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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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