What is $\bigcup_{i=1}^\infty \bigcap_{j\ge i}^\infty A_i$? What is this set? I am having trouble understanding the notation. Is it first taking all intersections, then taking all unions? What is that $j$? I don't see $j$ as subscript of $A$.
$$\bigcup_{i=1}^\infty \bigcap_{j\ge i}^\infty A_i$$
 A: I will assume that it should read
$$\bigcup_{i\ge 1}\bigcap_{j\ge 1}A_j\;.$$
It works just like a double summation: for each value of the outer index $i$ we compute the inner intersection. For each $i\in\Bbb Z^+$ let
$$B_i=\bigcap_{j\ge i}A_j\;;$$
this is the set of things that are in every $A_j$ with $j\ge i$. Notice that $B_1\subseteq B_2\subseteq B_3\subseteq\ldots\;$: if something is in every $A_j$ with $J\ge i$, it is certainly in every $A_j$ with $j\ge i+1$. Then we take the union of this increasing family of sets: 
$$\bigcup_{i\ge 1}\bigcap_{j\ge 1}A_j=\bigcup_{i\ge 1}B_i\;,$$
the set of things that are in at least one of the sets $B_i$.
Added: For example, $S$ be the set of all infinite sequences of zeroes and ones. let $A_j$ be the set of infinite sequences $\langle b_n:n\in\Bbb Z^+\rangle\in S$ such that $b_j=0$. Then for any $i\in\Bbb Z^+$, $B_i$ is the set of sequences $\langle b_n:n\in\Bbb Z^+\rangle\in S$ such that $b_n=0$ for each $n\ge i$. A sequence $\langle b_n:n\in\Bbb Z^+\rangle\in S$ belongs to
$$\bigcup_{i\ge 1}\bigcap_{j\ge 1}A_j=\bigcup_{i\ge 1}B_i\tag{1}$$
if and only if there is some $i\in\Bbb Z^+$ such that $b_n=0$ for each $n\ge i$, so the set in $(1)$ is the set of sequences that are $0$ from some point on: they have only finitely many ones.
A: This is known as the limit infimum of $A_j$:
$$\liminf_{j \to \infty} A_j = \{x\mid \forall^\infty (j \mid x \in A_j)\} = \bigcup_{m=1}^\infty \bigcap_{j=m}^\infty A_j$$
where $\forall^\infty$ means "for all ... except finitely many". Similarly we have limit supremum:
$$\limsup_{j \to \infty} A_j = \{x\mid \exists^\infty (j, x \in A_j)\} = \bigcap_{m=1}^\infty \bigcup_{j=m}^\infty A_j$$
and $\exists^\infty$ means "there exists infinitely many..."
