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For a continuous function $f: X \to Y$, the preimage of every closed set in $Y$ is closed in $X$.

Let $g: (0,1) \to [0,1]$ be a continuous surjection.

Isn't the preimage of $[0,1]$ = $(0,1)$ open?

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    And what Matt said: your argument that $(0,1)$ is compact in itself doesn't work because where you write "...hence compact" you use Heine-Borel which is a theorem about subsets of $\mathbb R^n$ but here you consider $(0,1)$ as the whole space, not as a subset of $\mathbb R$.2012-09-05

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As Matt points out, it is closed, as it is the whole space.

"Closed and bounded" is not the same as compact in general. Observe (for example) that $\mathcal{U}=\bigl\{(1/n,1-1/n):n\in\Bbb Z,n>2\bigr\}$ is an open cover of $(0,1)$, but has no finite subcover. Thus, you're right--such a $g$ cannot be continuous.

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    Thanks for all the comments guys. Much appreciated!2012-09-06