compact} }$
:$y \preceq x$ and $y$ is compact by definition of compact closure
:$y \in x^\preceq$ and $y$ is compact by definition of lower closure of element
:$y \in x^\preceq$ and $y \in \map K L$ by definition of compact subset
:$y \in x^\preceq \cap \map K L$ by definition of intersection
Hence the result, by definition of set equality.
== Sources ==
* 
* 
Category:Way Below Relation
Category:Lower Closures
Compact Closure is Set of Finite Subsets in Lattice of Power Set

== Theorem ==
Let $X$ be a set.
Let $L = \struct {\powerset X, \cup, \cap, \preceq}$ be the lattice of power set of $X$ where $\mathord\preceq = \mathord\subseteq \cap \powerset X \times \powerset X$
Let $x \in \powerset X$.
Then $x^{\mathrm{compact} } = \map {\operatorname{Fin} } x$
where $\map {\operatorname{Fin} } x$ denotes the set of all finite subsets of $x$.
== Proof ==
=== $\subseteq$ ===
Let $y \in x^{\mathrm{compact} }$.
By definition of compact closure:
:$y \preceq x$ and $y$ is compact.
By definition of $\preceq$:
:$y \subseteq x$
By Element is Finite iff Element is Compact in Lattice of Power Set:
"$y$ is a finite set.
Thus by definition of $\operatorname{Fin}$:
:$y \in \map {\operatorname{Fin} } x$
=== $\supseteq$ ===
Let $y \in \map {\operatorname{Fin} } x$.
By definition of $\operatorname{Fin}$:
:$y \subseteq x$ and $y$ is finite.
By definition of $\preceq$:
:$y \preceq x$
By Element is Finite iff Element is Compact in Lattice of Power Set:
:$y$ is compact.
Thus by definition of compact closure:
:$y \in x^{\mathrm{compact} }$
== Sources ==
* 
* 
Category:Way Below Relation
Category:Power Set
Compact Closure is Subset of Way Below Closure

== Theorem ==
Let $L = \struct {S, \preceq}$ be an ordered set.
Let $x \in S$.
Then $x^{\mathrm {compact} } \subseteq x^\ll$
where
:$x^{\mathrm {compact} }$ denotes the compact closure of $x$,
:$x^\ll$ denotes the way below closure of $x$.
== Proof ==
Let $y \in x^{\mathrm {compact} }$.
By definition of compact closure:
:$y \preceq x$ and $y$ is compact.
By definition of compact:
:$y \ll y$
where $\ll$ denotes the way below relation.
By Preceding and Way Below implies Way Below and definition of reflexivity:
:$y \ll x$
Thus by definition of way below closure:
:$y \in x^\ll$
== Sources ==
* 
* 
Category:Way Below Relation
Compact Closure of Element is Principal Ideal on Compact Subset iff Element is Compact

== Theorem ==
Let $L = \struct {S, \vee, \preceq}$ be a bounded below algebraic join semilattice.
Let $P = \struct {\map K L, \precsim}$ be an ordered subset of $L$
where $\map K L$ denotes the compact subset of $L$.
Let $x \in S$.
Then $x^{\mathrm{compact} }$ is principal ideal in $P$ $x$ is a compact element.
== Proof ==
=== Sufficient Condition ===
Assume that
:$x^{\mathrm{compact} }$ is principal ideal in $P$.
By definitions of compact subset and compact closure:
:$x^{\mathrm{compact} } \subseteq \map K L$
By definition of principal ideal:
:$\exists y \in x^{\mathrm{compact} }: y$ is upper bound for $x^{\mathrm{compact} }$ in $P$.
By definition of ordered subset:
:$y$ is upper bound for $x^{\mathrm{compact} }$ in $L$.
By definition of supremum:
:$\map {\sup_L} {x^{\mathrm{compact} } } \preceq y$
By definition of algebraic:
:$L$ satisfies the axiom of $K$-approximation.
By the axiom of $K$-approximation:
:$x \preceq y$
By definition of compact closure:
:$y \preceq x$ and $y$ is a compact element.
Thus by definition of antisymmetry:
:$x$ is a compact element.
=== Necessary Condition ===
Let $x$ be a compact element.
By definition of compact subset:
:$x \in \map K L$
We will prove that
:$x^{\mathrm{compact} } = x^\precsim$
Let $y \in \map K L$.
:$y \in x^{\mathrm{compact} }$
:$y \preceq x$ by definitions of compact closure and compact subset
:$y \precsim x$ by definition of ordered subset
:$y \in x^\precsim$
Hence $x^{\mathrm{compact} }$ is a principal ideal in $P$.
== Sources ==
* 
* 
Category:Principal Ideals of Preordered Sets
Category:Way Below Relation
Compact Complement Space is not T2, T3, T4 or T5

