## Cute Proofs: Continuous Image of Compact Set is Compact.

### October 3, 2010

While not an especially difficult proof, this kind of this is standard when we’re using compact sets.  What we’re going to do is take an open cover, pull it back, use compactness, then push it forward.  Let’s get down to it!

Theorem: If $f: X\rightarrow Y$ is continuous and $X$ is a compact space, then $f(X)$ is also compact.

It suffices to show that for every open cover of $f(X)$ there exists a finite subcover.  Let’s take $\{U_{\alpha}\}_{\alpha}$ is an arbitrary cover of $f(X)$.  Note that $\bigcup_{\alpha} f^{-1}(U_{\alpha}) \supseteq X$ and each $f^{-1}(U_{\alpha})$ is open since $f$ is continuous.  Since $X$ is compact, we have that $\{f^{-1}(U_{1}), f^{-1}(U_{2}), \dots,f^{-1}( U_{n})\}$ (with potentially reordering or renaming the indices) is a finite cover of our set $X$.  Therefore, we have that

$\{f(f^{-1}(U_{1})), f(f^{-1}(U_{2})), \dots, f(f^{-1}(U_{n}))\} \\ = \{U_{1}, U_{2}, \dots, U_{n}\}$

(with equality holding because $f$ is surjective onto its image) is a cover for $f(X)$ and is therefore a finite subcover of our original cover.  This proves that $f(X)$ is compact.  $\hfill \Box$

This is a nice theorem, because we can push things into spaces that are strange and preserve compactness knowing only that our map is continuous and our original set is compact.

Advertisements

### 7 Responses to “Cute Proofs: Continuous Image of Compact Set is Compact.”

1. Rajesh Banik said

thank u sir for proving this theorem wonderfully……

2. Carlos said

The word “continuous” is mentioned in the title, but not in the statement of the theorem. I would suggest adding it.

One more thing, in the proof you say that $f$ is defined on all of $Y$. I quite don’t get that. The domain of $f$ is $X$, not $Y$.

• James said

It seems I’ve succumb to the Topologist’s illness in assuming (without explicitly saying so) that the map in question is continuous. I’ve added it for clarity.

I’m also not entirely sure why I included the statement in question. I’ve removed it, and I think nothing is lost from the proof. Thank you!

3. Yotas said

You have a mistake, the last equality not necessarily holds. Each set of the left part of the equality is a subset of the right side of the equality.

• James said

In general, you’re correct. I’ve left out the fact that $f$ is surjective (of course, every map is surjective onto its image) and, therefore, equality should hold.

Here’s the relevant mathworld page that I’ll link to above as well.

http://mathworld.wolfram.com/Pre-Image.html

It’s been a while since I’ve looked at this stuff, so if I made another error, lemme know! :]