# $\liminf$ and $\limsup$ of sets (Bartle's exercise 2H)

This question is from Bartle's "The Elements of Integration and Lebesgue Measure".

Let $$(A_n)$$ be a sequence of subsets of a set $$X$$. Show that $$\emptyset\subseteq\liminf A_n\subseteq\limsup A_n\subseteq X$$. Give an example of a sequence $$(A_n)$$ of sets with $$\liminf A_n = \emptyset$$ and $$\limsup A_n = X$$. Give an example of a sequence $$(A_n)$$ of sets which is not monotone but such that $$\liminf A_n = \limsup A_n$$.

I'm working with the definitions $$\liminf A_n = \displaystyle\bigcup_{n\in\Bbb{N}}\bigcap_{m\ge n} A_m$$ and $$\limsup A_n = \displaystyle\bigcap_{n\in\Bbb{N}}\bigcup_{m\ge n} A_m$$.

I have shown the first part but I couldn't come up with the examples, can someone help me with them?

## 1 Answer

Let $$A_n=\emptyset$$ for $$n$$ even and $$X$$ for $$n$$ odd. Then $$\lim \inf A_n=\emptyset$$ and $$\lim \sup A_n=X$$.

Let $$A_n=\{n,2n,2n+1,2n+2,...\}$$ with $$X=\mathbb N$$. Then $$\lim \inf A_n=\emptyset=\lim \sup A_n$$ but $$(A_n)$$ is not monotone.

Note that $$n \in A_n$$ but $$n \notin A_{n+1}$$. Also $$n+1 \in A_{n+1}$$ but $$n \notin A_{n}$$ if $$n>1$$.