Show lim sup (an) = lim inf (an) for increasing for increasing sequence of subsets.

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.

• For any sequence $(A_n)_n$ of sets we have $x\in \lim \sup A_n$ iff $x \in A_n$ for infinitely many $n$ and $x\in \lim \inf A_n$ iff $x\in A_n$ for all but finitely many $n.$ – DanielWainfleet Oct 9 at 18:55

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$$\liminf_nA_n=\bigcup_{n=1}^\infty A_n.$$
• Great thanks! Just to check, that last line should be $\liminf_nA_n=\bigcup_{n=1}^\infty A_n$ right ? Since $\liminf_nA_n=\bigcup_{n=1}^\infty \bigcap_{k\geq n}^\infty A_k$ and you pointed out $\bigcap_{k\geqslant n}A_k=A_n$ – LDgh Oct 9 at 19:13