continuous images of compact spaces are compact



topology (point-set topology, point-free topology)

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The image under a continuous function of a compact topological space is itself compact (cor. 1 below.)

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


In fact the following more general statement holds


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

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

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

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


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

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

Y =f(X) =f(iJf 1(U i)) =iJU i \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 iY} iJI\{U_i \subset Y\}_{i \in J \subset I} is still an open cover of YY, and in particular a finite subcover of the original cover.


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

f:Xim(f)Y f \;\colon\; X \longrightarrow im(f) \hookrightarrow Y

with im(f)im(f) regarded as a topological subspace of YY. Now by construction Xim(f)X \to \im(f) is surjective, and so lemma 1 implies that im(f)im(f) is compact.

Revised on May 19, 2017 08:40:24 by Urs Schreiber (