Letting {A$_{n}$} be a sequence of sets. We are given the definition $$\overline{\lim}(A_{n})= \cap_{n\geq 1}\cup_{k\geq n}A_{k}$$ The question wants us to prove that x $\in\overline{\lim}(A_{n})$ implies that for each N there exists at least one n $\geq$ N such that x $\in A_{n}$.
My dilemma with this is where I was planning on making my substitution. I was planning on allowing B$_{n}=\cup^{\infty}_{k=n}A_{k}$, and having $\overline{\lim}A_{n} = \cap^{\infty}_{n=1}$.
I keep getting lost in my own work and over complicating it with the substitutions.
I was told that there is a result similar to this in Rudin's text but I can't seem to find it.
suppose $x\in \bar{lim}A_n$. This is equivalent to saying that for every $n$, $x\in \bigcup_{k\geq n}A_k$. Let $N>0$, $x\in \bigcup_{k\geq N}A_k$ this implies there exists $k\geq N$ such that $x\in A_k$.
Conversely, suppose that for every $N$, there exists $k>N$ such that $x\in A_k$, this implies that $x\in \bigcup_{n\geq N}A_k$, henceforth $x\in\cap_N\bigcup_{k\geq N}A_k =\bar{lim}A_n$.