Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I know that the continuous images of compact sets are compact, but if we know a mapping f that maps a particular compact set into a compact set, is the mapping continuous? What if f is a real function?

share|cite|improve this question
Let $f(x)=2x$ for $0\le x\le 1/2$, $f(x)=0$ for $1/2<x\le 1$. Then $f$ maps $[0,1]$ to $[0,1]$. – André Nicolas Jan 21 '12 at 2:41
This is too strong to be true. Certainly there are "wild" functions mapping $[0,1]$ onto $[0,1]$. – Srivatsan Jan 21 '12 at 2:41
Thank you! I was actually thinking about a problem in Baby Rudin: chapter 4, problem 6: Suppose E is compact, prove that f is continuous on E if and only if {(x,f(x)), $x \in E$} is compact. I am not sure if the conclusion is correct. Suppose this conclusion is true, now if E compact, f(E) compact, then {(x,f(x)), $x \in E$} should be compact, then f is continuous? What went wrong in my argument? – Yang Jan 21 '12 at 2:56
@Yang: $E$ and $f(E)$ can be compact without the graph of $f$ being compact. E.g., restrict Neal's example to the closed unit interval $[0,1]$. Then the graph is $\{(x,1):x\in\mathbb Q\}\cup\{(x,0):x\in(\mathbb R\setminus Q)\}$. This is not even a closed subset of $\mathbb R$, because for example $(1/(n\pi),0)$ is in the graph but converges to $(0,0)$ as $n\to \infty$, and $(0,0)$ is not in the graph. It is true that the graph is contained in the compact set $E\times f(E)$, but to be compact it must be closed (I'm assuming $T_2$, but we're probably talking metric spaces in Baby Rudin). – Jonas Meyer Jan 21 '12 at 3:41
See the question Characterising Continuous functions for interesting generalizations. – Jonas Meyer Jan 21 '12 at 3:46

Absolutely not. Consider the real function $f:\mathbb{R}\to\mathbb{R}$ defined by $f(x)=1$, $x\in\mathbb{Q}$, otherwise $f(x)=0$. Every compact set (in fact, every subset of $\mathbb{R}$) is mapped onto one of the compact sets $\emptyset$, $\{0\}$, $\{1\}$, or $\{0,1\}$, but the function $f$ is nowhere continuous.

(Edit: corrected small error, and remove "non-empty" -- thank you, Pierre)

share|cite|improve this answer
Dear Neal: The empty set is compact. – Pierre-Yves Gaillard Jan 21 '12 at 5:02
Argh! Fixing for even greater generality. – Neal Jan 21 '12 at 13:17

The exercise in Rudin is correct. Suppose $\lim x_n=x$. Consider the sequence $(x_n,f(x_n))$ since the set $\{(x,f(x)): x\in E\}$ is compact there exists a convergent subsequence $(x_{n_k},f(x_{n_k}))$ which converges to $(x,y)$. Since the point $(x,y)$ is in the set $\{(x,f(x)): x\in E\}$, then $y=f(x)$. The problem in your argument is that $E$ and $f(E)$ compact does not imply that the graph $\{(x,f(x)):x\in E\}$ is compact.

share|cite|improve this answer
I think you have only shown that $f(x_{n_k})$ converges to $f(x)$. How do you know then that $f(x_n)$ must converge to $x$? – user38268 Apr 15 '12 at 11:31
@BenjaminLim The argument above shows that $limsup f(x_n)=liminf f(x_n)$. – azarel Apr 15 '12 at 14:50

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.