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# 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*

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new correction: Error in proof of Proposition 2 by alex2907

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## Corrections

Broken by Mindspa ✓

Broken by Mindspa ✓

discussion by matte ✓

alternative proof by matte ✓