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I found one exercise in my book

Let $X$ be a compact subset of a metric space $M$. Prove that $X$ is closed.

In the definitions, the book only mentions compact space and never compact set.

An open cover of a metric space $M$ is a collection $U$ of open subsets of $M$ such that $M=\bigcup U$. A subcover of $U$ is a subcollection $U^*$ of $U$ such that $M=\bigcup U^*$. A metric space $M$ is said to be compact if every open cover of $M$ has a finite subcover.

So I wonder: When it says $X$ is a compact subset, are the open subsets in the cover open subsets of $X$, or open subsets of $M$? (If $X=[0,1]$ and $M=\mathbb{R}$, then $X$ is an open subset of $X$ but not an open subset of $M$.)

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1 Answer 1

up vote 2 down vote accepted

For $Y\subseteq X$, this means that the subset $Y$ is a compact space when considered as a space with the subspace topology coming down from $X$.

To jog your memeory, recall that the subspace topology works this way: the open sets of $Y$ are just the intersections of $Y$ with open sets of $X$.

This turns out to be equivalent to the following condition: given any open covering of $Y$ by open sets of $X$, there is a finite subcollection covering $Y$.

I think you'll be able to prove that equivalence rather easily. I am also unsure of how the terminology of "covers" jives with what you learned. I can clear up whatever questions you have about what I meant in the comments. Just ask :)

I hope this takes care of your question about definitions. If you need further help with the question you're working on, I can suggest another problem/answer posted at this site: Does a compact subspace have to be closed in an arbitrary metric space?.

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I think the equivalence is true for any set $Y \subseteq X$. –  Vectk Jun 17 '13 at 18:19
    
@Ink I think you're probably right. You know what I think I'm remembering? The proof that closed subsets of compact Hausdorff spaces are compact. –  rschwieb Jun 17 '13 at 18:27
    
@Ink Thanks for jarring a little rust off my brain... –  rschwieb Jun 17 '13 at 18:30
    
@rschwieb Okay that's clear. Thank you! –  PJ Miller Jun 17 '13 at 19:44

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