Compact-Hausdorff space in general topology

Question

Why Is in general topology every Compact-Hausdorff space normal? Is every normal space Compact-Hausdorff space?

Answer 1

This is classic and in most text books. First show that $X$ is regular:

Let $A$ be closed in $X$ and $p \notin A$. For each $a \in A$, $a \neq p$ and so Hausdorffness gives us open sets $U_a$ and $V_a$ such that

$$a \in U_a, p \in V_a \text{ and } U_a \cap V_a = \emptyset\tag{1}$$

Then $\{U_a: a \in A\}$ is an open cover of $A$, which is compact too (being closed in $X$) and so there are finitely many $\{a_1, \ldots, a_n\}$ such that $$A \subseteq \bigcup_{i=1}^n U_{a_i}$$

and define $U= \bigcup_{i=1}^n U_{a_i}$, an open neighbourhood of $A$ and $V= \bigcap_{i=1}^n V_{a_i}$, which is an open (because we have a finite intersection) neighbourhood of $p$. And $U \cap V = \emptyset$, or otherwise $x \in U \cap V$ implies there is some $1 \le i \le n$ with $x \in U_{a_i}$ but then $x \in V \subseteq V_{a_i}$ as well, contradicting $(1)$. So we have disjoint neighbourhoods for a closed $A$ and a point not in $A$.

For two disjoint closed sets $A$ and $B$ we apply the regularity we just proved to $A$ and each $b \in B$ and find a finite subcover for $B$ and their union and the corresponding intersection of open neighbourhoods of $A$ can be shown very similarly to be disjoint neighbourhoods of $A$ and $B$ again.