# Does a sequence of pairwise disjoint event converge to the empy set, $\emptyset$?

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.

• 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. Commented Oct 19, 2020 at 13:20
• 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). Commented Oct 19, 2020 at 13:21
• 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. Commented Oct 19, 2020 at 13:24
• 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. Commented Oct 19, 2020 at 13:25
• @JMoravitz I think of the limit of sequence of sets as when the lim sup and the lim inf are the same... Commented Oct 19, 2020 at 15:09

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 $$A_n$$.