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Let $(A_n)_{n=1}^{\infty}$ be a pairwise disjoint collection.

$\lim A_n = \emptyset$? (see here and there)

What about a set of extended real numbers $A_n=(n,n+1]$?

It seems that

$\limsup A_n = \bigcap_{k=1}^{\infty} \bigcup_{n=k}^{\infty} A_n$

$= \bigcap_{k=1}^{\infty} \bigcup_{n=k}^{\infty} (n, n+1]$

$= \bigcap_{k=1}^{\infty} ((k, k+1] \cup (k+1, k+2] \cup ...)$

$= \bigcap_{k=1}^{\infty} ((k, \infty])$

... = {$\infty$} ?

Community
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BCLC
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1 Answers1

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$\infty$ is not in those unions (i.e., it is not in $\bigcup\limits_{n=k}^\infty(n,n+1]$ for any $k$), because it is not in any of the individual sets being unioned (i.e., it is not in $(n,n+1]$ for any $n$).

Jonas Meyer
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  • Wait I think I get it. Is it for the same reason the union here http://en.wikipedia.org/wiki/Sigma_additivity#An_additive_function_which_is_not_.CF.83-additive is (0,1) instead of [0,1)? – BCLC Jul 11 '14 at 05:36
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    @BCLC: Yes. In that case, every positive number is eventually in the sets, but $0$ is not in any of them, so it is not in their union. – Jonas Meyer Jul 11 '14 at 05:50
  • @BCLC: I rolled back the edit which was incorrect, but will try to edit for greater clarity in a bit. – Jonas Meyer Oct 02 '18 at 19:44