# closed subsets of compact spaces are compact proof clarification

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.

• I can't discern whst your actual question is. – Lord Shark the Unknown Jun 2 '17 at 6:17

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.

• 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 – jnyan Jun 2 '17 at 6:46
• 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 – jnyan Jun 2 '17 at 6:50

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.

• 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. – jnyan Jun 2 '17 at 8:54
• 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. – Juanito Jun 2 '17 at 9:18
• 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. – jnyan Jun 2 '17 at 9:28
• The union has a finite sub cover because X is compact, where are we using the compactness of A here? – Juanito Jun 2 '17 at 9:29
• where does it prove then that cover of A has a subcover? since i am trying to prove A is compact – jnyan Jun 2 '17 at 9:31