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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

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

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

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

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

... = {$\infty$} ?

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    $\begingroup$ Note that $\infty\not\in (n,n+1]$ for every $n\in\mathbb{N}$, so $\infty\not\in \bigcup_{n=k}^\infty (n,n+1]$. $\endgroup$
    – Brandon
    Jul 11, 2014 at 5:11
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    $\begingroup$ Note that $\infty$ is not an element of $\cup_{n=k}^\infty(n,n+1]$ for any $k$. $\endgroup$ Jul 11, 2014 at 5:11
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    $\begingroup$ @Brandon Are you saying that the penultimate step is incorrect? $\endgroup$
    – BCLC
    Jul 11, 2014 at 5:15
  • $\begingroup$ @Chellapillai Are you saying that the penultimate step is incorrect? $\endgroup$
    – BCLC
    Jul 11, 2014 at 5:17
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    $\begingroup$ @BCLC: Yes.${}$ $\endgroup$ Jul 11, 2014 at 5:20

1 Answer 1

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$\begingroup$

$\infty$ 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 $n$).

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  • $\begingroup$ Wait I think I get it. Is it for the same reason the union here en.wikipedia.org/wiki/… is (0,1) instead of [0,1)? $\endgroup$
    – BCLC
    Jul 11, 2014 at 5:36
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    $\begingroup$ @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. $\endgroup$ Jul 11, 2014 at 5:50
  • $\begingroup$ @BCLC: I rolled back the edit which was incorrect, but will try to edit for greater clarity in a bit. $\endgroup$ Oct 2, 2018 at 19:44

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