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Let $X$ be a topological space. It is known that $X$ is compact if and only if every family of closed sets satisfying the finite intersection property has nonempty intersection.

Is there some kind of compactness property that is equivalent to the following:

$(P)$ every countable family of closed sets $\left\{F_{i}:i=1,2,3,\ldots \right\} $ such that $F_{1}\supseteq F_{2}\supseteq F_{3}\supseteq\ldots$ has nonempty intersection.

Clearly, if $X$ is compact, then $(P)$ is satisfied. Another question: is the converse also true (for some particular classes of topological spaces)?

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    $\begingroup$ I think that you have the inclusions going the wrong way: as you’ve written it, the intersection is equal to $F_1$. $\endgroup$ – Brian M. Scott Mar 11 '13 at 20:03
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    $\begingroup$ You probably want $\ldots F_3 \subset F_2 \subset F_1$, otherwise $F_1$ is their intersection (and is non-empty). $\endgroup$ – Henno Brandsma Mar 11 '13 at 20:03
  • $\begingroup$ @BrianM.Scott Thank you. I've made the corrections to the initial post. $\endgroup$ – digital-Ink Mar 11 '13 at 20:32
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A space is called countably compact iff every countable open cover has a finite subcover. This translates to (taking complements): every countable family of closed sets with the finite intersection property, has non-empty intersection.

Your property is equivalent to this: if it holds, and we have a countable family of such closed sets $G_n$, we can define $F_n = \cap_{i=1}^n G_i$ which is non-empty, and decreasing. By this property you state, it has non-intersection, which is also the intersection of the $G_n$. Of course, if $X$ is countably compact, your decreasing family has FIP, and so non-empty intersection.

So your property is just a reformulation of countable compactness. This is a well-studied property.

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I think that you want $(P)$ to say that if $\{F_n:n\in\Bbb N\}$ is a family of non-empty closed sets in $X$ such that $F_n\supseteq F_{n+1}$ for each $n\in\Bbb N$, then $\bigcap_{n\in\Bbb N}F_n\ne\varnothing$.

This property is equivalent to countable compactness of $X$: every open cover of $X$ has a countable subcover. To see this, let $\mathscr{U}=\{U_n:n\in\Bbb N\}$ be a countable family of open sets in $X$ such that no finite subfamily of $\mathscr{U}$ covers $X$. For each $n\in\Bbb N$ let $$F_n=X\setminus\bigcup_{k\le n}U_k\;;$$

then $F_n\supseteq F_{n+1}$ for each $n\in\Bbb N$, and the sets $F_n$ are closed and non-empty. Clearly $\bigcap_{n\in\Bbb N}F_n\ne\varnothing$ iff $\bigcup_{n\in\Bbb N}U_n=X$, i.e., iff $\mathscr{U}$ is an open cover of $X$ with no finite subcover.

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