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

  • 1
    $\begingroup$ 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

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 $\alpha < \omega_1$.

  • $\begingroup$ very cool example $\endgroup$ – 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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.