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$X$ is a topological space. Let $A_n$ be a non-increasing sequence of subsets of this space:

$ A_n\supseteq A_{n+1} $

and all $A_n$ are compact sets. Is it true that $A_\infty = \bigcap_n A_n$ is empty if and only if $A_N$ is empty for some $N$? If yes, how to prove it? Moreover, is $A_\infty$ compact?

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    @Theo, Gortaur: I merely wanted to be sure it was not a typo. There are plenty of non-English speakers who might make such mistake. :-)2011-06-01

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You need to assume that $A_1$ is compact and that the sets $A_{n}$ are closed (which is of course automatic under your assumption if $X$ is Hausdorff). A silly counter-example when the $A_n$ aren't closed: Take $A_n = [n,\infty)$ in $X = \mathbb{R}$ with the trivial topology.

If the $A_n$ are closed sets, note that $\bigcap A_n = \emptyset$ implies that $U_n = A_1 \smallsetminus A_n$ is an open cover of $A_1$ by passing to complements. Applying compactness of $A_1$ we see that finitely many of the $U_n$ already cover $A_1$. Passing to complements again and using that the sets are nested $A_n \supseteq A_{n+1}$, we see that $A_N$ must be empty for $N$ large enough.

Of course, if we are assuming each $A_n$ closed and $A_1$ compact, then $A_{\infty}$ is compact since closed subsets of a compact set are compact.

See also the Wikipedia page on the finite intersection property which yields one of the many equivalent characterizations of compactness.

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    Ok, I see. I haven't apologized, I just wrote the reasoning (though I have no reason to write it itself:) )2011-06-01