Compact T2 Spaces are T4 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 T4 (Normal Hausdorff) Topological Spaces page that a topological space $X$ is said to be T₄ or Normal Hausdorff if $X$ is both normal and T₁, that is, for every pair of disjoint closed sets $E$ and $F$ there exists open sets $U$ and $V$ with $E \subseteq U$, $F \subseteq V$ such that $U \cap V = \emptyset$ alongside with the T₁ propery.

On the Compact T2 Spaces are T3 Spaces page we proved that if $X$ is a T₂ space that is compact then it is also a T₃ space and we noted that in fact every T₂ space that is compact is further a T₄ space. We will now prove this assertion.

Theorem 1: Let $X$ be a topological space. If $X$ is T2 and compact then $X$ is T4.

• 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 normal.

• Let $E$ and $F$ be disjoint closed sets in $X$. We have already proven that $X$ is a T₃ space, so, for every $x \in E$ there exists open sets $U_x$ and $V_x$ such that $x \in U_x$ and $F \subseteq V_x$. Consider the following collection:

(1)

\begin{align} \quad \mathcal F = \{ U_x : x \in E \} \end{align}

• Then $\mathcal F$ is an open cover of $E$. Since $E$ is closed in $X$ and $X$ is compact we have that $E$ is also compact. So this open cover has a finite subcover, $\mathcal F^* = \{ U_{x_1} U_{x_2}, …, U_{x_n} \}$. Let $U = \bigcup_{i=1}^{n} U_{x_1}$. Then $A \subseteq U$.

• Now let $V = \bigcap_{i=1}^{n} V_{x_i}$. Then $V$ is open in $X$, $F \subseteq V$, and by construction, $U \cap V = \emptyset$. So $X$ is normal.

• Since $X$ is both T₁ and normal we have that $X$ is a T₄ space. $\blacksquare$