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Let $(M,d)$ be a metric space and let $\{S_n\}_n$ be a countable collection of non-empty closed and bounded subsets of $M$

Are there any additional conditions on the collection$\{S_n\}_n$ to ensure that $S:=\limsup S_n=\bigcap_n\bigcup_{m\ge n}S_m$ is closed and bounded?

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    Well, you clearly need some further conditions; consider $S_n = [0,n]$. Did you mean to make the restriction that $S_{n+1} \subseteq S_n$?2012-08-27

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For boundedness, a sufficient condition (not necessary) is that $\{ S_n \}_n$ are all contained in some bounded set. In general, we don't have a such condition.

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    Even in my first post, it was only for boundedness.2012-08-27