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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 $n>m$.

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(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.)

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Is this homework? If yes, please tag it as such and list what you have tried already. –  Johannes Kloos Apr 19 '12 at 8:57

2 Answers 2

up vote 2 down vote accepted

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 $n$.

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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

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 $n>m$.

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well, martini was few minutes earlier than me. –  Thomas E. Apr 19 '12 at 10:15

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