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Let $K \subset Y \subset X$ where $Y, X$ are metric spaces. $K$ is compact relative to $X \implies K$ is compact relative to $Y.$

Proof:

We'll assume the following theorem:

For any metric spaces $Y \subset X, \ E \subset Y$ open relative to $Y \iff E = Y \cap G$ for some open $G \subset X.$

Suppose $K$ is compact relative to $X.$ Assume $\{V_\alpha\}$ is an open cover of $K$ in $Y.$ Then $V_\alpha \subset Y.$ By the theorem above, $\forall V_\alpha \subset Y, \exists G_\alpha \subset X$ that's open relative to $X$ with $V_\alpha = Y \cap G_\alpha.$ Thus $K \subset \bigcup V_\alpha = \bigcup(Y \cap G_\alpha) = Y \cap (\bigcup G_\alpha).$ Since $K \subset Y,$ we have $K \subset \bigcup G_\alpha$ and so $\{G_\alpha\}$ is an open cover of $K.$ By hypothesis, $K$ is compact relative to $X$ and so $\{G_{\alpha_1}, G_{\alpha_2}, \ldots, G_{\alpha_n}\} \subset \{G_\alpha\}$ in $X$ and $K \subset \bigcup G_{\alpha_i}$. Thus, $K \subset Y \cap (\bigcup G_{\alpha_i}) = \bigcup(Y \cap G_{\alpha_i}) = \bigcup V_{\alpha_i}.$ Thus $\{V_\alpha\}$ has the finite subcover of $K$ and so $K$ is compact relative to $Y.$


I have a question about the part in bold in the proof above. I think for $\{G_\alpha\}$ to be an open cover of $K,$ among other conditions $G_\alpha$ ($\forall \alpha$) must be open relative to $Y.$ I am not sure I see it in the proof. By what condition or rule, is $G_\alpha$ ($\forall \alpha$) open relative to $Y$ if that's required?

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  • $\begingroup$ It doesn't really make sense to ask for $G_\alpha$ to be open rel $Y$ since it need not be a subset of $Y$. $G_\alpha$ is simply an open subset of $X$ $\endgroup$ – H. H. Rugh Aug 27 '17 at 14:03
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What's the doubt? Each $G_\alpha$ is an open subset of $X$ and $K\subset\bigcup_{\alpha}G_\alpha$. Therefore, $(G_\alpha)_\alpha$ is an open cover of $K$ in $X$.

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  • $\begingroup$ Ok, I see it now. $\endgroup$ – user474507 Aug 27 '17 at 14:05
  • $\begingroup$ One more question. We want(?) $K \subset X.$ Can we say "since $K \subset Y,$ we have $K \subset \bigcup G_\alpha,$ but since $G_\alpha$ is open relative to $X, \ \bigcup G_\alpha$ open relative to $X$ and so $\bigcup G_\alpha \subset X$ meaning $K \subset X$"? $\endgroup$ – user474507 Aug 27 '17 at 17:12
  • $\begingroup$ @user474507 “We want(?)”?! It is stated at the beginning of the problem that $K\subset Y\subset X$. What else is needed? $\endgroup$ – José Carlos Santos Aug 27 '17 at 17:14
  • $\begingroup$ @ José Carlos Santos, aha, I forgot to even look at the theorem. Ok, thanks. If something else is needed I'll let you know :) $\endgroup$ – user474507 Aug 27 '17 at 17:18

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