# compactness is preserved under a continuous map

Theorem [1, 2] Suppose $f\colon X\to Y$ 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 $f\colon X\to f(X)$ 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

 $\displaystyle X$ $\displaystyle\subseteq$ $\displaystyle f^{-1}f(X)$ $\displaystyle=$ $\displaystyle f^{-1}\big{(}\bigcup_{\alpha\in I}(V_{\alpha})\big{)}$ $\displaystyle=$ $\displaystyle\bigcup_{\alpha\in I}f^{-1}(V_{\alpha}).$

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 $ff^{-1}(A)=A$ for any $A\subseteq Y$ (see this page (http://planetmath.org/InverseImage)). Thus

 $\displaystyle f(X)$ $\displaystyle=$ $\displaystyle f\big{(}\bigcup_{i\in J}f^{-1}(V_{\alpha})\big{)}$ $\displaystyle=$ $\displaystyle ff^{-1}\bigcup_{i\in J}f^{-1}(V_{\alpha})$ $\displaystyle=$ $\displaystyle\bigcup_{i\in J}V_{\alpha}.$

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

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

 $f^{-1}\left(\,\bigcap_{i\in I}A_{i}\!\right)=\bigcap_{i\in I}f^{-1}(A_{i})\neq\varnothing$

and so $\bigcap_{i\in I}A_{i}\neq\varnothing$. Therefore $Y$ is compact. $\Box$

## References

• 1 I.M. Singer, J.A.Thorpe, Lecture Notes on Elementary Topology and Geometry, Springer-Verlag, 1967.
• 2 J.L. Kelley, General Topology, D. van Nostrand Company, Inc., 1955.
• 3 G.J. Jameson, Topology and Normed Spaces, Chapman and Hall, 1974.
Title compactness is preserved under a continuous map CompactnessIsPreservedUnderAContinuousMap 2013-03-22 13:55:50 2013-03-22 13:55:50 yark (2760) yark (2760) 13 yark (2760) Theorem msc 54D30 ContinuousImageOfACompactSpaceIsCompact ContinuousImageOfACompactSetIsCompact ConnectednessIsPreservedUnderAContinuousMap