Compact Subspace of Hausdorff Space is Closed/Proof 1

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

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

Then $C$ is closed in $H$.

Proof
From Subspace of Hausdorff Space is Hausdorff, any subspace of a Hausdorff space is itself Hausdorff.

Let $a \in A \setminus C$.

We are going to prove that there exists an open set $U_a$ such that $a \in U_a \subseteq A \setminus C$.

For any single point $x \in C$, the Hausdorff condition ensures the existence of disjoint open set $U \left({x}\right)$ and $V \left({x}\right)$ containing $a$ and $x$ respectively.

Suppose there were only a finite number of points $x_1, x_2, \ldots, x_r$ in $C$.

Then we could take $\displaystyle U_a = \bigcap_{i \mathop = 1}^r U \left({x_i}\right)$ and get $a \in U_a \subseteq A \setminus C$.

Now suppose $C$ is not finite.

The set $\left\{{V \left({x}\right): x \in C}\right\}$ is an open cover for $C$.

As $C$ is compact, it has a finite subcover, say $\left\{{V \left({x_1}\right), V \left({x_2}\right), \ldots, V \left({x_r}\right)}\right\}$.

Let $\displaystyle U_a = \bigcap_{i \mathop = 1}^r U \left({x_i}\right)$.

Then $U_a$ is open because it is a finite intersection of open sets.

Also, $a \in U_a$ because $a \in U \left({x_i}\right)$ for each $i = 1, 2, \ldots, r$.

Finally, if $b \in U_a$ then for any $i = 1, 2, \ldots, r$ we have $b \in U \left({x_i}\right)$.

Hence $b \notin V \left({x_i}\right)$, so $b \notin C$, since $\displaystyle C = \bigcup_{i \mathop = 1}^r V \left({x_i}\right)$.

Thus $U_a \subseteq A \setminus C$.

Then $\displaystyle A \setminus C = \bigcup_{a \mathop \in A \mathop \setminus C} U_a$, so $A \setminus C$ is open and so $C$ is closed.