Compact Subspace of Hausdorff Space is Closed/Proof 1

Proof
From Subspace of Hausdorff Space is Hausdorff, a 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 $\map U x$ and $\map V x$ 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 $\displaystyle U_a = \bigcap_{i \mathop = 1}^r \map U {x_i}$ and get $a \in U_a \subseteq A \setminus C$.

Now suppose $C$ is not finite.

The set $\set {\map V x: x \in C}$ is an open cover for $C$.

As $C$ is compact, it has a finite subcover, say $\set {\map V {x_1}, \map V {x_2}, \dotsc, \map V {x_r} }$.

Let $\displaystyle U_a = \bigcap_{i \mathop = 1}^r \map U {x_i}$.

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

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

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

Because $\displaystyle C \subseteq \bigcup_{i \mathop = 1}^r \map V {x_i}$:
 * $b \notin \map V {x_i}$, so $b \notin C$

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.

It follows that $C$ is closed.