$X$ 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?

  • $\begingroup$ 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? $\endgroup$ – Asaf Karagila Jun 1 '11 at 10:13
  • $\begingroup$ @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"? $\endgroup$ – t.b. Jun 1 '11 at 10:24
  • $\begingroup$ @Asaf, I thought that it was right usage of English. I was interested, is $A_\infty$ a compact or not. $\endgroup$ – Ilya Jun 1 '11 at 11:08
  • $\begingroup$ @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. :-) $\endgroup$ – Asaf Karagila Jun 1 '11 at 11:49

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.

  • $\begingroup$ 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. $\endgroup$ – Ilya Jun 1 '11 at 11:09
  • $\begingroup$ @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 $X$? $\endgroup$ – Ilya Jun 1 '11 at 11:13
  • $\begingroup$ @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 :) $\endgroup$ – t.b. Jun 1 '11 at 11:18
  • $\begingroup$ @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 $X$). $\endgroup$ – t.b. Jun 1 '11 at 11:21
  • $\begingroup$ 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" $\endgroup$ – Ilya Jun 1 '11 at 11:30

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