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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3For those who arrive later, the case $n=1$ is discussed here. – lhf May 31 '11 at 12:50
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No. This is a special case of Brouwer's theorem of invariance of domain.
Chris Eagle
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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
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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. – Chris Eagle May 31 '11 at 13:12
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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). – Samuel May 31 '11 at 13:26
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3@Samuel: That's the opposite direction of what's being considered here. – Nate Eldredge May 31 '11 at 14:18
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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. – lhf May 31 '11 at 22:53
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