I was wondering if someone would be so kind to provide a very simple explanation of lim sup and lim inf of s sequence of sets. For a sequence of subsets $A_n$  of a set $X$, the $\limsup A_n= \bigcap_{N=1}^\infty \left( \bigcup_{n\ge N} A_n \right)$ and $\liminf A_n = \bigcup_{N=1}^\infty \left(\bigcap_{n \ge N} A_n\right)$. But I am having a hard time imagining what that really means unions of intersections and intersections of unions I think maybe causing the trouble. I read the version on Wikipedia but that didn't resolve this either.
Any help would be much appreciated.
A member of
$$
\bigcup_{N=1}^\infty \bigcap_{n\ge N} A_n
$$
is a member of at least one of the sets
$$
\bigcap_{n\ge N} A_n,
$$
meaning it's a member of either $A_1\cap A_2 \cap A_3 \cap \cdots$ or $A_2\cap A_3 \cap A_4 \cap \cdots$ or $A_3\cap A_4 \cap A_5 \cap \cdots$ or $A_4\cap A_5 \cap A_6 \cap \cdots$ or $\ldots$ etc.  That means it's a member of all except finitely many of the $A$.

A member of
$$
\bigcap_{N=1}^\infty \bigcup_{n\ge N} A_n
$$
is a member of all of the sets
$$
\bigcup_{n\ge N} A_n,
$$
so it's a member of $A_1\cup A_2 \cup A_3 \cup \cdots$ and of $A_2\cup A_3 \cup A_4 \cup \cdots$ and of $A_3\cup A_4 \cup A_5 \cup \cdots$ and of $A_4\cup A_5 \cup A_6 \cup \cdots$ and of $\ldots$ etc.  That means no matter how far down the sequence you go, it's a member of at least one of the sets that come later.  That means it's a member of infinitely many of them, but there might also be infinitely many that it does not belong to.
