1
$\begingroup$

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?

  • 0
    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

1 Answers 1

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.

  • 0
    Not true: if $S_n=\{0,n\}$ for $n\in\Bbb N$, then $\limsup_nS_n=\{0\}$, but $\bigcup_{n\in\Bbb N}S_n=\Bbb N$.2012-08-27
  • 0
    Your sets $S_n$ are not contained in **one** bounded set.2012-08-27
  • 0
    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.2012-08-27
  • 0
    I didn't say it's equivalent, but it's a sufficient condition. I should remove "at least"...2012-08-27
  • 2
    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.2012-08-27
  • 0
    I said: "For boundedness"...2012-08-27
  • 0
    In your edit, yes; I was responding to the previous comment, since I had not yet seen the edit. (And in any case a weak sufficient for boundedness seems quite uninteresting in this context.)2012-08-27
  • 0
    Even in my first post, it was only for boundedness.2012-08-27