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By reading over some basic topology questions today with my friend today, me and my friend encountered an question about the definition about compact set and descending of the set.

The definitions of the term used in the following sentences are mostly from Hatchers. 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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  • Kai, can you say a bit about what you and your friend tried on this problem so we can get the question reopened - as we don't want it to disappear, because it's an interesting point. – Rob Arthan Jun 13 '24 at 22:16

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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