Compact T2 Spaces are T3 Spaces

# Compact T2 Spaces are T3 Spaces

Recall from the T2 (Hausdorff) Topological Spaces page that a topological space $X$ is said to be T2 or Hausdorff if for all $x, y \in X$, $x \neq y$ there exists open neighbourhoods $U$ of $x$ and $V$ of $y$ such that $U \cap V = \emptyset$.

Also recall from the T3 (Regular Hausdorff) Topological Spaces page that a topological space $X$ is said to be T3 or Regular Hausdorff if $X$ is both regular and T1, that is, for every $x \in X$ and for all closed sets $F$ not containing $x$ there exists open sets $U$ and $V$ such that $x \in U$, $F \subseteq V$, and $U \cap V = \emptyset$ alongside with the T1 propery.

We will now look at a very nice theorem which will tell us that every T2 space that is compact is a T3 space.

We will now look at a very nice theorem which will tell us that every T2 space that is compact is a T3 space. (We will see later that in fact every T2 space that is compact is further a T4 space).

 Theorem 1: Let $X$ be a topological space. If $X$ is T2 and compact then $X$ is T3.
• Proof: Let $X$ be a T2 space that is compact. Since $X$ is a T2 space, it is also a T1 space and so all that remains to show is that $X$ is regular.
• Let $x \in X$ and let $F$ be a closed set in $X$ that does not contain $x$. Since $X$ is a T2 space, for this $x \in X$ and for every $y \in F$ there exists open neighbourhoods $U_x$ of $x$ and $U_y$ of $y$ such that $U_x \cap U_y = \emptyset$.
• Consider the following collection of open sets:
(1)
\begin{align} \quad \mathcal F = \{ U_y : y \in F \} \end{align}
• Then $\mathcal F$ is an open cover of $F$. Furthermore, since $X$ is compact and $F$ is closed in $X$ we have that $F$ is compact in $X$. So, for the open cover of $F$ above there exists a finite subcover:
(2)
\begin{align} \quad \mathcal F^* = \{ U_{y_1}, U_{y_2}, …, U_{y_n} \} \end{align}
• Let $V = \bigcup_{i=1}^{n} U_{y_i}$. Then $V$ is open in $X$, contains $y$, but does not contain $x$. Let $U = \bigcap_{i=1}^{n} U_{x_i}$. Then $U$ is open (as it is a finite intersection of open sets) that contains $x$ and not $y$. Moreover, $U \cap V = \emptyset$.
• So, for all $x \in X$ and for every closed set $F$ that does not contain $x$ there exists open sets $U$ and $V$ such that $x \in U$ and $F \subseteq V$, so $X$ is regular.
• Since $X$ is both T1 and regular we have that $X$ is T3. $\blacksquare$