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 $ | ${f}^{-1}f(X)$ | ||
$=$ | ${f}^{-1}\left({\displaystyle \bigcup _{\alpha \in I}}({V}_{\alpha})\right)$ | |||
$=$ | $\bigcup _{\alpha \in I}}{f}^{-1}({V}_{\alpha}).$ |
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
$f(X)$ | $=$ | $f\left({\displaystyle \bigcup _{i\in J}}{f}^{-1}({V}_{\alpha})\right)$ | ||
$=$ | $f{f}^{-1}{\displaystyle \bigcup _{i\in J}}{f}^{-1}({V}_{\alpha})$ | |||
$=$ | $\bigcup _{i\in J}}{V}_{\alpha}.$ |
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. $\mathrm{\square}$
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 |
---|---|
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 |