In Hausdorff spaces, compact sets are closed

Question

I've been stuck on this exercise for awhile. It was suggested that I fix a point in the compact set, but I didn't know what to do with that. Can anyone provide an answer?

Answer 1

I just so happened to have just done this exercise. Note that user 313 is right about just googling this sort of thing. I will provide the proof nonetheless.

Proof.

Let $H$ be Hausdorff and $C$ compact. Note that by showing that $H \setminus C$ is open, this shows that $C$ is closed. Hence showing $H \setminus C$ is open is our goal.

Fix $a \in (H \setminus C)$

$\forall_{x \in C}$, $\exists_{U_x, \; V_x}$ s.t. $x \in U_x$ and $a \in V_x$, $U_x \cap V_x = \emptyset$

$\bigcup_{x \in C} U_x$ covers $C$, so $\exists \; \{U_x, \cdots , U_{x_k}\}$ that covers $C$

Take corresponding $\{V_x, \cdots , V_{x_k}\}$. $\left ( \bigcap_{i=1}^{k}V_{x_i} \right ) \cap \left ( \{U_x, \cdots , U_{x_k}\} \supset C \right ) = \emptyset$

$a \in \bigcap V_{x_i}$ is open. And $\bigcap V_{x_k} \subset (H \setminus C)$.

Hence we have shown that $H\setminus C$ is open as desired which implies that $C$ is closed.