"Infinitely often" interpretation of lim sup of sequence of sets One type of interpretation of the lim sup and lim inf of a sequence of sets is as follows:
$\text{lim sup}_n A_n = \{ \omega: \omega \in A_n \text{ for infinitely many } A_n \}$
$\text{lim inf}_n A_n = \{ \omega: \omega \in A_n \text{ for all but finitely many } A_n \}$
I don't understand the difference between these two statements. Since there are infinitely many $A_n$, I am tempted to conclude that the latter statement automatically implies that $\omega \in A_n$ for infinitely many $A_n$?
Is the difference basically that elements lim sup are not necessarily in the lim inf, because they might occur in every single $A_n$, whereas elements of the lim inf are definitely in the lim sup?
 A: "Infinitely many" is a consequence of "all but finitely many", but it's not the same. For example, there are infinitely many even numbers; it is not the case that all but finitely many whole numbers are even, because $1, 3, 5, \ldots$ are not even. On the other hand, all but finitely many whole numbers are greater than twelve; $1, 2, 3, \ldots, 12$ are not, but the list of counterexamples stops there.
Another way of thinking about "all but finitely many" is "eventually". In other words, $\omega \in A_n$ for all $n$ past a certain point. "Infinitely many" just means it always reappears when it leaves; "all but finitely many" means it eventually stops leaving at all.
A: Another way to think of set theoretic limits is using indicator functions:
$$
x \in \textstyle\limsup_nA_n \iff \limsup_n \mathbf{1}_{A_n}(x) = 1,
$$
which holds if and only if the binary sequence $\{\mathbf{1}_{A_n}(x)\}_{n=1}^{\infty}$ has infinitely many ones, i.e., $x$ is a member of infinitely many $A_n$.
Similarly, we have
$$
x \in \textstyle\liminf_nA_n \iff \liminf_n \mathbf{1}_{A_n}(x) = 1,
$$
here, we note that since $\{\mathbf{1}_{A_n}(x)\}_{n=1}^{\infty}$ is composed of $1$s and $0$s, the above statement is equivalent to $\lim_n \mathbf{1}_{A_n}(x) = 1$, which holds if and only if sequence eventually becomes $1$ forever.
A: 
Since there are infinitely many $A_n$, I am tempted to conclude that the latter statement automatically imply that $ω∈A_n$ for infinitely many $A_n$?

You are correct. To rephrase what you said, $w \in$ lim inf$_n A_n \implies w \in$ lim sup$_n A_n$. This is true since lim inf$_n A_n \subset$ lim sup$_n A_n$. Think of partial ordering via set inclusion: as we may intuitively ascertain, lim inf$_n A_n$ is "less than" lim sup$_n A_n$, by inclusion.
A good way to start wrapping your head around these terminologies is by defining them more concretely as follows:
lim sup$_n A_n := \bigcap_{n \geq 1} \bigcup_{k \geq n} A_k$ 
lim inf$_n A_n := \bigcup_{n \geq 1} \bigcap_{k \geq n} A_k$ 
