Compact Sets in Hausdorff Topological Spaces

Recall from the Compactness of Sets in a Topological Space page that if $X$ is a topological space and $A \subseteq X$ then $A$ is said to be compact in $X$ if every open cover of $A$ has a finite subcover.

On the Closed Sets in Compact Topological Spaces page we proved a very nice theorem which showed us that if $X$ is a compact topological space then every closed set $A$ in $X$ is also compact in $X$.

We will now look at a similar theorem which says that every compact set in a Hausdorff topological space is closed.

Theorem 1: Let $X$ be a Hausdorff topological space and let $A \subseteq X$. If $A$ is compact in $X$ then $A$ is closed in $X$.

• Proof: Let $A$ be a compact set in $X$. Let $y \in A^c$ be a fixed point. Since $X$ is a Hausdorff space, for all $x \in A$ there exists open neighbourhoods $U_x$ of $x$ and $V_{(y, x)}$ of $y$ such that:

(1)

\begin{align} \quad U_x \cap V_{(y, x)} = \emptyset \end{align}

• Then we have that $\{ U_x \}_{x \in A}$ is an open cover of $A$. Since $A$ is compact in $X$ there exists a finite subcover, $\{ U_{x_1}, U_{x_2}, ..., U_{x_n} \}$ where $x_i \in A$ for all $i \in \{ 1, 2, ..., n \}$ and:

(2)

\begin{align} \quad A \subseteq \bigcup_{i=1}^{n} U_{x_i} \end{align}

• Correspondingly, the set $\{ V_{(y, x_1)}, V_{(y, x_2)}, ..., V_{(y, x_n)} \}$ of open sets are such that $U_{x_i} \cap V_{(y, x_i)} = \emptyset$ for all $i \in \{1, 2, ..., n \}$. Let $U$ denote the following intersection:

(3)

\begin{align} \quad V = \bigcap_{i=1}^{n} V_{(y, x_i)} \end{align}

• Then $V$ is a finite intersection of open sets and is hence open. Moreover, $V$ is an open neighbourhood of $y$. Additionally, $A \cap V = \emptyset$ since $\displaystyle{\bigcup_{i=1}^{n} U_{x_i}}$ covers $A$.

• Since $A \cap V = \emptyset$, this implies that $V \subseteq A^c$. So $V$ is an open neighbourhood of $y$ that is fully contained in $A^c$, i.e., $y \in \mathrm{int} (A^c)$. Hence $\mathrm{int}(A^c) = A^c$ which shows that $A^c$ is open in $X$. Therefore $A$ is closed in $X$.