# Compactness relatively to open subset

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?

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 $$K$$?
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 $$\phi(U) \to \mathbb R$$
• It is compact in $U$ Apr 29, 2020 at 13:27
• .......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. Apr 29, 2020 at 13:40