# How to intuitively understand why for a sequence of a sets $A_n$, $\liminf_{n}A_n \subseteq\limsup_{n}A_n$?

I am trying to understand exactly why for a sequence of sets $A_n$, $\liminf_{n}A_n \subseteq\limsup_{n}A_n$.

My understanding is that a member of $$\liminf_{n}A_n =\bigcup_{N=1}^\infty \bigcap_{n\ge N} A_n$$ is a member of at least one of the sets $$\bigcap_{n\ge N} A_n,$$ so that it is a member of all but finitely many $A_n$.

A member of $$\bigcap_{N=1}^\infty \bigcup_{n\ge N} A_n$$ is a member of every one of the sets $$\bigcup_{n\ge N} A_n,$$ So, my understanding is that it is a member of infinitely many of them, but that there may be infinitely many sets it is not a member of, since if I were to "put" some element inside $A_{100000}$, then all the union sets that came before have membership.

However, what confuses me is why the first case is a proper subset of the second case. It seems that in BOTH the limit infimum and limit supremum case, an element of each would be an element of infinitely many $A_n$. However, in the limit supremum case, it is possible that an element only comes up every, say 1 trillion times. Then, my element is in $A_1 \cup A_2 \cup \ldots$, and $A_2 \cup A_3 \cup \ldots$, BUT, it is not in the sequence of sets until the trillionth case.

In contrast, because the limit infimum deals with the intersection, the restrictions seem tighter in that an element needs to be in each and every one of them expect for finitely many.

In total, it seems that the limit infimum case appears more populated and less sparse than the limit supremum case. Can anyone tell me where my logic went wrong here? thanks!!

• Maybe it helps noting that $\cup_{n\geq N} A_n$ is decreasing in $N$. Thus you can write $\cap_{N=1} \cup_{n\geq N} A_n = \cap_{N=k} \cup_{n \geq N}$ for any $k$ you like. So some element is in the liminf, as you correctly stated, if we can find a $j$ s.t. it is in $\cap_{n\geq j} A_n$. Now pick $k=j$. limsup is then $\cap_{n\geq j} \cup_{N\geq n} A_n$. As $A_n \subset \cup_{N\geq n} A_n$, we see that limsup really is much bigger than liminf. – Furrer Sep 20 '16 at 22:54

Just as you said, the restrictions are tighter to be in the lim inf than the lim sup. That means that it's harder for a set to be in the lim inf, which makes the lim inf less populated and more sparse than the lim sup.

You have made most of the important steps. It is good to reiterate what it means for a set $A$ to lie in a set $B$. Showing that $A \subseteq B$ can be done by proving $x \in A \Rightarrow x \in B$.

So, you have figured out a necessary condition for an element to lie in $\liminf A_n$: if we have $x \in \liminf A_n$, then there is some $N$ such that $x \in A_n$ for all $n \geq N$. In short, this is the following implication: $$x \in \liminf A_n \Rightarrow \exists \; N \text{ s.t. } x \in A_n \text{ for all } n \geq N.$$

Moreover, you have also found a sufficient condition for an element to lie in $\limsup A_n$, namely if $x$ lies in infinitely many of the $A_n$, then it lies in $\limsup A_n$. Hence, you have the following implication:

$$\exists \text{ infinitely many } n \text{ s.t. } x \in A_n \Rightarrow x \in \limsup A_n.$$

What is left for you to do is to combine these two implications into an implication of the form

$$x \in \liminf A_n \Rightarrow \dots \Rightarrow x \in \limsup A_n,$$

as this would prove $\liminf A_n \subseteq \limsup A_n$. What would you still need to prove to obtain this implication?