== Theorem ==
Let $T = \struct {\R, \tau}$ be the compact complement topology on $\R$.
Then $T$ is not a $T_2$ (Hausdorff) space, $T_{2 \frac 1 2}$ (completely Hausdorff) space, $T_3$ space, $T_4$ space or $T_5$ space.
== Proof ==
We have that a Compact Complement Space is a $T_1$ space.
From $T_1$ Space is $T_0$ Space, $T$ is a $T_0$ space.
We have:
:Compact Complement Topology is Irreducible
:Irreducible Hausdorff Space is Singleton
From $T_{2 \frac 1 2}$ (completely Hausdorff) space is $T_2$ (Hausdorff) space, $T$ is not a $T_{2 \frac 1 2}$ (completely Hausdorff) space.
From Regular Space is $T_2$ Space, $T$ is not a regular space.
By definition, a regular space is a space that is both a $T_0$ space and a $T_3$ space.
But $T$ is a $T_0$ space and not a regular space.
So it follows that $T$ can not be a $T_3$ space.
Next we have that a Normal Space is a $T_3$ Space.
But as $T$ is not a $T_3$ space, $T$ can not be a normal space.
By definition, a normal space is a space that is both a $T_1$ space and a $T_4$ space.
But $T$ is a $T_1$ space and not a normal space.
So it follows that $T$ can not be a $T_4$ space.
Finally we have that a $T_5$ Space is a $T_4$ Space.
But as $T$ is not a $T_4$ space, $T$ can not be a $T_5$ space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $22$. Compact Complement Topology: $2$
Category:Completely Hausdorff Spaces
Category:Hausdorff Spaces
Category:T3 Spaces
Category:T4 Spaces
Category:T5 Spaces
Category:Compact Complement Topology
Compact Complement Topology is Coarser than Euclidean Topology

== Theorem ==
Let $T = \struct {\R, \tau}$ be the compact complement topology on $\R$.
Then $\tau$ is coarser than the usual (Euclidean) topology in $\R$.
== Proof ==
Let $U \in \tau$.
Then $V = \R \setminus U$ is compact.
So by definition $V$ is closed in the Euclidean topology.
That is, $U = \R \setminus V$ is open in the Euclidean topology.
That is, every open set in the compact complement topology is also open in the Euclidean topology.
Hence the result, by definition of coarser topology.
== Sources ==
* : Part $\text {II}$: Counterexamples: $22$. Compact Complement Topology: $6$
Category:Compact Spaces
Category:Compact Complement Topology
Compact Complement Topology is Compact

== Theorem ==
Let $T = \struct {\R, \tau}$ be the compact complement topology on $\R$.
Then $T$ is a compact space.
== Proof ==
Let $\CC$ be an open cover of $\R$.
Let $U \in \CC$.
Then by definition $\R \setminus U$ is compact in the usual (Euclidean) topology.
Since each $U$ is open in the Euclidean topology, a finite number must cover $\R \setminus U$.
== Sources ==
* : Part $\text {II}$: Counterexamples: $22$. Compact Complement Topology: $4$
Category:Compact Spaces
Category:Compact Complement Topology
Compact Complement Topology is Connected

