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Let $X$ be a compact Hausdorff space with infinitely many points, then how to construct a strictly decreasing (infinite) sequence of closed subsets of $X$?

Lao-tzu
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2 Answers2

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Let $A_1$ be an infinite closed subset of $X$.

For two distinct elements $a$ and $b$ of $A_1$ , find disjoint open sets containing $a$ and $b$ respectively. The collection of all such sets is an open cover of the compact set $A_1$ and thus has a finite subcover.

One element, $E$, from this finite subcover is open, contains infinitely many elements of $A_1$, but the closure of $E\cap A_1$ is not all of $A_1$.

Let $A_2$ be the closure of $E\cap A_1$.

Continue ...

David Mitra
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Hint: Every infinite Hausdorff space includes a (countably) infinite discrete subspace (see here) $A$. Since $A$ is discrete, it follows that $x \notin \overline{ A \setminus \{ x \} }$ for all $x \in A$. Now expand on this to construct a strictly decreasing sequence of closed subsets of $X$.

Community
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user642796
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  • I don't think it includes a (countably) infinite discrete subspace, since $X$ is compact, any infinite subset has a limit point hence cannot be discrete. – Lao-tzu Feb 09 '14 at 13:19
  • @Lao-tzu: I mean discrete as a subspace. Consider $[0,1]$ and the subset $A = { \frac{1}{n} : \geq 1 }$. As a subspace of $[0,1]$ $A$ is discrete, even though it has a limit point. (It is just discrete, not closed discrete.) – user642796 Feb 09 '14 at 13:27
  • OK, thank you, I'll think a little. – Lao-tzu Feb 09 '14 at 13:35