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Let $U$ be an open set. If $f\colon U\subset{R}^m\to N$ is a local homeomorphism and exists $y\in N$ such that $\operatorname{card}\big(f^{-1}(\{y\})\big)$ is infinite prove that $f$ is not closed.

I proved that $f^{-1}(\{y\})$ is closed and discrete, I proved also that $f$ must be continuous and open. But I don't know how to use the fact that $U$ is open in $\mathbb{R}^m$ and the fact that $\operatorname{card}\big(f^{-1}(\{y\})\big)$ is infinite.

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up vote 3 down vote accepted

Let $\{x_n\}$ be a sequence of preimages of $y$. Each of them has a neighborhood $U_n$ in which $f$ is a homeomorphism. Pick a sequence $y_n$ in $N\setminus\{y\}$ such that $y_n\to y$ and for each $n$ there exists $x_n'\in U_n$ such that $f(x_n')=y_n$. Then:

  • the set $\{x_n'\}$ is closed in $U$
  • the set $\{f(x_n')\}$ is not closed in $f(U)$.
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Is the sequence $y_n$ in $N$ or in $U$? – Mercy King Jun 22 '12 at 0:19
@Mercy. Should have been in $N$. Thanks, I fixed it. – user31373 Jun 22 '12 at 0:21

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