== Theorem ==
Let $T = \struct {\R, \tau}$ be the compact complement topology.
Then $T$ is a connected space.
== Proof ==
Follows from:
:Compact Complement Topology is Irreducible
:Irreducible Space is Connected
== Sources ==
* : Part $\text {II}$: Counterexamples: $22$. Compact Complement Topology: $3$
Category:Connected Spaces
Category:Compact Complement Topology
Compact Complement Topology is First-Countable

== Theorem ==
Let $T = \struct {\R, \tau}$ be the compact complement topology on $\R$.
Then $T$ is a first-countable space.
== Proof ==
Let $\CC$ be an open cover of $\R$.
Let $p \in \R$.
Consider the set:
:$\BB_p = \set {\openint {-\infty} {-n} \cup \openint {p - \dfrac 1 n} {p + \dfrac 1 n} \cup \openint n \infty: n \in \N}$
From Countable Local Basis in Compact Complement Topology, $\BB_p$ is a countable local basis for $T$.
Hence the result, by definition of first-countable space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $22$. Compact Complement Topology: $5$
Category:First-Countable Spaces
Category:Compact Complement Topology
Compact Complement Topology is Irreducible

== Theorem ==
Let $T = \struct {\R, \tau}$ be the compact complement topology on $\R$.
Then $T$ is an irreducible space.
== Proof ==
Let $U_1, U_2 \in \tau$ be open in $T$.
Let $\relcomp \R {U_1} = V_1$ and $\relcomp \R {U_2} = V_2$.
By definition of compact complement topology, $V_1, V_2 \subseteq \R$ such that $V_1$ and $V_2$ are both compact.
$V_1$ and $V_2$ are both bounded by definition of compact in $\R$, so their union is likewise bounded: above by $\max \set {\sup V_1, \sup V_2}$ and below by $\min \set {\inf V_1, \inf V_2}$.
So $V_1 \cup V_2$ can not equal $\R$ as $\R$ is not bounded.
So:
So any two open sets in $T$ are not disjoint and so $T$ is irreducible space by definition.
== Sources ==
* : Part $\text {II}$: Counterexamples: $22$. Compact Complement Topology: $3$
Category:Irreducible Spaces
Category:Compact Complement Topology
Compact Complement Topology is Locally Connected

== Theorem ==
Let $T = \struct {\R, \tau}$ be the compact complement topology on $\R$.
Then $T$ is a locally connected space.
== Proof ==
Follows from:
:Compact Complement Topology is Irreducible
:Irreducible Space is Locally Connected
== Sources ==
* : Part $\text {II}$: Counterexamples: $22$. Compact Complement Topology: $3$
Category:Locally Connected Spaces
Category:Compact Complement Topology
Compact Complement Topology is Second-Countable

== Theorem ==
Let $T = \struct {\R, \tau}$ be the compact complement topology on $\R$.
Then $T$ is a second-countable space.
== Proof ==
== Sources ==
* : Part $\text {II}$: Counterexamples: $22$. Compact Complement Topology: $5$
Category:Second-Countable Spaces
Category:Compact Complement Topology
Compact Complement Topology is Separable

== Theorem ==
Let $T = \struct {\R, \tau}$ be the compact complement topology on $\R$.
Then $T$ is a separable space.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Part $\text {II}$: Counterexamples: $22$. Compact Complement Topology: $6$
Category:Separable Spaces
Category:Compact Complement Topology
Category:Compact Complement Topology is Separable
Compact Complement Topology is Sequentially Compact

== Theorem ==
Let $T = \struct {\R, \tau}$ be the compact complement topology on $\R$.
Then $T$ is a sequentially compact space.
== Proof ==
We have:
:Compact Complement Topology is Compact
:Compact Space is Countably Compact
:Compact Complement Topology is First-Countable
:First-Countable Space is Sequentially Compact iff Countably Compact
== Sources ==
* : Part $\text {II}$: Counterexamples: $22$. Compact Complement Topology: $5$
Category:Sequentially Compact Spaces
Category:Compact Complement Topology
Compact Complement Topology is T1

