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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.

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  • $\begingroup$ 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 $n.$ $\endgroup$ – DanielWainfleet Oct 9 at 18:55
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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$$\liminf_nA_n=\bigcup_{n=1}^\infty A_n.$$

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  • $\begingroup$ 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 $\bigcap_{k\geqslant n}A_k=A_n$ $\endgroup$ – LDgh Oct 9 at 19:13
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    $\begingroup$ Sure! It was a typo. $\endgroup$ – José Carlos Santos Oct 9 at 20:07

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