Continuous Functions on Compact Sets of Metric Spaces

We have looked quite a bit at continuous functions so far, and now we will put out attention on functions that are continuous on compact sets $X$.

The following Theorem tells us that if $(S, d_S)$ and $(T, d_T)$ are metric spaces and $f : S \to T$ is a continuous function then for every subset $X$ which is compact in $S$ we will have that $f(X)$ is compact in $T$, that is, compact sets are mapped to compact sets under continuous $f$.

Theorem 1: Let $(S, d_S)$ and $(T, d_T)$ be metric spaces and let $f : S \to T$ be continuous. If $X$ is a compact subset of $S$ then the image $f(X)$ is a compact subset of $T$.

• Proof: Let $X \subseteq S$ be a compact subset of $S$ and let $f : S \to T$ be continuous.

• Consider any open covering of $f(X)$, $\mathcal F = \{ A_{\alpha} : \alpha \in \Gamma \}$ for $\Gamma$ as some indexing set. Then since $\mathcal F$ is an open covering of $f(X)$ we have that:

(1)

\begin{align} \quad f(X) \subseteq \bigcup_{\alpha \in \Gamma} A_{\alpha} \end{align}

• Since $\mathcal F$ is an OPEN covering, we have by definition that $A_{\alpha}$ is open in $T$ for all $\alpha \in \Gamma$. Since $f$ is continuous we have that $f^{-1}(A_{\alpha})$ is open in $X$ for all $\alpha \in \Gamma$ and furthermore:

(2)

\begin{align} \quad X \subseteq f^{-1} \left ( \bigcup_{\alpha \in \Gamma} A_{\alpha} \right ) = \bigcup_{\alpha \in \Gamma} f^{-1}(A_{\alpha}) \end{align}

• Therefore $\{ f^{-1}(A_{\alpha}) : \alpha \in \Gamma \}$ is an open covering of $X$. Since $X$ is compact in $S$ this open covering has a finite open subcovering, say $\{ f^{-1}(A_{\alpha_1}), f^{-1}(A_{\alpha_2}), ..., f^{-1}(A_{\alpha_n}) \}$ where $\alpha_1, \alpha_2, ..., \alpha_n \in \Gamma$ and also:

(3)

\begin{align} \quad X \subseteq \bigcup_{i=1}^{n} f^{-1}(A_{\alpha_i}) \end{align}

• But then:

(4)

\begin{align} \quad f(X) \subseteq f \left ( \bigcup_{i=1}^{n} f^{-1}(A_{\alpha_i}) \right ) = \bigcup_{i=1}^{n} f(f^{-1}(A_{\alpha_i})) = \bigcup_{i=1}^{n} A_{\alpha_i} \end{align}

• Therefore $\{ A_{\alpha_1}, A_{\alpha_2}, ..., A_{\alpha_n} \}$ is a finite open subcovering of $f(X)$. Therefore every open covering $\{ A_{\alpha} : \alpha \in \Gamma \}$ of $f(X)$ has a finite open subcovering, so $f(X)$ is compact in $T$. $\blacksquare$

Corollary 1: Let $(S, d_S)$ and $(T, d_T)$ be metric spaces and let $f : S \to T$ be continuous. If $X$ is a compact subset of $S$ then the image $f(X)$ is: closed, bounded, and every infinite subset of $f(X)$ has an accumulation point.

• Proof: By Theorem 1, $f(X)$ is a compact subset of $T$, and as we saw on the Boundedness of Compact Sets in a Metric Space, Closedness of Compact Sets in a Metric Space, and Every Infinite Subset of a Compact Set in a Metric Space Contains an Accumulation Point page we have the Corollary 1 follows immediately. $\blacksquare$

The following diagram illustrates the results of Theorem 1 and Corollary 1.