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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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Yes but it's also closed since it's the whole space. – Rudy the Reindeer Sep 5 '12 at 13:26
What about this: "$g: [0,1] \to (0,1)$ (g is onto) cannot be continuous because a continuous function maps a compact set to a compact set"? Is this wrong because $(0,1)$ is closed since its the whole space, and it is bounded, and hence compact? – Legendre Sep 5 '12 at 13:29
Your argument that $(0,1)$ is compact uses the Heine-Borel theorem, which applies to $\mathbb{R}^n$, not arbitrary subspaces of it. It is true that $(0,1)$ is not compact in itself (under the topology induced from $\mathbb{R}$) so your argument about $g$ is correct. – Matthew Pressland Sep 5 '12 at 13:31
Re your comment question: It is not wrong. There is no continuous function $g:[0,1]\to\(0,1)$ which is onto. – Thomas Andrews Sep 5 '12 at 13:47
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 $\mathbb R$. – Rudy the Reindeer Sep 5 '12 at 13:47
up vote 1 down vote accepted

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. I guess I am suffering from serious confusion the theorems in Topology. Revision time! :p – Legendre Sep 5 '12 at 13:36
Another way to see that it $(0,1)$ isn't compact is to observe that (for example) $$x\mapsto\cfrac{2x-1}{4(x-x^2)}$$ is a homeomorphism $(0,1)\to\Bbb R$. Since $\Bbb R$ isn't compact, then neither can $(0,1)$ be. – Cameron Buie Sep 5 '12 at 13:43
Another way is to note that $f(x) = \frac{1}{x}$ is continuous on $(0,1)$, but unbounded. But on compact spaces, all continuous functions are bounded. (I'm not sure if this is circular or not). – Jason DeVito Sep 5 '12 at 13:55
It isn't circular, Jason. It's certainly true that all real-valued continous functions on compact spaces are bounded (or more generally, that all metric-space-valued continuous functions on compact spaces are bounded). Thus, there can't be an unbounded continuous function $X\to\Bbb R$ for compact spaces $X$. – Cameron Buie Sep 5 '12 at 14:12
Thanks for all the comments guys. Much appreciated! – Legendre Sep 6 '12 at 16:51

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