# Compact T2 Spaces are T3 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 T3 (Regular Hausdorff) Topological Spaces page that a topological space $X$ is said to be T_{3} or Regular Hausdorff if $X$ is both regular and T_{1}, 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 T_{1} propery.

We will now look at a very nice theorem which will tell us that every T_{2} space that is compact is a T_{3} space.

We will now look at a very nice theorem which will tell us that every T_{2} space that is compact is a T_{3} space. (We will see later that in fact every T_{2} space that is compact is further a T_{4} space).

Theorem 1: Let $X$ be a topological space. If $X$ is T_{2} and compact then $X$ is T_{3}. |

**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 regular.

- Let $x \in X$ and let $F$ be a closed set in $X$ that does not contain $x$. Since $X$ is a T
_{2}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:

- 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:

- 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 T
_{1}and regular we have that $X$ is T_{3}. $\blacksquare$