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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?

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    Let $f(x)=2x$ for $0\le x\le 1/2$, $f(x)=0$ for $1/2. Then $f$ maps $[0,1]$ to $[0,1]$.2012-01-21
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    This is too strong to be true. Certainly there are "wild" functions mapping $[0,1]$ onto $[0,1]$.2012-01-21
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    @Srivatsan: For instance, a bijection $f:[0,1]\to[0,1]$ such that $f[V]$ is dense in $[0,1]$ for every non-empty open set $V\subseteq[0,1]$.2012-01-21
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    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?2012-01-21
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    @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).2012-01-21
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    See the question [Characterising Continuous functions](http://math.stackexchange.com/q/4412) for interesting generalizations.2012-01-21

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