Compact Subspace of Hausdorff Space is Closed

Theorem
Let $\left({H, \vartheta}\right)$ be a Hausdorff space.

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

Then $C$ is closed in $H$.

Proof
Consider any element $x \in H \setminus C$ of the relative complement of $C$ in $H$.

Define:
 * $\displaystyle \mathcal O = \left\{{V \in \vartheta: \exists U \in \vartheta: x \in U, \, U \cap V = \varnothing}\right\}$

By the definition of a Hausdorff space, it follows that $\mathcal O$ is an open cover for $C$.

By the definition of a compact space, there exists a finite subcover $\mathcal F$ of $\mathcal O$ for $C$.

Define:
 * $\forall V \in \mathcal F: \mathcal A_V = \left\{{U \in \vartheta: x \in U, \, U \cap V = \varnothing}\right\}$

By the definition of $\mathcal O$, it follows that $\mathcal A_V$ is non-empty.

Since the finite cartesian product of non-empty sets is non-empty, there exists a family $\left\langle{U_V}\right\rangle_{V \in \mathcal F}$ such that:
 * $\forall V \in \mathcal F: x \in U_V \in \vartheta, \, U_V \cap V = \varnothing$

Define:
 * $\displaystyle U = \bigcap_{V \mathop \in \mathcal F} U_V$

By General Intersection Property of Topological Space, it follows that $U \in \vartheta$.

Clearly, $x \in U$.

We have that:

Hence, because $\subseteq$ is transitive, we have that $U \subseteq H \setminus C$.

By the definition of the interior, it follows that $x \in \left({H \setminus C}\right)^{\circ}$.

Since $x \in H \setminus C$ is arbitrary, it follows by the definition of a subset that $H \setminus C \subseteq \left({H \setminus C}\right)^{\circ}$.

By Set Interior is Largest Open Set, $\left({H \setminus C}\right)^{\circ} \subseteq H \setminus C$.

Therefore, by Equality of Sets, $H \setminus C = \left({H \setminus C}\right)^{\circ}$.

Hence, by Interior of Open Set, $H \setminus C \in \vartheta$.

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