Closed Sets in Compact Topological Spaces

# Closed Sets in Compact 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.

We will now look at a very nice theorem which says that if $X$ is a compact topological space, then any closed subset of $A$ of $X$ will also be compact in $X$.

 Theorem 1: Let $X$ be a compact topological space and let $A \subseteq X$. If $A$ is closed in $X$ then $A$ is compact in $X$.
• Proof: Let $\mathcal F = \{ U_i \}_{i \in I}$ be an open covering of $A$. Then we have that:
(1)
\begin{align} \quad A \subseteq \bigcup_{i \in I} A_i \end{align}
• Since $A$ is closed, $A^c = X \setminus A$ is open. Notice that $\{ U_i \}_{i \in I} \cup \{ X \setminus A \}$ is therefore an open covering of all of $X$. Since $X$ is a compact space, there exists a finite open covering of $X$: $\{ U_1, U_2, ..., U_n \} \cup \{ X \setminus A \}$ such that:
(2)
\begin{align} \quad X \subseteq \left ( \bigcup_{i=1}^{n} U_i \right ) \cup (X \setminus A) \end{align}
• But then $\{ U_1, U_2, ..., U_n \}$ is a finite subcover of $A$. Therefore, $A$ is compact in $X$. $\blacksquare$