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Let $\phi: U \rightarrow \mathbb{R}^m$, $U\subseteq \mathbb{R}^n$ open, such that $\phi(U)$ is open and $\phi$ is a homeomorphism onto its image.

Let $K\subseteq \phi(U)$ be compact. Show that $\phi^{-1}(K)$ is compact in $\mathbb{R}^n$

My try: for every open cover of $K$ by open sets of $\mathbb{R}^m$ we can extract a finite subcover. Because $\phi(U)$ is open this means that for every open cover of $K$ by open sets of $\phi(U)$ we can extract a finite subcover, so $K$ is a compact as a subspace of $\phi(U)$. Then, because $\phi: U \rightarrow \phi(U)$ is a homeomorphism, $\phi^{-1}(U)$ is compact as a subspace of $U$ which is itself open in $\mathbb{R}^n$.

Hence for every open cover of $\phi^{-1}(K)$ by sets open in $\mathbb{R}^n$ contained in $U$. How do I check all the other covers?

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You are almost done. Let $\mathcal{A}$ be an open cover of $K$. Then $\{U\cap W\mid W\in\mathcal{A}\}$ is an open cover of $K$ which consists of open subsets of $U$. By your argument, it has a finite subcover $\{U\cap W_i \mid i=1,2,\cdots, n\}$. Could you see that $\{W_1,\cdots, W_n\}$ is a desired open subcover of $K$?

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Theorem:Continuous image of a compact set is compact. See the proof is here

Since $\phi:U\to \phi(U)$ is a homeomorphism, hence $\phi^{-1}:\phi(U) \to U (\subseteq \mathbb{R})$ is also a continuous bijection . Hence $\phi^{-1}(K)$ is compact in $\mathbb R$ since you can look at $\phi^{-1}$ as a continouos function from $\phi(U) \to \mathbb R$

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    $\begingroup$ It is compact in $U$ $\endgroup$
    – Leonardo
    Apr 29, 2020 at 13:27
  • $\begingroup$ .......and hence compact in $\mathbb R$ . I have edited as well . Now I think your doubt is clarified. @warm_fish Thank you for your question. $\endgroup$ Apr 29, 2020 at 13:40

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