Preservation of Compactness under Continuous Maps

# Preservation of Compactness under Continuous Maps

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 covering of $A$ has a finite subcover.

As we will see in the following theorem, compactness is preserved under continuous maps, i.e., if $X$ and $Y$ are topological spaces, and $A \subseteq X$ is compact in $X$, and $f : A \to Y$ is a continuous map, then must like how we saw that connectedness is preserved under continuous maps on the Preservation of Connectivity under Continuous Maps page, the image $f(A)$ will also be compact in $Y$.

Theorem 1: Let $X$ and $Y$ be topological spaces, $A \subseteq X$, and $f : A \to Y$ be a continuous map. If $A$ is compact in $X$ then $f(A)$ is compact in $Y$. |

**Proof:**Let $\mathcal F = \{ U_i \}_{i \in I}$ be any open covering of $f(A)$ in $Y$. Then:

\begin{align} \quad f(A) \subseteq \bigcup_{i \in I} U_i \end{align}

- Taking the inverse images of both sides above and we see that:

\begin{align} \quad A & \subseteq f^{-1} \left ( \bigcup_{i \in I} U_i \right ) \\ \quad A & \subseteq\bigcup_{i \in I} f^{-1}(U_i) \end{align}

- Since $f$ is continuous and $U_i$ is open in $Y$ for all $i \in I$ we have that $f^{-1}(U_i)$ is open in $X$ for all $i \in I$. From above, we see that then $\{ f^{-1}(U_i) \}_{i \in I}$ is an open cover of $A$. Since $A$ is compact in $X$, this open cover has a finite subcover, say $\{ f^{-1}(U_{i_1}), f^{-1}(U_{i_2}), ..., f^{-1}(U_{i_n}) \}$ where $i_n \in I$ where:

\begin{align} \quad A \subseteq \bigcup_{k=1}^{n} f^{-1}(U_{i_k}) \end{align}

- Taking the image of both sides above and we have that:

\begin{align} \quad f(A) & \subseteq f \left ( \bigcup_{k=1}^{n} f^{-1}(U_{i_k}) \right ) \\ \quad f(A) & \subseteq\bigcup_{k=1}^{n} f(f^{-1}(U_{i_k})) \\ \quad f(A) & \subseteq \bigcup_{k=1}^{n} U_{i_k} \end{align}

- Thus $\mathcal F^* = \{ U_{i_1}, U_{i_2}, ..., U_{i_n} \}$ is a finite subcover of $\mathcal F$. Hence $f(A)$ is compact in $Y$. $\blacksquare$