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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. –  lhf May 31 '11 at 12:50

1 Answer 1

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 $n=2$? –  lhf May 31 '11 at 13:04
    
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. –  Chris Eagle May 31 '11 at 13:12
    
Another way: the continuous image of a compact set (the closed ball) must be compact (which the open ball is not). –  Samuel May 31 '11 at 13:26
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@Samuel: That's the opposite direction of what's being considered here. –  Nate Eldredge May 31 '11 at 14:18
    
@Chris, I meant a simpler proof for the case of an open disk in the plane, not a general open set. –  lhf May 31 '11 at 22:53

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