== Theorem ==
Let $T = \struct {S, \tau}$ be a compact complement space.
Then $T$ is a $T_1$ (Fréchet) space.
== Proof ==
We have that a Finite Topological Space is Compact.
So any finite subspace of $T$ is compact.
Let $\tau^*$ be the set defined as:
:$\tau^* = \leftset {S \subseteq \R: S = \O \text { or } \relcomp \R S}$ is finite $\rightset {}$
where $\relcomp \R S$ denotes the complement of $S$ in $\R$.
Then $\tau^*$ is a subset of $\tau$ by definition of the compact complement topology.
But $\tau^*$ is the finite complement topology.
The result follows from Finite Complement Topology is Minimal $T_1$ Topology.
== Sources ==
* : Part $\text {II}$: Counterexamples: $22$. Compact Complement Topology: $1$
Category:T1 Spaces
Category:Compact Complement Topology
Compact Complement Topology is Topology

== Theorem ==
Let $T = \struct {\R, \tau}$ be the compact complement space.
Then $\tau$ is a topology on $T$.
== Proof ==
By definition, we have that $\O \in \tau$.
We also have that $\R \in \tau$ as $\relcomp \R \R = \O$ which is trivially compact.
Now suppose $A, B \in \tau$.
Let $H = A \cap B$.
Then:
But as $A, B \in \tau$ it follows that $\relcomp \R A$ and $\relcomp \R B$ are both compact.
Hence from Union of Compact Sets is Compact their union is also compact and so $\relcomp \R H$ is compact.
So $H = A \cap B \in \tau$ as its complement is compact.
Now let $\UU \subseteq \tau$.
Then from De Morgan's Laws: Complement of Union:
:$\ds \relcomp \R {\bigcup \UU} = \bigcap_{U \mathop \in \UU} \relcomp \R U$
But as:
:$\forall U \in \UU: \relcomp \R U \in \tau$
each of the $\relcomp \R U$ is compact.
Hence from Intersection of Compact Sets is Compact, so is their intersection.
So $\ds \relcomp \R {\bigcup \UU}$ is compact which means $\ds \bigcup \UU \in \tau$.
So $\tau$ is a topology on $T$.
== Sources ==
* : Part $\text {II}$: Counterexamples: $22$. Compact Complement Topology
Category:Compact Complement Topology
Compact Complement Topology is not Ultraconnected

== Theorem ==
Let $T = \struct {\R, \tau}$ be the compact complement topology on $\R$.
Then $T$ is not an ultraconnected space.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Part $\text {II}$: Counterexamples: $22$. Compact Complement Topology: $3$
Category:Ultraconnected Spaces
Category:Compact Complement Topology
Category:Compact Complement Topology is not Ultraconnected
Compact Complement Topology is not Ultraconnected/Proof 1

== Theorem ==
== Proof ==
By definition, closed sets of $T$ are compact sets of $T$.
So, for example, $\left[{0 \,.\,.\, 1}\right]$ and $\left[{2 \,.\,.\, 3}\right]$ are disjoint compact sets and therefore closed sets of $T$.
Hence the result by definition of ultraconnected.
Category:Compact Complement Topology is not Ultraconnected
Compact Convergence Implies Local Uniform Convergence if Weakly Locally Compact

