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Assume $U$ is open in $\mathbb{R}^m$ and $V$ open in $\mathbb{R}^n$, $U\cong V$. Does it imply $m=n$?

Fan
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    Not if $U = \varnothing$. Otherwise yes, look up "invariance of domain". – Daniel Fischer Oct 28 '14 at 18:44
  • @DanielFischer, is that assuming that we are looking at the standard topology on $\mathbb R^n$, $\mathbb R^m$? What if it's not the standard topology? – Robert Cardona Oct 28 '14 at 18:47
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    @RobertCardona Yes, it assumes the standard topologies. Otherwise, since for $n,m > 0$ the two sets are equipotent, any bijection of the two sets plus a topology on one gives a homeomorphism, so it's not an interesting question for arbitrary topologies. It may be interesting again for specific non-standard topologies on the spaces. – Daniel Fischer Oct 28 '14 at 18:54

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