5

Let $K_n$ be a nested sequence of non-empty compact sets in a Hausdorff space.

Prove that if an open set $U$ contains contains their (infinite) intersection, then there exists an integer $m$ such that $U$ contains $K_n$ for all $n>m$.

...

(I know that compact sets are closed in Hausdorff spaces. I can also prove that the infinite intersection of non-empty compact sets is non-empty, closed and compact in a Hausdorff space. I don't know how to use the fact that U is open.)

CJQ
  • 53

2 Answers2

6

let $U \supseteq \bigcap_n K_n$ be open. Set $L_n = K_n \setminus U$, then the $L_n$ are a nested sequence of compact sets. Now $\bigcap_n L_n = \bigcap_n K_n \setminus U = \emptyset$, therefore there is an $m$ such that $L_n = \emptyset$ for $n> m$, which means that $K_n \subseteq U$ for these $n$.

martini
  • 84,101
  • Thanks! Now that (both of) you have provided the solution, it really isn't a difficult question at all. – CJQ Apr 19 '12 at 10:20
5

Assume the contrary, that $V_{n}:=K_{n}\cap U^{c}\neq \emptyset$ for all $n$. Each $V_{n}$ is compact, since $U$ is open, and non-empty, so the intersection $\bigcap_{n} V_{n}$ is non-empty as well. But $\bigcap_{n}V_{n}=(\bigcap_{n}K_{n})\cap U^{c}\subset U\cap U^{c}=\emptyset$, which is a contradiction. Hence there exists $m$ such that $V_{n}=\emptyset$ for all $n>m$, whence $K_{n}\subset U$ for all $n>m$.

T. Eskin
  • 8,303