== Theorem ==
Let $T = \struct {S, \tau}$ be a weakly locally compact topological space.
Let $M = \struct {A, d}$ be a metric space.
Let $\sequence {f_n}$ be a sequence of mappings $f_n: X \to M$.
Let $f_n$ converge compactly to $f: X \to M$.
Then $f_n$ converges locally uniformly to $f$.
== Proof ==
Because $T$ is weakly locally compact, every point of $S$ has a compact neighborhood in $T$.
We are given $f_n$ converge compactly to $f$, so $f_n$ converges uniformly to $f$ on every compact subset of $X$
So every point of $S$ has a (compact) neighborhood on which $f_n$ converges uniformly to $f$.
Thus $f_n$ converges locally uniformly to $f$.
== Also see ==
* Local Uniform Convergence Implies Compact Convergence
Category:Uniform Convergence
Category:Compact Convergence
Compact Convex Sets with Nonempty Interior are Homeomorphic

== Theorem ==
Let $n \in \N_{> 0}$.
Let $T, T' \subseteq \R^n$ be compact convex subsets of real Euclidean $n$-space.
Then, $T$ is homeomorphic to $T'$.
== Proof ==
By Boundary of Compact Convex Set with Nonempty Interior is Homeomorphic to Sphere:
:$\partial T \sim \Bbb S^{n - 1}$
:$\partial T' \sim \Bbb S^{n - 1}$
Thus, by Homeomorphism Relation is Equivalence:
:$\partial T \sim \partial T'$
Hence, by Cones on Homeomorphic Spaces are Homeomorphic:
:$C \partial T \sim C \partial T'$
But, by Compact Convex Set with Nonempty Interior is Homeomorphic to Cone on Boundary:
:$T \sim C \partial T$
:$T' \sim C \partial T'$
Therefore, by Homeomorphism Relation is Equivalence:
:$T \sim T'$
Category:Homeomorphisms (Topological Spaces)
Compact Element iff Principal Ideal

== Theorem ==
Let $L = \struct {S, \vee, \preceq}$ be a bounded below join semilattice.
Let $P = \struct {\map {\operatorname {Ids} } L, \precsim}$ be an inclusion ordered set
where
:$\map {\operatorname {Ids} } L$ denotes the set of all ideals in $L$,
:$\mathord\precsim = \mathord\subseteq \cap \paren {\map {\operatorname {Ids} } L \times \map {\operatorname {Ids} } L}$
Let $x \in \map {\operatorname {Ids} } L$
Then $x$ is compact element $x$ is principal ideal in $L$.
== Proof ==
By Ideals are Continuous Lattice Subframe of Power Set:
:$P$ is continuous lattice subframe of $\struct {\powerset S, \subseteq'}$
where 
:$\powerset S$ denotes the power set of $S$,
:$\mathord\subseteq' = \mathord\subseteq \cap \struct {\powerset S \times \powerset S}$
=== Sufficient Condition ===
Assume that
:$x$ is compact element.
By Compact Element iff Existence of Finite Subset that Element equals Intersection and Includes Subset:
:$\exists F \in \map {\operatorname {Fin} } S: x = \bigcap \set {I \in \map {\operatorname {Ids} } L: F \subseteq I} \land F \subseteq x$
where $\map {\operatorname {Fin} } S$ denotes the set of all finite subsets of $S$.
We will prove that
:$\exists y \in x: y$ is upper bound for $x$.
Define $y = \sup_L F$.
By Directed in Join Semilattice with Finite Suprema:
:$F \ne \O \implies y \in x$
By Supremum of Empty Set is Smallest Element:
:$F = \O \implies y = \bot_L$
where $\bot_L$ is the smallest element in $L$.
By Bottom in Ideal:
:$F = \O \implies y \in x$
Thus $y \in x$
Let $z \in x$.
We will prove that
:$F \subseteq y^\preceq$
Let $u \in F$.
By definitions of supremum and upper bound:
:$u \preceq y$
Thus by definition of lower closure of element:
:$u \in y^\preceq$
By Lower Closure of Element is Ideal:
:$y^\preceq$ is ideal in $L$.
Then
:$y^\preceq \in \set {I \in \map {\operatorname {Ids} } L: F \subseteq I}$
By definition of intersection:
:$z \in y^\preceq$
Thus by definition of lower closure of element:
:$z \preceq y$
Hence $x$ is principal ideal.
=== Necessary Condition ===
Assume that
$x$ is principal ideal.
By definition of principal ideal:
:$\exists y \in x: y$ is upper bound for $x$.
We will prove that
:$\exists F \in \map {\operatorname {Fin} } S: F \subseteq x \land x = \bigcap \set {I \in \map {\operatorname {Ids} } L: F \subseteq I}$
Define $F = \set y$
By Singleton is Finite:
:$F$ is finite.
Thus by definition of $\operatorname {Fin}$:
:$F \in \map {\operatorname {Fin} } S$
Thus by definitions of subset and singleton:
:$F \subseteq x$
We will prove that
:$\forall z: z \in x \iff z \in \bigcap \set {I \in \map {\operatorname {Ids} } L: F \subseteq I}$
Thus by definition of set equality:
:$x = \bigcap \set {I \in \map {\operatorname {Ids} } L: F \subseteq I}$
Thus by Compact Element iff Existence of Finite Subset that Element equals Intersection and Includes Subset:
:$x$ is compact element.
== Sources ==
* 
* 
Category:Join and Meet Semilattices
Category:Way Below Relation
Compact First-Countable Space is Sequentially Compact

