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If $n\in\mathbb{N}$, and X is a compact metric space, show that every continuous injective function $f:X\to\mathbb{R}^n$ is a homeomorphism of X. $(f\subseteq\mathbb{R}^n)$

So I persumed that I have to show that f is an open map from $X\to\mathbb{R}^n$: Let U be an open set in X. Then $U^c$ is closed in X, which is compact, so $U^c$ is also compact. Since $f$ is continuous $f(U^c)$ is compact in $\mathbb{R}^n$.

Then I got stuck. Does this lead to $f(U^c)$ being closed in $\mathbb{R}^n$? is it sufficient enough?

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  • $\begingroup$ You mean a homeomorphism onto its image right? so between $X$ and $f(X) \subset \Bbb R^n$. $\endgroup$ – Ruben Jun 24 '19 at 17:13
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If $U$ is compact and $f$ continuous, then $f(U)$ is also compact. In addition, if $f$ is injective (that is, bijective from $U$ to $f(U)$), then $f$ has a continuous inverse. I think you can easily find proofs about the relevant theorems in any intro topology tutorial or, e.g., chapter 2 of Baby Rudin.

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