Closed Subsets of Compact Sets in Metric Spaces

# Closed Subsets of Compact Sets in Metric Spaces

Recall from the Compact Sets in a Metric Space page that if $(M, d)$ is a metric space then a set $S \subseteq M$ is said to be compact in $M$ if every open covering of $S$ has a finite subcovering.

We will now look at a very useful theorem which shows us that if $S \subseteq T \subseteq M$ and if $S$ is closed and $T$ is compact then $S$ is compact as well.

Theorem 1: Let $(M, d)$ be a metric space and let $S, T \subseteq M$ be such that $S \subseteq T$. Furthermore, let $S$ be closed and let $T$ be compact in $M$. Then $S$ is compact in $M$. |

**Proof:**Let $\mathcal F$ be any open covering of $S$. Since $S$ is closed, $S^c = M \setminus S$ is open, and so in fact, $\mathcal F \cup \{ M \setminus S \}$ is an open covering of $T$:

\begin{align} \quad T \subseteq \left ( \bigcup_{A \in \mathcal F} A \right ) \cup (M \setminus S) \end{align}

- Now since $T$ is compact in $M$ there exists a finite subcollection $\mathcal F^* \subseteq \mathcal F \cup \{ M \setminus S \}$ such that:

\begin{align} \quad T \subseteq \bigcup_{A \in \mathcal F^*} A \end{align}

- Now $\mathcal F^*$ may contain $M \setminus S$. Regardless, $\mathcal F^*$ is a finite open subcovering of $T$ and since $S \subseteq T$, $\mathcal F^* \setminus \{ M \setminus S \} \subseteq \mathcal F$ is a finite open subcovering of $S$.

- So for every open covering of $S$ there exists a finite open subcovering of $S$ and by definition $S$ is compact. $\blacksquare$