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?


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.

  • $\begingroup$ Note that the Hausdorffness is not needed. $\endgroup$ – user87690 Dec 5 '14 at 15:52
  • 3
    $\begingroup$ @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
  • 1
    $\begingroup$ @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

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 $K \subseteq U$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.