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My question is regarding Compactness and sequential compactness

How to show that the only compact subsets of $X$ are finite sets?

In case of specific sets like $\mathbb{N}$, we can choose cover containing open sets of the form $\mathbb N\setminus \{np: p$ is prime$, n\in \mathbb{N}\}$. Then no finite subcollection of it will cover $\mathbb N$.

But how to start the proof in case of arbitrary sets like $X$?

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user100943
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  • The question is meaningless until you specify the topology on $X$. On any infinite set there are compact topologies. – Brian M. Scott Oct 21 '13 at 02:24
  • @ B.M. Scott: the topology is cocountable as mentioned in the link – user100943 Oct 21 '13 at 02:27
  • That’s what I thought initially, but then your example suggested otherwise, because it’s unnecessarily complicated. $\Bbb N$ with the co-countable topology is discrete, so you could have used the much simpler open sets ${n}$. – Brian M. Scott Oct 21 '13 at 02:29
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    @user100943 : you should state the question instead of expecting everyone to click on the link. Call me lazy, but you'll probably get more and better answers that way. – Stefan Smith Oct 21 '13 at 02:30

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Suppose that $X$ has the co-countable topology, and let $A$ be an infinite subset of $X$. Since $A$ is infinite, it has a countably infinite subset $A_0=\{a_n:n\in\Bbb N\}$. Let $A_1=A\setminus A_0$, and for each $n\in\Bbb N$ let $U_n=\{a_n\}\cup(X\setminus A_0)$; $X\setminus U_n=A_0\setminus\{a_n\}$, which is countable, so $U_n$ is open. Let $\mathscr{U}=\{U_n:n\in\Bbb N\}$. If $x\in A_1$, then $x\in U_0$, and if $x\in A_0$, then $x=a_n$ for some $n\in\Bbb N$, so $x\in U_n$. Thus, $\mathscr{U}$ is an open cover of $A$. However, if $n\in\Bbb N$, the only member of $\mathscr{U}$ that contains $a_n$ is $U_n$, so no finite subset of $\mathscr{U}$ covers $A$. In fact, no proper subset of $\mathscr{U}$, finite or infinite, covers $A$: if you remove $U_n$ from $\mathscr{U}$, the remaining sets do not cover the point $a_n$.

Brian M. Scott
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  • @ B.M. Scott: I think $U_0$ should be replaced by $U_n$. – user100943 Oct 23 '13 at 03:03
  • @user100943: You mean in the statement that $x\in U_0$ if $x\in A_1$? There’s no reason to do so. It happens to be true that in that case $x\in U_n$ for all $n\in\Bbb N$, but we don’t care: we just need to be sure that $x$ is in at least one member of $\mathscr{U}$, and $U_0$ does as well as any. – Brian M. Scott Oct 23 '13 at 03:06
  • @ B.M. Scott: That means $0\in \mathbb N$. But why $X\setminus U_n=A_0\setminus {a_n}$? Isn't $(X\setminus U_n)\cap A=A_0\setminus {a_n}$? – user100943 Oct 23 '13 at 03:14
  • @user100943: Yes, I follow the convention that $\Bbb N$ includes $0$. The other is a mistake: I forget that I was working in $X$ rather than in $A$ with its relative topology. I’ve fixed it now. – Brian M. Scott Oct 23 '13 at 03:19
  • @B.M.: OK. Thanks! – user100943 Oct 23 '13 at 03:20
  • @user100943: You’re welcome! I’m glad that you caught the oversight. – Brian M. Scott Oct 23 '13 at 03:20