Contents

Idea

The image under a continuous function of a compact topological space is itself compact (cor. below.)

This is a generalization of the extreme value theorem in analysis.

Statement

In fact the following more general statement holds

Lemma

Let $f \colon (X,\tau_X) \longrightarrow (Y,\tau_Y)$ be a continuous function between topological spaces such that

1. $(X,\tau_X)$ is a compact topological space;

2. $f \colon X \to Y$ is a surjective function.

Then also $(Y,\tau_Y)$ is compact.

Proof

Let $\{U_i \subset Y\}_{i \in I}$ be an open cover of $Y$. We need show that this has a finite sub-cover.

By the continuity of $f$ the pre-images $f^{-1}(U_i)$ form an open cover $\{f^{-1}(U_i) \subset X\}_{i \in I}$ of $X$. Hence by compactness of $X$, there exists a finite subset $J \subset I$ such that $\{f^{-1}(U_i) \subset X\}_{i \in J \subset I}$ is still an open cover of $X$. Finally, by surjectivity of $f$ it follows that

\begin{aligned} Y & = f(X) \\ & = f\left( \underset{i \in J}{\cup} f^{-1}(U_i) \right) \\ & = \underset{i \in J}{\cup} U_i \end{aligned}

where we used that images of unions are unions of images.

This means that also $\{U_i \subset Y\}_{i \in J \subset I}$ is still an open cover of $Y$, and in particular a finite subcover of the original cover.

Corollary

If $f \colon X \longrightarrow Y$ is a continuous function out of a compact topological space $X$ which is not necessarily surjective, then we may consider its image factorization

$f \;\colon\; X \longrightarrow im(f) \hookrightarrow Y$

with $im(f)$ regarded as a topological subspace of $Y$. Now by construction $X \to \im(f)$ is surjective, and so lemma implies that $im(f)$ is compact.

Last revised on November 20, 2017 at 07:09:14. See the history of this page for a list of all contributions to it.