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In my text, we define a Monotone Class $\mathcal{M}$ of a non-empty set $X$ to be a collection of subsets of $X$ that is closed under monotone limits: that is, $(1)$ if $A_{i} \uparrow A$ with $A_{i} \in \mathcal{M}$, then $A \in \mathcal{M}$ and $(2)$ if $A_{i} \downarrow A$ with $A_{i} \in \mathcal{M}$, then $A \in \mathcal{M}$.

The exercise for which I am seeking solution verification says the following:

If $(\mathcal{M}_{i})$ is an increasing sequence of monotone classes, then is $\mathcal{M} = \bigcup^{\infty}_{i=1} \mathcal{M}_{i}$ a monotone class? Prove or give a counterexample.

I want to construct a counterexample as follows.

Let $X = \mathbb{N}$. Put $\mathcal{M}_{1} = \{\{1\}\}$, $\mathcal{M}_{2} = \{\{1\}, \{1,2\}\}$, and, in general, $\mathcal{M}_{i} = \{\{1\}, \{1,2\},\ldots, \{1,2,\ldots, i\} \}$. Then, $(\mathcal{M}_{i})$ is an increasing sequence of monotone classes. Now, let $A_{i}=\{1, \ldots,i\}$. Then, $(A_{i})$ is increasing, and $A_{i}$ belongs to $\mathcal{M}$ since $A_{i} \in \mathcal{M}_{i}$, for each $i$. Notice that $A_{i} \uparrow A$, where $A = \bigcup A_{i} = \mathbb{N}$. But, $A \notin \mathcal{M}$ since $A \notin \mathcal{M}_{i}$ for any $i$.

Is my counterexample constructed correctly? (Note: I have noticed that some defintoins of a monotone class require that the entire set be in the class. If my construct ion above is correct, then we can just add the entire set $X$ to each $\mathcal{M}_{i}$.) Thank you!

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up vote 2 down vote accepted

This is correct. However, if you use the definition requiring that $X \in \mathcal{M}_i$, then your counterexample no longer works since $\mathbb{N}$ would be in $\mathcal{M}$. However if you modify your solution a bit, you could get a counterexample that works for both definitions. Here is such a counterexample.

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