# Decreasing sequence of sets

$X$ 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?

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

• @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 $X$? – Ilya Jun 1 '11 at 11:13
• @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 $X$). – t.b. Jun 1 '11 at 11:21