# Given a pairwise disjoint collection, $\liminf A_n = \emptyset$?!

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

$\liminf A_n = \emptyset$?!

Spin-off from here.

My attempt:

$\liminf A_n$

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

$= \bigcup_{k=1}^{\infty} (A_k \cap A_{k+1} \cap ...)$

$= (A_1 \cap A_{2} \cap ...) \cup (A_2 \cap A_{3} \cap ...) \cup ...$

$= \emptyset \cup \emptyset \cup ... = \emptyset$

Is this right?

How does one best intuitively explain $\liminf A_n = \emptyset$, $\limsup A_n = \emptyset$ and/or $\lim A_n = \emptyset$ if $(A_n)_{n=1}^{\infty}$ is a pairwise disjoint collection? As I recall, $\limsup A_n$ means something like $A_n$ infinitely often and $\liminf A_n$ means something like $A_n$ almost always, but I'm lost after that.

• yes! $\liminf A_n = \emptyset$. – Mohammad Khosravi Jul 11 '14 at 5:15
• @MohammadKhosravi Intuitive explanation please? Hahaha – BCLC Jul 11 '14 at 5:17
• Do you consider this as something intuitively: In fact, when $x\in \liminf A_n$, it means for some $N$, we have that $x\in A_n$, for all $n\ge N$, so if $\liminf A_n = \emptyset$ then for any $N\in{\mathbb N}$ and $x\in \Omega$, there is a set $A_n$ such that $x\notin A_n$ meanwhile $n\ge N$. As here $A_n$s are mutually disjoint, this is no way possible. :D – Mohammad Khosravi Jul 11 '14 at 5:40
• @MohammadKhosravi Thanks! – BCLC Nov 8 '15 at 11:01