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Consider a topological space $(X,\mathcal{T})$, and a sequence of sets $(K_n)_{n\in\omega}$, such that $(\forall n\in\omega)(K_n\text{ is compact})$, and $(\forall m<n)(K_m\supset K_n)$. Is it true that $\cap_{n\in\omega}K_n\neq\emptyset$?

Kai
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1 Answers1

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As pointed out elsewhere you need to assume the $K_n$ are all non-empty. However, that assumption is not strong enough: under the cofinite topology on $\Bbb{R}$, every subset of $\Bbb{R}$ is compact. So if you take $K_n = \{n, n + 1, n + 2, \ldots\}$, you get a counter-example to your claim.

If you assume $(X, {\cal T})\,$ is Hausdorff, your claim holds. (If $\bigcap_n K_n\,$ were empty, the sets $K_1 \setminus K_n\,$ would be open, because of Hausdorffness, and would cover $K_1$. Hence a finite subset of these open sets would cover $K_1$, implying that the $K_n$ are eventually all empty).

All the above argument relies on is that the $K_i$ are all closed. So it works if we take that as an assumption or if we assume that $X$ is a $KC$-space, one in which all compact subsets are closed.

Rob Arthan
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  • We can use a weaker assumption: https://topology.pi-base.org/properties/P000100 For example, the one point compactification of the rationals has this property but isn't T_2. – Steven Clontz Sep 25 '23 at 23:06
  • @StevenClontz: thanks for pointing that out. – Rob Arthan Sep 25 '23 at 23:07
  • If the assumption $K_m \supset K_n$ is a strict inclusion, non emptiness follows. – chi Sep 26 '23 at 07:07
  • @chi: not so: read the first paragraph of my answer. – Rob Arthan Sep 26 '23 at 08:19
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    @RobArthan the point is, if $K_m = {}$ then you can't possibly have $K_m \supsetneq K_n$. – leftaroundabout Sep 26 '23 at 08:31
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    That's a good point, but the result is stronger to assume $\subset$ allows equality, with the caveat that sets are nonempty. – Steven Clontz Sep 26 '23 at 11:16
  • @chi: if you mean non-emptiness of all the $K_n$ follows, then you are right, although I preferred the more general reading of $\subset$ allowing equality. I thought you meant non-emptiness of the intersection. – Rob Arthan Sep 26 '23 at 19:44
  • @leftaroundabout: I see now that chi's commment was unclear. See my second response to it – Rob Arthan Sep 26 '23 at 19:45
  • Indeed my comment was misleading. I meant the emptiness of $K_n$, and I forgot that the thesis was also about non emptiness (of the intersection). I was focusing on the part "you need to assume that $K_n$ are all nonempty", and did not realize the ambiguity. – chi Sep 26 '23 at 20:41
  • @chi: OK. No worries! We are all agreed now, I think. – Rob Arthan Sep 26 '23 at 21:05
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    Perhaps you should mention in your answer that spaces in which all compact sets are closed are known as KC-spaces. And of course the claim is true if all $K_n$ are assumed to be closed ... – Paul Frost Sep 29 '23 at 15:44
  • @PaulFrost: I have done as you suggested. Thanks for improving the answer. – Rob Arthan Sep 29 '23 at 19:22