# Compact T2 Spaces are T4 Spaces

Recall from the T2 (Hausdorff) Topological Spaces page that a topological space $X$ is said to be T_{2} 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 T_{4} or Normal Hausdorff if $X$ is both normal and T_{1}, 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 T_{1} propery.

On the Compact T2 Spaces are T3 Spaces page we proved that if $X$ is a T_{2} space that is compact then it is also a T_{3} space and we noted that in fact every T_{2} space that is compact is further a T_{4} space. We will now prove this assertion.

Theorem 1: Let $X$ be a topological space. If $X$ is T_{2} and compact then $X$ is T_{4}. |

**Proof:**Let $X$ be a T_{2}space that is compact. Since $X$ is a T_{2}space, it is also a T_{1}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 T
_{3}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:

- 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 T
_{1}and normal we have that $X$ is a T_{4}space. $\blacksquare$