## You are here

Homecompactness is preserved under a continuous map

## Primary tabs

# 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.

*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), 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). 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:

*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.

## Mathematics Subject Classification

54D30*no label found*

- Forums
- Planetary Bugs
- HS/Secondary
- University/Tertiary
- Graduate/Advanced
- Industry/Practice
- Research Topics
- LaTeX help
- Math Comptetitions
- Math History
- Math Humor
- PlanetMath Comments
- PlanetMath System Updates and News
- PlanetMath help
- PlanetMath.ORG
- Strategic Communications Development
- The Math Pub
- Testing messages (ignore)

- Other useful stuff

## Recent Activity

new question: Prove a formula is part of the Gentzen System by LadyAnne

Mar 30

new question: A problem about Euler's totient function by mbhatia

new problem: Problem: Show that phi(a^n-1), (where phi is the Euler totient function), is divisible by n for any natural number n and any natural number a >1. by mbhatia

new problem: MSC browser just displays "No articles found. Up to ." by jaimeglz

Mar 26

new correction: Misspelled name by DavidSteinsaltz

Mar 21

new correction: underline-typo by Filipe

Mar 19

new correction: cocycle pro cocyle by pahio

Mar 7

new image: plot W(t) = P(waiting time <= t) (2nd attempt) by robert_dodier

new image: expected waiting time by robert_dodier

new image: plot W(t) = P(waiting time <= t) by robert_dodier

## Corrections

Broken by Mindspa ✓

Broken by Mindspa ✓

discussion by matte ✓

alternative proof by matte ✓