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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 $C = \lim_{n\to \infty} C_n$.

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    $\begingroup$ 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. $\endgroup$ – Kabo Murphy Mar 11 at 12:23
  • $\begingroup$ You are correct I made a mistake in writing the claim I will correct right away thanks for pointing that out. $\endgroup$ – TheBridge Mar 11 at 13:10
  • $\begingroup$ 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. $\endgroup$ – TheBridge Mar 11 at 13:20
  • $\begingroup$ 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 $\endgroup$ – TheBridge Mar 11 at 14:49
  • $\begingroup$ I don't think $X \in \mathcal C$ is a requirement for a monotone class. See en.wikipedia.org/wiki/Monotone_class_theorem $\endgroup$ – Kabo Murphy Mar 11 at 23:14
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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 $\alpha < \omega_1$.

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  • $\begingroup$ very cool example $\endgroup$ – TheBridge Mar 23 at 13:43
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Let C be the collection of countable ordinals.
C is its own closure.
$\tilde C$ is the set of denumerable (infinitely countable) ordinals.

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  • $\begingroup$ 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 $\endgroup$ – TheBridge Mar 12 at 15:44
  • $\begingroup$ @TheBridge. What don't you understand? $\endgroup$ – William Elliot Mar 12 at 22:14
  • $\begingroup$ 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 $\endgroup$ – TheBridge Mar 13 at 7:44
  • $\begingroup$ Could you handle an example of well ordered subsets of X that are ordered by a well ordering of X? $\endgroup$ – William Elliot Mar 14 at 0:34
  • $\begingroup$ At least I can try, if the proof is complete or the reference detailed enough $\endgroup$ – TheBridge Mar 14 at 16:27

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