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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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    In the last part, you say "Moreover, if $A_\infty$ is compact?" do you mean "Moreover, is $A_\infty$ compact?" or the same question under the assumption that the intersection is compact?2011-06-01
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    @Asaf: I don't understand your question. If the intersection is empty then it certainly is compact and my answer applies. If the intersection is non-empty then certainly no $A_n$ is empty. The only way I can make sense of this part of the question is: "Is the intersection of countably many nested compact sets compact"?2011-06-01
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    @Asaf, I thought that it was right usage of English. I was interested, is $A_\infty$ a compact or not.2011-06-01
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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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    Thanks a lot. I'm just trying to extend my results from Euclidean space to more general space (like Polish) since I used only compactness and continuity.2011-06-01
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    @Theo, could you please tell me: if $X$ is a *compact* Borel space and $f:X\to [0,1]$ is continuous on $X$, does it mean that $(x\in X:f(x) = 1)$ is compact in $X$?2011-06-01
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    @Gortaur: Well, I'd say compactness and continuity arguments tend to extend well to both the locally compact and the metrizable setting maybe using completeness. Locally compact and separable (in particular Polish by Urysohn's metrization theorem) is even more convenient. No need to explain, things that are easy for one person to answer need not be easy at all for someone with a different background :)2011-06-01
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    @Gortaur: Yes, as $\{1\} \subset [0,1]$ is closed, so is $f^{-1}(\{1\}) = \{f = 1\}$ by continuity. Closed subsets $F$ of compact spaces $X$ are compact (If $\{U_{i} \cap F\}$ is an open cover of $F$ with $U_i \subset X$ open then $\{X \smallsetminus F\} \cup \{U_i\}$ is an open cover of $X$).2011-06-01
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    Sorry, I haven't got you sentence: "No need to explain, things that are easy for one person to answer need not be easy at all for someone with a different background"2011-06-01
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    @Gortaur: I read your first comment as a sort of apology for asking two questions in a row that I could answer immediately. Sorry if I got that wrong. I tried to say that explaining the source of your questions is not necessary and added that the fact that it was easy for me doesn't mean that it should be easy for you. It was only meant to be kind and a sign of respect and encouragement from me.2011-06-01
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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