Help on a families of sets question I have been given the problem of
Let $ X$ be a subest and $A_n$ be a sequence of subsets of $X$
$ \bigcup_{k=1}^\infty [\bigcap_{k=n}^\infty A_n $ ]= {$x | x \in A_n $ for all but finitely many n}
I am not too sure of the families of sets of the right hand side, should it be $A_k$ instead of $A_n$ ?and if not what exactly is the family which is considered.
Please help if its possible
Thank You.
 A: Notation: $T = \{x | x \in A_n $ for all but finitely many n $\}$  
$$x \in  \bigcup_{k=1}^\infty [\bigcap_{k=n}^\infty A_k] \implies x \in [\bigcap_{k=m}^\infty A_k] \; \text{for some $m \in  \Bbb N$} \implies x \in A_k \; \forall k \ge m$$
Therefore the set $C = \{n \ | \ x \not \in A_n\}$ is finite since $|C| \lt m$ and hence the number of sets which do not contain $x$ are finite. So $x \in A_n$ for all but finitely many $n$. 
$$ \implies x \in T \implies \bigcup_{k=1}^\infty [\bigcap_{k=n}^\infty A_k] \subseteq T --- (1)$$
Now suppose $x \in T$. Then there are only finitely many sets $A_k$ that do not contain $x$. Let $M \in \Bbb N$ such that  $M \gt  \text{Max} \{ n \ | \ x \not \in A_n \}$. Then; $n \ge M \implies x \in A_n \implies x \in [\bigcap_{k=M}^\infty A_k] \implies x \in \bigcup_{k=1}^\infty [\bigcap_{k=n}^\infty A_k]$
This implies that $T \subseteq \bigcup_{k=1}^\infty [\bigcap_{k=n}^\infty A_k] --- (2)$
Statements $(1)$ and $(2)$ prove our set equality. 
A: We start with a sequence of sets $\langle A_{n}\vert n\in\mathbb{N}\rangle$. Let $L$ be defined by $$L=\bigcup_{k\in\mathbb{N}}\bigcap_{j\geq k} A_{j}$$ (note that $L$ is sometimes denoted $\liminf_{j}A_{j}$), and let $R$ be the set 
$$R=\bigg\{x\bigg\vert  \big\vert \{n\in\mathbb{N}\vert x\notin A_{n}\}\big\vert <\aleph_{0} \bigg\}.$$
We first show that $R\subseteq L$.  Suppose that $x\in R$.  Then there is some $k\in\mathbb{N}$ such that $\forall j\geq k \, \left(x\in A_{j}\right)$. It follows that $x\in\bigcap_{j\geq k}A_{j}$.
For the converse direction, suppose that $x\in L$.  Then for some $k$, we have $x\in \bigcap_{j\geq k}A_{j}$.  It follows that 
$$\{n\in\mathbb{N}\vert x\notin A_{n}\}\subseteq \{0,1,\ldots , k-1\}.$$ 
But this set is finite, so we are done.
