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in Rudin's mathematical analysis book, in the proof of "closed subset of compact spaces are compact" the author takes the open cover $U$ of a closed subset $A$ and takes union of that open cover and complement of $A$, which will cover the compact space $X$.

and since $X$ is compact the union of $A'$ and $U$ will have a finite subcover, which will also cover $A$.

my question is I can keep the subcover as union of $U$ and take subcovers of the open set $A'$, which is still a subcover of $X$, but this doesnt prove that cover $U$ has a subcover and that $A$ is compact too.

I dont understand what is wrong with above argument.

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    $\begingroup$ I can't discern whst your actual question is. $\endgroup$ – Lord Shark the Unknown Jun 2 '17 at 6:17
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As $X\setminus A$ and finitely members of $U$ cover all of $X$, they also cover $A$. As $X\setminus A$ is disjoint form $A$, the finitely many members of $U$ alone cover $A$, as desired.

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  • $\begingroup$ In the first statement of yours, dont you think here we are using the fact that $A$ is compact by saying finitely members of U cover complement of $X\A$ , which is A $\endgroup$ – jnyan Jun 2 '17 at 6:46
  • $\begingroup$ i mean X\A is open and when you say finitely members of U cover all of X, what you mean is only set left to be covered now is complement of X\A, which is A again. and you say finitely members of U cover A, which in turn means A is compact $\endgroup$ – jnyan Jun 2 '17 at 6:50
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In your argument, you are not using the full set of assumptions you have been provided. In particular, you are not using the fact that $A^c$ is open as $A$ is closed!

How do you expect to prove what you want to proof, without using all the resources you have at your disposal!

Read the actual proof again, and I hope this helps you understand why the selection of the cover of $U$ alongwith $A^c$ was a clever choice in the first place! :)

Also note that when we are building the cover for $X$, we are not using arbitrary covers, we are using a special cover which covers the whole of $X-A$ with a single open set. Of course, any cover, even his special one, must have a finite subcover, for $X$ to be compact.

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  • $\begingroup$ i get that. i understand the need for closedness of set. but I feel that compactness of A is used to prove the same fact. Please look at the comments to other answer. $\endgroup$ – jnyan Jun 2 '17 at 8:54
  • $\begingroup$ Also note that when we are building the cover for $X$, we are not using arbitrary covers, we are using a special cover which covers the whole of $X\A$ with a single open set. Of course, any cover, even his special one, must have a finite subcover, for $X$ to be compact. $\endgroup$ – Juanito Jun 2 '17 at 9:18
  • $\begingroup$ once i have covered X\A, with a single open set, and A with an open cover, this union should have a finite subcover. this is where i have problem. if one says, the cover of A has finite subcover, it again assumes compactness of A. $\endgroup$ – jnyan Jun 2 '17 at 9:28
  • $\begingroup$ The union has a finite sub cover because X is compact, where are we using the compactness of A here? $\endgroup$ – Juanito Jun 2 '17 at 9:29
  • $\begingroup$ where does it prove then that cover of A has a subcover? since i am trying to prove A is compact $\endgroup$ – jnyan Jun 2 '17 at 9:31

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