Compact Subsets of T3 Spaces

From ProofWiki
Jump to navigation Jump to search



Theorem

Let $T = \struct {S, \tau}$ be a $T_3$ space.

Let $A \subseteq S$ be compact in $T$.


Then for each $U \in \tau$ such that $A \subseteq U$:

$\exists V \in \tau: A \subseteq V \subseteq V^- \subseteq U$

where $V^-$ denotes the closure of $V$.


Proof

Let $A \subseteq S$ be compact in $T$.

Let $U \in \tau$ such that $A \subseteq U$.


Since $T$ is $T_3$:

Each open set contains a closed neighborhood around each of its points:
$\forall x \in U: \exists N_x: \relcomp S {N_x} \in \tau: \exists V_x \in \tau: x \in V_x \subseteq N_x \subseteq U$


Note that $\set {V_x: x \in A}$ forms an open cover for $A$.

By compactness of $A$, $\set {V_x: x \in A}$ has a finite subcover.

Therefore there is a finite subset $I \subseteq A$ such that:

$\set {V_x: x \in I}$ is an open cover for $A$.


Let $V = \ds \bigcup_{i \mathop \in I} V_x$, $N = \ds \bigcup_{i \mathop \in I} N_x$.

By definition of a topology, the set $\ds \bigcup_{i \mathop \in I} V_x$ is open since it is a union of open sets.

By Finite Union of Closed Sets is Closed in Topological Space, the set $\ds \bigcup_{i \mathop \in I} N_x$ is closed since it is a finite union of closed sets.

By Set Union Preserves Subsets, since $V_x \subseteq N_x \; \forall x \in I \subseteq U$:

$V \subseteq N$

By Set Closure is Smallest Closed Set in Topological Space:

$V \subseteq V^- \subseteq N$

Thus:

\(\ds A\) \(\subseteq\) \(\ds V\) Definition of Open Cover
\(\ds \) \(\subseteq\) \(\ds V^-\)
\(\ds \) \(\subseteq\) \(\ds N\)
\(\ds \) \(\subseteq\) \(\ds U\) Union of Subsets is Subset: $\forall x \in I \subseteq U: N_x \subseteq U$

satisfies our claim.

$\blacksquare$


Sources