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Let $(A_n)$ be any sequence of pairwise disjoint events, is

$$ \lim_{n \to \infty} A_n = \emptyset. $$

always true?

Conceptually I would say yes, since if we partition $\Omega$ an infinite amount of times, I guess that the events would get smaller and smaller.

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    $\begingroup$ How are you defining limits of sets? The concept makes little sense generally and it is usually only used for chains of the form $A_1\supseteq A_2\supseteq A_3\supseteq\dots$ or $A_1\subseteq A_2\subseteq A_3\subseteq\dots$. I suppose if you were to try to define $\lim\limits_{n\to\infty}A_n = \{x~:~\exists N~(\forall n>N~x\in A_n)\}$ then yes, the result is empty. $\endgroup$
    – JMoravitz
    Commented Oct 19, 2020 at 13:20
  • $\begingroup$ The expression $\lim_{n \rightarrow \infty} A_n$ is not well defined if the sets $A_n$ are disjoint. (as far as I know) One usually defines such limits only for increasing or decreasing sets, i.e. sequences of sets which fulfill $A_n \subseteq A_{n+1}$ (or $\supseteq$, respectively). $\endgroup$ Commented Oct 19, 2020 at 13:21
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    $\begingroup$ The expression that @JMoravitz defined is also known as the $\liminf$ of a set-sequence and may be written as $\bigcup_{n \in \mathbb{N}} \bigcap_{m \geq n} A_m$. If the sets are pairwise disjoint it is (trivially) empty. $\endgroup$ Commented Oct 19, 2020 at 13:24
  • $\begingroup$ This isn't really a very great definition of convergence of sets however... consider applying that definition to the sequence of sets $\{1,2\},\{1,3\},\{1,2\},\{1,3\},\dots$... with that definition you'd have the sequence "converges to $\{1\}$" which might not match our intuition. The sequence of numbers $2,3,2,3,2,3,\dots$ we would have said didn't converge. $\endgroup$
    – JMoravitz
    Commented Oct 19, 2020 at 13:25
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    $\begingroup$ @JMoravitz I think of the limit of sequence of sets as when the lim sup and the lim inf are the same... $\endgroup$
    – fire-bee
    Commented Oct 19, 2020 at 15:09

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One can say that a sequence of sets $(A_n)_{n\geqslant 1}$ converges to $A$ if $\limsup_{n\to\infty}A_n=\liminf_{n\to\infty}A_n=A$, where $$ \limsup_{n\to\infty}A_n=\bigcap_{N\geqslant 1}\bigcup_{n\geqslant N}A_n $$ and $$ \liminf_{n\to\infty}A_n=\bigcup_{N\geqslant 1}\bigcap_{n\geqslant N}A_n. $$ One can show that $\omega$ belongs to $\limsup_{n\to\infty}A_n$ if and only if the set $\{n\in\mathbb N, \omega\in A_n\}$ is infinite and that $\omega$ belongs to $\liminf_{n\to\infty}A_n$ if and only if there exists an integer $N$ such that $\omega\in A_n$ for all $n\geqslant N$.

The inclusion $\liminf_{n\to\infty}A_n\subset \limsup_{n\to\infty}A_n$ always hold. In the particular case where $(A_n)_{n\geqslant 1}$ is pairwise disjoint, $\limsup_{n\to\infty}A_n$ is empty, because no $\omega$ belongs to more than one $A_n$.

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  • $\begingroup$ Thanks. Do you know of any book with this or any other proof? $\endgroup$
    – fire-bee
    Commented Oct 19, 2020 at 20:44

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