Basic Theorems Regarding Compact Sets in a Metric Space

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 for every open covering of $S$ there exists a finite subcovering of $S$.

We will now look at some theorems regarding compact sets in a metric space.

Theorem 1: Let $(M, d)$ be a metric space and let $S, T \subseteq M$. Then if $S$ is closed and $T$ is compact in $M$ then $S \cap T$ is compact in $M$.

• Proof: Let $S$ be closed and let $T$ be compact in $M$. Notice that:

(1)

\begin{align} \quad S \cap T \subseteq T \end{align}

• Furthermore, $S \cap T$ is closed. This is because $S$ is given as closed, and since $T$ is compact we know that $T$ is closed (and bounded). So the finite intersection $S \cap T$ is closed. But any closed subset of a compact set is also compact as we proved on the Closed Subsets of Compact Sets in Metric Spaces page, so $S \cap T$ is compact in $M$. $\blacksquare$

Theorem 2: Let $(M, d)$ be a metric space and let $S_1, S_2, ..., S_n \subseteq M$ be a finite collection of compact sets in $M$. Then $\displaystyle{\bigcup_{i=1}^{n} S_i}$ is also compact in $M$.

• Proof: Let $S_1, S_2, ..., S_n \subseteq M$ be a finite collection of compact sets in $M$. Consider the union $S = \bigcup_{i=1}^{n} S_i$ and let $\mathcal F$ be any open covering of $S$, that is:

(2)

\begin{align} \quad S \subseteq \bigcup_{A \in \mathcal F} A \end{align}

• Now since $S_i \subseteq S$ for all $i \in \{1, 2, ..., n \}$ we see that $\mathcal F$ is also an open covering of $S$ and so there exists a finite subcollection $\mathcal F_i \subseteq \mathcal F$ that also covers $S_i$, i.e.:

(3)

\begin{align} \quad S_i \subseteq \bigcup_{A \in \mathcal F_i} A \end{align}

• Let $\mathcal F^* = \bigcup_{i=1}^{n} \mathcal F_i$. Then $\mathcal F^*$ is finite since it is equal to a finite union of finite sets. Furthermore:

(4)

\begin{align} \quad S = \bigcup_{i=1}^{n} S_i \subseteq \bigcup_{i=1}^{n} \left ( \bigcup_{A \in \mathcal F_i} A \right ) = \bigcup_{A \in \mathcal F^*} A \end{align}

• So $\mathcal F^* \subseteq \mathcal F$ is a finite open subcovering of $S$. So for all open coverings $\mathcal F$ of $S$ there exists a finite open subcovering of $S$, so $\displaystyle{S = \bigcup_{i=1}^{n} S_i}$ is compact in $M$. $\blacksquare$

Theorem 3: Let $(M, d)$ be a metric space and let $\mathcal C$ be an arbitrary collection of compact sets in $M$. Then $\displaystyle{\bigcap_{C \in \mathcal C} C}$ is also compact in $M$.

• Proof: Let $\mathcal C$ be an arbitrary collection of compact sets in $M$. Notice that for all $C \in \mathcal C$ that:

(5)

\begin{align} \quad \bigcap_{C \in \mathcal C} C \subseteq C \end{align}

• Furthermore, since each $C \in \mathcal C$ is compact, then each $C$ is closed (and bounded). An arbitrary intersection of closed sets is closed, and so $\displaystyle{\bigcap_{C \in \mathcal C}}$ is a closed subset of the compact set $C$. Therefore by the theorem referenced earlier, $\displaystyle{\bigcap_{C \in \mathcal C} C}$ is compact in $M$. $\blacksquare$