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Does there exist a continuous bijection from open n ball to closed n-ball? One with a simple argument can show that no such function exists for n=1.But, what about n>1?

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    For those who arrive later, the case $n=1$ is discussed [here](http://math.stackexchange.com/questions/42308/continuous-bijection-from-0-1-to-0-1).2011-05-31

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No. This is a special case of Brouwer's theorem of invariance of domain.

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    Is there a simpler proof for the special case in the question? In particular, for $n=2$?2011-05-31
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    Not that I can see as yet. You only need the original Jordan curve theorem, rather than the $n$-dimensional generalization, but that's not much gain.2011-05-31
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    Another way: the continuous image of a compact set (the closed ball) must be compact (which the open ball is not).2011-05-31
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    @Samuel: That's the opposite direction of what's being considered here.2011-05-31
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    @Chris, I meant a simpler proof for the case of an open disk in the plane, not a general open set.2011-05-31
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    @lhf: yes, I gathered that.2011-05-31