# 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, $f\colon X\to Y$ 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, $\displaystyle f(A)\subseteq\bigcup_{\alpha\in I}V_{\alpha}$. Therefore, $\displaystyle A\subseteq f^{-1}\bigg{(}f(A)\bigg{)}\subseteq f^{-1}\left(% \bigcup_{\alpha\in I}V_{\alpha}\right)=\bigcup_{\alpha\in I}f^{-1}(V_{\alpha})$.

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

Title continuous image of a compact set is compact ContinuousImageOfACompactSetIsCompact 2013-03-22 15:53:14 2013-03-22 15:53:14 Wkbj79 (1863) Wkbj79 (1863) 16 Wkbj79 (1863) Theorem msc 54D30 CompactnessIsPreservedUnderAContinuousMap