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Let $E$ be a Hausdorff topological space and $(K_{n})_{n \in \mathbb{N}}$ be a decreasing sequence of compact subsets of $E$. Let $U \subset E$, $U$ open with $\bigcap_{n \in \mathbb{N}} K_{n} \subset U$. Then the following assertion is true?

$$\exists n_{0} \in \mathbb{N}\, \text{s.t. } \forall n\geq n_{0}, \,K_{n}\subset U$$

Is there a counter example?

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Suppose that $$(\exists n_{0} \in \mathbb{N}) (\forall n\geq n_{0}) K_{n}\subset U$$ is not true.

This means that $K_n\setminus U\ne\emptyset$ for each $n$.
(Here we also use the fact that the given system is decreasing.)

Then the system $(K_n\setminus U)_{n\in\mathbb N}$ is system of compact1 sets which has finite intersection property. By compactness we get that the intersection $$\bigcap_{n\in\mathbb N} (K_n\setminus U)= \left(\bigcap_{n\in\mathbb N} K_n\right)\setminus U$$ is non-empty.

Thus we get $\bigcap\limits_{n\mathbb N} K_n \not\subseteq U$.

This proves the claim from your question. (More precisely, we proved the contraposition.)


1 As pointed out in comments, this needs some clarification. Since these sets are compact, they are also closed. (This is the place where we use that we are working in a Hausdorff space.)

So we have a system of closed sets with finite intersection property. At the same time, all these sets are subsets of $K_1\setminus U$. This is a closed set of $K_1$, hence it is compact.

So we are working with a system of closed subsets of the compact space $K_1\setminus U$ which has a finite intersection property and we can use an equivalent characterization of compactness.

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  • $\begingroup$ Note that the Hausdorffness is not needed. $\endgroup$ – user87690 Dec 5 '14 at 15:52
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    $\begingroup$ @user87690 It is. In a countable cofinite set we can easily find counterexamples. Exercise: find the place where we used Hausdorffness.... $\endgroup$ – Henno Brandsma Dec 5 '14 at 16:03
  • $\begingroup$ @HennoBrandsma: It took me a couple moments to work out the non-Hausdorff counterexample. Consider $\mathbb{N}$ with the cofinite topology, in which every subset is compact. Let $K_n = \{n, n+1, \dots\}$ so that $\bigcap_n K_n = \emptyset$. Take $U = \emptyset$ which is open. Then trivially no $K_n$ is contained in $U$. $\endgroup$ – Nate Eldredge Dec 5 '14 at 16:16
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    $\begingroup$ @HennoBrandsma: (Replacement of a previous comment) Ah, the issue is that the finite intersection property (FIP) characterization of compactness is based on a collection of closed sets with the FIP (not a collection of compact sets). Without assuming Hausdorffness, the $K_n \setminus U$ need not be closed. $\endgroup$ – Nate Eldredge Dec 5 '14 at 16:18
  • $\begingroup$ @NateEldredge Indeed, this is what I had in mind. It's the simplest (I think) example of a decreasing sequence of compact sets with empty intersection. We really need that compact sets are closed. $\endgroup$ – Henno Brandsma Dec 5 '14 at 16:18
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Sure. $K = \cap_n K_n$ is non-empty and compact (closed subset of compact $K_0$, and compact sets are closed due to Hausdorffness). We know that $K \subseteq U$. If $K_m \subset U$, by decreasingness for all $k \ge m$ also $K_k \subseteq U$. So if the statement fails we know that $K_n \setminus U$ is non-empty for all $n$.

The same fact that showed that $K$ was non-empty, shows that $\cap_n (K_n\setminus U) = (\cap_n K_n) \setminus U = K \setminus U$ is non-empty, contradicting $K \subseteq U$.

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