Let $X$ be a set and let $(A_n)$ be a sequence of subsets of $X$.
Show that;
If $A_n$ is increasing then $\liminf (A_n) = \limsup (A_n) = \bigcup^{\infty}_{n=1} A_n$
So what I know from the question that I think I'll need for a solution is that
$\limsup a_n := \bigcap^{\infty}_{n=1} \bigcup_{k{\geq}n} A_k$
$\liminf a_n := \bigcup^{\infty}_{n=1} \bigcap_{k{\leq}n} A_k$
and the sequence is increasing so $A_k \leq A_{k+1}$
I'm not to sure how to put this together into a solution, help please ?
I'm also required to show for a decreasing sequence $\liminf (A_n) = \limsup (A_n) = \bigcap^{\infty}_{n=1} A_n$ but I'm sure the solutions will be similar.
Best Answer
Since $A_1\subset A_2\subset A_3\subset\cdots$, then, for each $n\in\mathbb N$,$$\bigcup_{k\geqslant n}A_k=\bigcup_{k=1}^\infty A_k$$and therefore$$\limsup_nA_n=\bigcap_{n=1}^\infty\bigcup_{k=1}^\infty A_k=\bigcup_{k=1}^\infty A_k.$$On the other hand, for each $n\in\mathbb N$,$$\bigcap_{k\geqslant n}A_k=A_n$$and therefore$$\liminf_nA_n=\bigcup_{n=1}^\infty A_n.$$