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Let $U\subset \mathbb{R}^n$ (open set) , $V\subset \mathbb{R}^n$ and $f:U\longrightarrow V$ a homeomorphism then we can say that $V$ is a open set in $\mathbb{R}^n$ ?

Any hints would be appreciated.

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Google for the theorem of «Invariance of Domain» and its consequences. –  Mariano Suárez-Alvarez Sep 14 '12 at 19:32

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Yes, this result goes by the name Invariance of Domain and is due to Brouwer. It's a non-trivial result.

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To add to the above comment: An homeomorphism is in particular an open function, so maps open sets to open sets. –  kjetil b halvorsen Sep 14 '12 at 19:56
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@Manos: It is assumed that the map $f:U \rightarrow V$ is a homoemorphism, not that $f$ is the restriction of a homoemorphism $\tilde{f}:\mathbb{R}^n \rightarrow \mathbb{R}^n$. The fact that $f$ is a homeomorphism tells you nothing about whether $V$ is open in $\mathbb{R}^n$. It only tells you that if $U'$ is open in $U$, then $f(U)$ is open in $V$. –  Michael Joyce Sep 14 '12 at 20:01
    
Got it got it :) –  Manos Sep 14 '12 at 20:06

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