# Monotone limits of sets do not exhaust the collection defined by closure by those limits

Here is a problem for which I would like to have a direct (constructive ideally) proof or a reference.

First a definition :

Let $$X$$ be a set, we say that a collection $$\mathcal C\subseteq \mathcal P (X)$$ ($$\mathcal P (X)$$ is the collection of all subsets of $$X$$) has closure (the collection) $$\bar {\mathcal C}$$, if $$\bar {\mathcal C}$$ is the minimal collection of sets of $$X$$ that fulfills the two following conditions :

• $$\mathcal C \subseteq \bar{\mathcal C}$$

• $$\bar {\mathcal C}$$ is stable for any monotone sequences of sets $$\{C_n\}_{n\geq 0} \subset \bar {\mathcal C}$$ (i.e. for all sequences such that $$C_n \nearrow$$ or $$\searrow$$ with $$C = \lim_{n\to \infty} C_n$$ then $$C \in \bar {\mathcal C}$$)

Now here is my question :

Prove that the collection $$\tilde C$$ can be strictly included in $$\bar {\mathcal C}$$, where $$\tilde C$$ consists of the sets $$C$$ such that there exists an increasing (or decreasing) sequence $$C_n$$ such that $$C = \lim_{n\to \infty} C_n$$.

• In standard terminology $\overline {\mathcal C}$ is called the monotone class generated by $\mathcal C$. The statement about strict inclusion is false. There are examples where there is strict inclusion, but if $\mathcal C$ contains a single set then we get equality. – Kabo Murphy Mar 11 at 12:23
• You are correct I made a mistake in writing the claim I will correct right away thanks for pointing that out. – TheBridge Mar 11 at 13:10
• By the way as I do not ask here for $X$ to be in the collection $\mathcal{C}$, it is not a monotone class (or a Dynkin system). But I agree that it is close in spirit. – TheBridge Mar 11 at 13:20
• I may add that the closure here, seen as an operator on the collection $\mathcal C$, is idempotent which is the (one of) very reason to define it as it is (i.e. in a non constructive way). Regards – TheBridge Mar 11 at 14:49
• I don't think $X \in \mathcal C$ is a requirement for a monotone class. See en.wikipedia.org/wiki/Monotone_class_theorem – Kabo Murphy Mar 11 at 23:14

First, there are obviously examples where $$\tilde C=\bar C$$. For example, if $$C=\mathcal P(X)$$ then it is immediate that $$\tilde C=\bar C=C$$.

However, there are important cases where $$\tilde C$$ is a strict subset of $$\bar C$$, which is what I think the question is after. Suppose that $$X=\mathbb R$$ and $$C\subseteq\mathbb R$$ is the collection of open sets. In this case, $$C$$ is already closed under taking limits of increasing sequences of sets, so $$\tilde C$$ is the collection of decreasing limits of sequences of sets in $$C$$. On the other hand, $$\bar C$$ is the Borel sigma-algebra. So, the question is, are there Borel sets which are not limits of decreasing sequences of open sets.

Here is one example. The rational numbers $$\mathbb Q$$ is Borel. For each $$x\in\mathbb Q$$, the singleton $$\{x\}$$ is the limit of the decreasing sequence $$(x-1/n,x+1/n)$$. Similarly, every finite subset of $$\mathbb R$$ is a limit of a decreasing sequence of open sets. As $$\mathbb Q$$ is countable, it is the limit of an increasing sequence of limits of decreasing sequences of open sets. So, if $$C$$ is the collection of open sets then $$\mathbb Q\in\bar C$$. However, $$\mathbb Q\not\in\tilde C$$. To see this, if there was a decreasing sequence $$U_n$$ of open sets with $$\bigcap_nU_n=\mathbb Q$$, this would contradict the Baire category theorem. Enumerating $$\mathbb Q=\{q_1,q_2,\ldots\}$$, the sets $$\{q_n\}\cup(\mathbb R\setminus U_n)$$ would be nowhere dense closed sets with union the whole of $$\mathbb R$$.

The Borel hierarchy takes this much much further. If we let $$C$$ be the collection of open subsets of $$\mathbb R$$ then we can define $$C_\alpha$$ inductively for all ordinals $$\alpha$$, $$C_\alpha=\begin{cases} C,&\textrm{if }\alpha=0.\\ \tilde C_\beta,&\textrm{if }\alpha=\beta+1.\\ \bigcup_{\beta < \alpha}C_\beta,&\textrm{if }\alpha > 0\textrm{ is a limit ordinal}. \end{cases}$$ This is clearly increasing, $$C_\alpha \subseteq C_\beta$$ for $$\alpha\le\beta$$. Also, as soon as you have equality $$C_\alpha=C_{\alpha+1}$$ then the sequence stabilises, so that $$C_\beta=C_\alpha=\bar C$$ for all $$\beta\ge\alpha$$. As the closure only involves countable sequences, it can be seen to stabilise once you get to the uncountable ordinals.

The question above is asking if $$C_2 > C_1$$. In fact, the Borel hierarchy tells us that $$C_\alpha < C_\beta$$ whenever $$\alpha < \beta$$ and $$\alpha$$ is a countable ordinal. In other words, it does not stabilise until you get to the first uncountable ordinal, $$\omega_1$$. So $$C_{\omega_1}=\bar C$$ and $$C_\alpha\subsetneq\bar C$$ for $$\alpha < \omega_1$$.

• very cool example – TheBridge Mar 23 at 13:43

Let C be the collection of countable ordinals.
C is its own closure.
$$\tilde C$$ is the set of denumerable (infinitely countable) ordinals.

• Hi it's a little dry (for me) as an answer, could you elaborate ? At the moment I can't tell, if it's true or false, so I can't accept your answer even if you are right.Thx – TheBridge Mar 12 at 15:44
• @TheBridge. What don't you understand? – William Elliot Mar 12 at 22:14
• Sorry but I am nut much acquainted with this the notion at hand and even after investigating a little what ordinal numbers are, I am not sure why a collection of ordinal numbers is its own closure neither why $\tilde C$ is the set of denumerable ordinals. Regards – TheBridge Mar 13 at 7:44
• Could you handle an example of well ordered subsets of X that are ordered by a well ordering of X? – William Elliot Mar 14 at 0:34
• At least I can try, if the proof is complete or the reference detailed enough – TheBridge Mar 14 at 16:27