In Royden 3rd P192,

Assertion 1: Let $K_n$ be a decreasing sequence compact sets, that is, $K_{n+1} \subset K_n$. Let $O$ be an open set with $\bigcap_1^\infty K_n \subset O$. Then $K_n \subset O$ for some $n$.

Assertion 2: From this, we can easily see that $\bigcap_1^\infty K_n$ is also compact.

I know this is trivial if $K_1$ is $T_2$ (Hausdorff). But is it true if we assume only $T_0$ or $T_1$?

Any counterexample is greatly appreciated.


Consider the natural numbers with the co-finite topology. Then this is $T_1$ and every subset of $\mathbb N$ is compact. In particular set $K_n=\{ k \in \mathbb N \mid k \geq n\}$ this is a decreasing sequence of compact sets and their intersection is empty. So for instance we may take $O=\emptyset$ and we have our desired counterexample (edit) for assertion 1.

  • $\begingroup$ Hi, thanks for your answer. But I think this is not a counterexample for assertion 1. Also since in this topology space, every subset is compact. Especially if we also treat $\emptyset$ as a compact set. So I think this is not a very nice counterexample for assertion 2. $\endgroup$ – XxXxX May 17 '12 at 13:43
  • $\begingroup$ I agree that it's not a counterexample for assertion 2. Are you saying that it's not a counterexample for assertion 1 either? $\endgroup$ – JSchlather May 17 '12 at 13:55
  • $\begingroup$ Yes. I think so. Given any open set $O$, clearly $\emptyset \subset O$. However, since $O$ is open, it must contain a base element $B = \{ k \in \mathbb{N} | k \geq M \}$. Then $K_M \subset O$, right? So assertion 1 still holds. $\endgroup$ – XxXxX May 17 '12 at 14:00
  • $\begingroup$ @XxXxX I took $O$ to be the empty set thought. Is there a stipulation that $O$ need be non-empty? $\endgroup$ – JSchlather May 17 '12 at 14:14
  • $\begingroup$ I agree with you now. I think there is no requirement on $O$ being nonempty. Thanks. Also I would appreciate any further counterexamples. Thanks for your help. $\endgroup$ – XxXxX May 17 '12 at 14:18

Here's a T_1 space for which Assertion 2 fails. Take the set of integers. Say that a set is open iff it is either a subset of the negative integers or else is cofinite. Then let K_n be the complement of {0, 1, ..., n}. Then each K_n is compact, but the intersection of K_n from n=1 to infinity is the set of negative integers, which is open and noncompact.

  • $\begingroup$ please put dollar signs around your math $\endgroup$ – Forever Mozart Dec 14 '14 at 5:22

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