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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?

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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$.

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