# Is the pre-image of a compact space compact?

I'm trying to think of an example of a continuous function $f:\mathbb{R} \to \mathbb{R}$ such that $f^{-1}(X)$ is not compact but $X \subset \mathbb{R}$ is compact. Any ideas?

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$f(x) = \arctan(x)$, $X=[-\pi/2,\pi/2]$. More generally, any bounded function will work: if $-M\leq f(x)\leq M$ for every $x\in \mathbb{R}$, then $f^{-1}([-M,M])=\mathbb{R}$ is not compact. – Arturo Magidin Mar 5 '12 at 17:12
@Arturo: this suggests, if I haven't missed something, that a continuous , bounded map:f:X-->Y ; Y metric, with f(X) closed, cannot have a continuous inverse (since continuous image takes compact to compact). – AQP Mar 5 '12 at 17:33
@AQP: Depends on what you mean by inverse. If you mean, a map $\mathbb{R}\to \mathbb{R}$ that is a two-sided inverse, you already knew that because a bounded map cannot be onto. – Arturo Magidin Mar 5 '12 at 17:36
@Arturo. Right, I missed that obvious point. – AQP Mar 5 '12 at 17:43

What about the the constant function $f(x)=0$?
Note that if $X$ is compact, it is closed, and so $f^{-1}(X)$ is closed. Now take your favourite set that is closed but not compact, call that $B$, and let $f(x) = \text{dist}(x,B)$. That is a continuous function on $\mathbb R$, and $B = f^{-1}(\{0\})$.
Consider the continuous function $\sin x$