compactness is preserved under a continuous map

Theorem [1, 2] Suppose $f:X\to Y$ is a continuous map between topological spaces $X$ and $Y$. If $X$ is compact and $f$ is surjective, then $Y$ is compact.

The inclusion map $[0,1]\hookrightarrow [0,2)$ shows that the requirement for $f$ to be surjective cannot be omitted. If $X$ is compact and $f$ is continuous we can always conclude, however, that $f(X)$ is compact, since $f:X\to f(X)$ is continuous (http://planetmath.org/IfFcolonXtoYIsContinuousThenFcolonXtoFXIsContinuous).

Proof of theorem. (Following [1].) Suppose $\{V_{\alpha }\mid \alpha \in I\}$ is an arbitrary open cover for $f(X)$. Since $f$ is continuous, it follows that

$$\{f^{-1}(V_{\alpha })\mid \alpha \in I\}$$

is a collection of open sets in $X$. Since $A\subseteq f^{-1}f(A)$ for any $A\subseteq X$, and since the inverse commutes with unions (see this page (http://planetmath.org/InverseImage)), we have

$X$ $\subseteq$ $f^{-1}f(X)$

$=$ $f^{-1}\left({\displaystyle \bigcup _{\alpha \in I}}(V_{\alpha })\right)$

$=$ ${\displaystyle \bigcup _{\alpha \in I}}{f}^{-1}(V_{\alpha }).$

Thus $\{f^{-1}(V_{\alpha })\mid \alpha \in I\}$ is an open cover for $X$. Since $X$ is compact, there exists a finite subset $J\subseteq I$ such that $\{f^{-1}(V_{\alpha })\mid \alpha \in J\}$ is a finite open cover for $X$. Since $f$ is a surjection, we have $f{f}^{-1}(A)=A$ for any $A\subseteq Y$ (see this page (http://planetmath.org/InverseImage)). Thus

$f(X)$ $=$ $f\left({\displaystyle \bigcup _{i\in J}}{f}^{-1}(V_{\alpha })\right)$

$=$ $f{f}^{-1}{\displaystyle \bigcup _{i\in J}}{f}^{-1}(V_{\alpha })$

$=$ ${\displaystyle \bigcup _{i\in J}}{V}_{\alpha }.$

Thus $\{V_{\alpha }\mid \alpha \in J\}$ is an open cover for $f(X)$, and $f(X)$ is compact. $\mathrm{\square }$

A shorter proof can be given using the characterization of compactness by the finite intersection property (http://planetmath.org/ASpaceIsCompactIfAndOnlyIfTheSpaceHasTheFiniteIntersectionProperty):

Shorter proof. Suppose $\{A_i\mid i\in I\}$ is a collection of closed subsets of $Y$ with the finite intersection property. Then $\{f^{-1}(A_i)\mid i\in I\}$ is a collection of closed subsets of $X$ with the finite intersection property, because if $F\subseteq I$ is finite then

$$\bigcap _{i\in F}{f}^{-1}(A_i)=f^{-1}\left(\bigcap _{i\in F}{A}_i\right),$$

which is nonempty as $f$ is a surjection. As $X$ is compact, we have

$$f^{-1}\left(\bigcap _{i\in I}{A}_i\right)=\bigcap _{i\in I}{f}^{-1}(A_i)\ne \mathrm{\varnothing }$$

and so ${\bigcap }_{i\in I}{A}_i\ne \mathrm{\varnothing }$. Therefore $Y$ is compact. $\mathrm{\square }$

Title	compactness is preserved under a continuous map
Canonical name	CompactnessIsPreservedUnderAContinuousMap
Date of creation	2013-03-22 13:55:50
Last modified on	2013-03-22 13:55:50
Owner	yark (2760)
Last modified by	yark (2760)
Numerical id	13
Author	yark (2760)
Entry type	Theorem
Classification	msc 54D30
Related topic	ContinuousImageOfACompactSpaceIsCompact
Related topic	ContinuousImageOfACompactSetIsCompact
Related topic	ConnectednessIsPreservedUnderAContinuousMap