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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 $S_{n+1} \subseteq S_n$? –  Paul Z Aug 27 '12 at 18:23
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1 Answer 1

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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Not true: if $S_n=\{0,n\}$ for $n\in\Bbb N$, then $\limsup_nS_n=\{0\}$, but $\bigcup_{n\in\Bbb N}S_n=\Bbb N$. –  Brian M. Scott Aug 27 '12 at 18:49
    
Your sets $S_n$ are not contained in one bounded set. –  Ilies Zidane Aug 27 '12 at 19:44
    
That is exactly the point: they are not contained in one bounded set, and yet $\limsup_n S_n$ is bounded. Thus, your claim that we must assume that the $S_n$ are all contained in one bounded set is false. –  Brian M. Scott Aug 27 '12 at 19:49
    
I didn't say it's equivalent, but it's a sufficient condition. I should remove "at least"... –  Ilies Zidane Aug 27 '12 at 19:55
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It is not a sufficient condition for anything but boundedness: if $S_n=[2^{-n},1]$, then $\limsup_nS_n=(0,1]$, which is not closed. –  Brian M. Scott Aug 27 '12 at 19:57
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