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If $X$ is a non-empty set, a family $\mathcal{K} \subseteq \mathcal{P}(X)$ is called a monotone class if for every increasing sequence of subsets of $X$, $\lbrace B_{n} \rbrace_{n\in \mathbb{N}}$, such that $(\forall n \in \mathbb{N}) B_{n} \in \mathcal{K}$, we have $\bigcup_{n=1}^{\infty} B_{n} \in \mathcal{K}$, and for every decreasing sequence of subsets of $X$, $\lbrace C_{n} \rbrace_{n\in \mathbb{N}}$, $\bigcap_{n=1}^{\infty} C_{n} \in \mathcal{K}$.

My professor told us today that monotone classes are closed under $\limsup_{n \to \infty}$ and $\liminf_{n \to \infty}$, and told us to try to prove that. I'm having some difficulties. Here's what I tried:

Let $B_{n} \in \mathcal{K}$ for all $n \in \mathbb{N}$. Then $\limsup_{n \to \infty} B_{n} = \bigcap_{n=1}^{\infty} \bigcup_{k=n}^{\infty} B_{k}$. Denote $C_{n} = \bigcup_{k=n}^{\infty} B_{k}$. Then $C_{n} \downarrow \limsup_{n \to \infty} B_{n}$, so if I prove that $C_{n} \in \mathcal{K}$ for all $n \in \mathbb{N}$, I'm done. However, I can't prove this, and I'm not even sure if it's true.

Is the statement true, and if so, how does the proof go?

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It's not so - I suspect that what the professor had in mind was more or less what Jose said.

If it were true then every monotone class would be closed under arbitrary finite unions, because the $\limsup$ of the sequence $A,B,A,B,\dots$ is $A\cup B$.

And if that were true it would be a theorem in those books. Say $I$ is the class of all (open, half-open or closed, bounded or unbounded, empty or nonempty) intervals. Then $I$ is a monotone class not closed under finite unions.

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  • $\begingroup$ I see, what a great counterexample! $\endgroup$ – Matija Sreckovic Dec 22 '17 at 16:49
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    $\begingroup$ Or just $X=\{1,2\}, \mathcal K=\{\{1\},\{2\}\}$. $\endgroup$ – hmakholm left over Monica Dec 22 '17 at 16:52
  • $\begingroup$ @HenningMakholm Touche. (I think this was the first time I ever thought about constructing a monotone class not closed under unions. Best excuse I can come up with.) $\endgroup$ – David C. Ullrich Dec 22 '17 at 17:09
  • $\begingroup$ I don't think you need an excuse -- I only came up with my example by trying to simplify yours. $\endgroup$ – hmakholm left over Monica Dec 22 '17 at 17:16

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