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For all sequences of nested (non-increasing with respect to set-inclusion) non-empty closed sets, their intersection is non-empty.

What are the minimum conditions on a topological space for it to satisfy this property in ZF (without any choice)?

Henno Brandsma
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Lorenzo
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1 Answers1

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This true iff every countable open cover has a finite subcover. The proof involves just taking complements and noting that if $U_n$ is any sequence of open sets then $U_1 \cup U_2\cup...\cup U_n$ is an increasing sequence of open sets whose union is the same as the union of the $U_i$'s.

Kavi Rama Murthy
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    It is customary these days for an answer to also have some mathematical justifications behind it, and not just a comment. Especially coming from someone with nearly 100k in reputation. – Asaf Karagila Jul 09 '19 at 08:57
  • Being elementary does not mean that it shouldn't be said. On the contrary, if the proof is so short, what's your excuse for not writing it in the first place? – Asaf Karagila Jul 09 '19 at 09:13
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    Such spaces are called countably compact – Wojowu Jul 09 '19 at 09:21
  • In this answer I show it is equivalent to the fact that every infinite subset $A$ of $X$ has an $\omega$-limit point. But in the proof I think I use (countable) dependent choice, and this might well be essential. So under ZF I'm not sure the equivalence will hold...@AsafKaragila – Henno Brandsma Jul 09 '19 at 09:53
  • @Henno: To my knowledge this is somewhere in the vicinity of countable choice, or at least "every infinite set is Dedekind-infinite". Requiring countable covers to have finite subcovers seem to circumvent this issue directly. – Asaf Karagila Jul 09 '19 at 13:16