If $(A_n)_{n\epsilon \Bbb N}$ is increasing, prove $lim inf A_n = lim sup (A_n) = \bigcup^\infty_{n=1} (A_n)$?

Question

So, for a set $X$ and a sequence of subsets of $X$, $(A_n)_{n\in \Bbb N}$, how can I prove that when the sequence of subsets is increasing, then $$lim inf (A_n) = lim sup (A_n) = \bigcup^\infty_{n=1} A_n$$