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Let $(X,d)$ be a metric space and $Y\subseteq X$ a metric space with the same metric. Show that $Y$ is compact iff every collection of open sets in $X$ that covers $Y$ has a finite subcover.

Seems a bit too simple so I feel I have overlooked something. I simply need a proof check. If it is correct, is there any details I can add to make it slightly more rigorous?

Proof (my attempt):

If $Y$ is compact, then every open cover of $Y$ has a finite subcover. Choose some open cover $O$ and a subcollection of $O$ that is a finite subcover for $Y$, call it $O_f$. Then, because $Y\subseteq X$, then $O_f$ is a collection of open sets in $X$ that covers $Y$. It is a finite subcover.

Conversely, suppose any collection of open sets on $X$ that covers $Y$ has a finite subcover. By definition, $Y$ is compact.

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    $\begingroup$ The point is that the open cover which you are calling $O$ may not be composed of open sets of $X$ (they are open sets in the induced topology, not necessarily open on $X$). Analogously, your "conversely" part suffers from the same issue. $\endgroup$ – Aloizio Macedo Feb 28 '18 at 0:17
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I think it is almost right, but in the first part I find a cuple of mistakes. First, it is not true that if $V\subseteq Y$ is open in $Y$, then it is open in $X$: the correct part is that there exists an open set $U\subseteq X$ so that $V=U\cap Y$.

I would approach this in the following way (and maybe I will be a little bit too technical):

If $Y$ is compact, let $\{U_i\}_{i\in I}$ a collection of open sets in $X$ such that $Y\displaystyle\subseteq\bigcup_{i\in I} U_i$. Then, if we name $V_i=U_i\cap X$, $\{V_i\}_{i\in I}$ is a collection of open sets of $Y$ that covers $Y$, so there exists a finite subset that covers it: $Y=V_{i_1}\cup...\cup V_{i_n}$, and then $Y\subseteq U_{i_1}\cup...\cup U_{i_n}$.

Conversely, if every collection of open sets in $X$ that covers $Y$ has a finte subscover, let $\{V_i\}$ be a collection of open sets in $Y$ that cover $Y$. If we prove that there is a finite number of $V_i$ that cover $Y$, then we will have finished. Since every $V_i$ is open in $Y$, there exists $U_i$ open in $X$ so that $V_i=U_i\cap Y$. Then $\{U_i\}$ is a collection of open sets in $X$ that covers $Y$, so there exists a finite subcover: $Y\subseteq U_{i_1}\cup...\cup U_{i_n}$, and $Y=(U_{i_1}\cup...\cup U_{i_n})\cap Y=V_{i_1}\cup...\cup V_{i_n}$. Therefore, $Y$ is compact.

You got mixed up the topology in $X$ with the topology it induces in $Y$, but the argument was not very different.

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  • $\begingroup$ I know this is a year old question. I was wondering if you could clarify, in the converse proof, when you construct $V_i = U_i \cap Y$ as a covering of $Y$, why does that imply the collection $\{U_i\}$ itself is also a covering of $Y$. $\endgroup$ – Hushus46 Mar 20 at 6:44

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