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Does there exist a continuous bijection from $(0,1)$ to $[0,1]$? Of course the map should not be a proper map.

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    @Asaf: I don't think that this really is a duplicate, even if the question is briefly addressed there.2011-05-31
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    @Theo: You are correct. :-)2011-05-31
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    What about such map from a non compact set to compact set in nice topologies?2011-05-31
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    @Alex: Take $f: [0,1) \to S^1 = \{z \in \mathbb{C}\,:\,|z| = 1\}$ with $f(x) = e^{2\pi i x}$ This is continuous and bijective, but has no continuous inverse.2011-05-31
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    what exactly is the reason for non existence? In the (0,1) case some sort of local compactness is the reason.Can the proof generalized to non existence of a map from open ball in R^n to a closed ball.2011-05-31
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    @Alex: I should have said *noncontinuous inverse*. The point is that the inverse $g$ is already determined by $f$. Now if $z_{n} = e^{2\pi i x_{n}} = f(x_n)$ with $x_n \nearrow 1$ hen $z_n \to 1$ while $g(z_n) = x_{n}$ and and $g(1) = 0$, so $g(z_n) = x_n$ doesn't converge to $g(1)$ and thus $g$ isn't continuous. As for the generalization to open and closed balls in $\mathbb{R}^n$, I think you should ask this as a separate question, because clearly other techniques are required than the three (very similar if not identical) arguments you received here.2011-05-31
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    @theo:please look at ncmathsadist 's answer to this question.What do you say on montonocity of 1-1 continuous function.Do nowhere monotonous functions exist?2011-05-31
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    @Alex: A continuous $1-1$-function from an interval to an interval is strictly monotonous, that's true. This follows from the intermediate value theorem and the fact that intervals and points are precisely the connected subsets of an interval in $\mathbb{R}$.2011-05-31
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    Alex, if you have any question on t.b.'s answer (or any of the ones you got) you may use the comment below the answer - I am sure he/she/they would be happy to assist!2011-12-13

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