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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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    See the question [Characterising Continuous functions](http://math.stackexchange.com/q/4412) for interesting generalizations.2012-01-21

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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)

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    Argh! Fixing for even greater generality.2012-01-21
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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.

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    @BenjaminLim The argument above shows that $limsup f(x_n)=liminf f(x_n)$.2012-04-15