continuous image of a compact set is compact

Theorem 1.

The continuous image of a compact set is also compact.

Proof.

Let $X$ and $Y$ be topological spaces, $f:X\to Y$ be continuous, $A$ be a compact subset of $X$, $I$ be an indexing set, and ${\{V_{\alpha }\}}_{\alpha \in I}$ be an open cover of $f(A)$. Thus, $f(A)\subseteq {\displaystyle \bigcup _{\alpha \in I}}{V}_{\alpha }$. Therefore, $A\subseteq f^{-1}\left(f(A)\right)\subseteq f^{-1}\left({\displaystyle \bigcup _{\alpha \in I}}{V}_{\alpha }\right)={\displaystyle \bigcup _{\alpha \in I}}{f}^{-1}(V_{\alpha })$.

Since $f$ is continuous, each $f^{-1}(V_{\alpha })$ is an open subset of $X$. Since $A\subseteq {\displaystyle \bigcup _{\alpha \in I}}{f}^{-1}(V_{\alpha })$ and $A$ is compact, there exists $n\in \mathbb{N}$ with $A\subseteq {\displaystyle \bigcup _{j=1}^n}{f}^{-1}\left(V_{{\alpha }_j}\right)$ for some ${\alpha }_1,\mathrm{\dots },{\alpha }_n\in I$. Hence, $f(A)\subseteq f\left({\displaystyle \bigcup _{j=1}^n}{f}^{-1}(V_{{\alpha }_j})\right)=f\left(f^{-1}\left({\displaystyle \bigcup _{j=1}^n}{V}_{{\alpha }_j}\right)\right)\subseteq {\displaystyle \bigcup _{j=1}^n}{V}_{{\alpha }_j}$. It follows that $f(A)$ is compact.∎