# Example of sequence of sets $(A_n)$ such that $\liminf{A_n}$ and $\limsup{A_n}$ have a specific property

I need to find an example of a set $X$ and a sequence of sets $(A_n)$ in $\mathcal{P}(X)$ such that $\displaystyle \emptyset \subsetneqq \bigcap_{n} {A_n} \subsetneqq \liminf_{n}{A_n} \subsetneqq \limsup_{n} {A_n} \subsetneqq \bigcup_{n}{A_n} \subsetneqq X$. Can someone give me an idea?

## 1 Answer

One way to do it is to build what you want by just thinking about what these expressions mean. The first thing you need is for the sequence $\{A_n\}$ of sets to have non-empty intersection, so let's start by stipulating that $1 \in A_n$ for all $n$.

Next, we need there to be some element (other than $1$) that's in all but finitely many $A_n$. So let's stipulate that $2 \in A_n$, $n \geq 2$. Then we have $\liminf A_n = \{1,2 \}$.

Next, we need there to be some element that's in infinitely many of the $A_n$ but not in all but finitely many of them. So let $3 \in A_n$, $n \geq 2$, and $n$ even. Then $\limsup A_n = \{1,2,3 \}$.

Our sequence looks like this so far: $$\{ 1\}, \{ 1,2,3\}, \{1,2 \}, \{ 1,2,3\}, \{ 1,2\},...$$

We're almost done, but now we have $\cup_n A_n = \{ 1,2,3\}$, so let's just add $4 \in A_1$ so that the union is strictly bigger than the $\limsup$. And now we're done: $$\{ 1, 4\}, \{ 1,2,3\}, \{1,2 \}, \{ 1,2,3\}, \{ 1,2\},...$$

• Thank you very much! Your answer made a lot of things clear in my mind! – dmvlt Mar 18 '17 at 17:13