To show that any compact Hausdorff space is $T_4$, one may first show
that it’s $T_3$. To see this, using set theory notations may be quite
*difficult*. For mediocre students, the contents of the following
seciton may sound *unnatural*.

## A sketch of the proof

Suppose that we want to separate a point $x \notin A$ and a closed set
$A$ in a compact $T_2$ space $X$ by two *disjoint* open sets $U$ and
$V$ so that $x \in U$ and $A \subseteq V$. A standard proof is to
apply the $T_2$ property of $X$ to $x$ and each $y \in A$ so as to
yield two *disjoint* open sets $U_y,V_y \in \mathcal{T}$ such that $x \in U_y$ and $y
\in V_y$. Since $A$ should be contained in an open set $V$, in
other words, an *open cover* of $A$ is needed, one might be tempted to
construct the following union of open sets.

However, one *can’t* ensure that the following *infinite* intersection
of open sets is open.

For instance, by the Nested Interval Theorem,

Thus, one applies the compactness of $X$ to get *finite* versions of
\eqref{Vinf} and \eqref{Uinf}.

## A snapshot of this fact

Since it’s so *difficult* to remember every detail of the proof, I
love illustrating it using a picture.

How can a compact regular space be regular? See my next post.