Compact Subspace of Hausdorff Space is Closed/Proof 2

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Theorem

Let $H = \struct {A, \tau}$ be a Hausdorff space.

Let $C$ be a compact subspace of $H$.


Then $C$ is closed in $H$.


Proof

For a subset $S \subseteq A$, let $S^{\complement}$ denote the relative complement of $S$ in $A$.


Consider an arbitrary point $x \in C^{\complement}$.

Define the set:

$\ds \OO = \set {V \in \tau: \exists U \in \tau: x \in U \subseteq V^\complement}$

By Empty Intersection iff Subset of Complement:

$U \subseteq V^\complement \iff U \cap V = \O$

Hence, by the definition of a Hausdorff space, it follows that $\OO$ is an open cover for $C$.


By the definition of a compact subspace, there exists a finite subcover $\FF$ of $\OO$ for $C$.

By the Principle of Finite Choice, there exists an $\FF$-indexed family $\family {U_V}_{V \mathop \in \FF}$ of elements of $\tau$ such that:

$\forall V \in \FF: x \in U_V \subseteq V^\complement$

Define:

$\ds U = \bigcap_{V \mathop \in \FF} U_V$

By General Intersection Property of Topological Space:

$U \in \tau$


Clearly, $x \in U$.

We have that:

\(\ds U\) \(\subseteq\) \(\ds \bigcap_{V \mathop \in \FF} \paren {V^\complement}\) Set Intersection Preserves Subsets
\(\ds \) \(=\) \(\ds \paren {\bigcup \FF}^\complement\) De Morgan's laws
\(\ds \) \(\subseteq\) \(\ds C^\complement\) Definition of Cover of Set and Relative Complement inverts Subsets

From Subset Relation is Transitive, we have that $U \subseteq C^\complement$.


Hence $C^\complement$ is a neighborhood of $x$.

From Set is Open iff Neighborhood of all its Points:

$C^\complement \in \tau$

That is, $C$ is closed in $H$.

$\blacksquare$