== Theorem ==
Let $T = \struct {S, \tau}$ be a compact first-countable topological space.
Then every infinite sequence in $S$ has a convergent subsequence; that is, $T$ is sequentially compact.
== Proof ==
Let $\sequence {x_n}_{n \mathop \in \N}$ be an infinite sequence in $S$.
 that $\sequence {x_n}_{n \mathop \in \N}$ does not have a convergent subsequence.
By Accumulation Point of Infinite Sequence in First-Countable Space is Subsequential Limit, it follows that $\sequence {x_n}_{n \mathop \in \N}$ does not have an accumulation point in $S$.
Thus, for each $x \in S$, we can select an open set $U_x$ such that $x \in U_x$ and $U_x$ only contains $x_n$ for finitely many $n \in \N$.
The set $\UU = \set {U_x : x \in X}$ is clearly an open cover for $S$.
By the definition of a compact space, there exists a finite subcover $\set {U_{x_1}, U_{x_2}, \ldots, U_{x_m} }$ of $\UU$ for $X$.
Then $U_{x_1} \cup U_{x_2} \cup \cdots \cup U_{x_m}$ contains all of $S$ since it is a cover.
However, each open set in this union only contains $x_n$ for a finite number of $n$, and this is a finite union of sets.
So the union only contains $x_n$ for finitely many $n$.
This implies that $x_n \in X$ for only finitely many $n \in \N$.
This is a contradiction, since $\sequence {x_n}$ is a sequence in $S$, and so $x_n \in S$ for all $n \in \N$.
Category:Sequences
Category:Compact Spaces
Category:First-Countable Spaces
Compact Hausdorff Space is Locally Compact

== Theorem ==
Let $T = \left({S, \tau}\right)$ be a Hausdorff space which is compact.
Then $T$ is locally compact.
== Proof 1 ==
== Proof 2 ==
Category:Compact Spaces
Category:Hausdorff Spaces
Category:Locally Compact Hausdorff Spaces
Category:Compact Hausdorff Space is Locally Compact
Compact Hausdorff Space is T4

== Theorem ==
Let $T = \struct {S, \tau}$ be a compact Hausdorff space.
Then $T$ is a $T_4$ space.
== Proof ==
We have that a Compact Subspace of Hausdorff Space is Closed.
We also have that a Closed Subspace of Compact Space is Compact.
We also have that Disjoint Compact Sets in Hausdorff Space have Disjoint Neighborhoods.
$T$ is a $T_4$ space when any two disjoint closed subsets of $S$ are separated by neighborhoods.
Hence the result.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $2$: Separation Axioms
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Compactness Properties and the $T_i$ Axioms
Category:T4 Spaces
Category:Hausdorff Spaces
Category:Compact Spaces