Short version: what can we say about subsets of $\omega_2$ which - in a generic extension where $\omega_2$ is the new $\omega_1$ - contain a club?

We could of course generalize beyond $\omega_2$, but already the questions seem hard.

Motivating example:

Even in much weaker theories than ZFC (although ZF itself is not enough to my knowledge - the obstacle being that $\omega_1$ might be singular!), the club filter on $\omega_2$ is not an ultrafilter. An easy proof of this is to consider the sets $$A=\{x\in\omega_2: cf(x)=\omega\},\quad B=\{y\in\omega_2: cf(y)=\omega_1\};$$ it is easy to see that, while the union of $A$ and $B$ is club, neither $A$ nor $B$ contains a club.

However, consider the forcing notion $\mathbb{P}$ consisting of partial maps $p: \omega_1\rightarrow\omega_2$ such that

  • $dom(p)=\lambda+1$ for some limit $\lambda<\omega_1$ (in particular, $dom(p)$ is countable and has a greatest element),

  • $p$ is increasing: $\alpha<\beta\in dom(p)\implies p(\alpha)<p(\beta)$,

  • $p$ is continuous: if $\lambda$ is a limit ordinal in $dom(p)$, then $p(\lambda)=\sup\{p(\beta): \beta<\lambda\}$, and

  • $cf(p(\alpha))=\omega$ for all $\alpha\in dom(p)$.

Let $C$ be the range of the union of the conditions in some generic filter $G$. Then in $V[G]$, $C$ is a club subset of $\omega_2^V$, and consists entirely of ordinals of countable $V$-cofinality.

From the perspective of $\mathbb{P}$, then, $A$ is "more clubby" than $B$.

This is the kind of situation I'm interested in: When does a subset of $\omega_2$ contain a club in some forcing extension, and what can we say about the induced filter on $\omega_2$?

Specifically, for $\mathbb{P}$ a set forcing, let $\mathcal{F}_\mathbb{P}$ be the filter on $\omega_2$ of subsets of $\omega_2$ which in $V^\mathbb{P}$ contain a club. (Note that this makes sense even if $\omega_2^V$ is not a cardinal in $V^\mathbb{P}$ - however, it does trivialize, yielding the cofinite filter, if $cf(\omega_2^V)^{V^\mathbb{P}}=\omega$.)

Of particular interest are the forcings which preserve $\omega_2$ but collapse $\omega_1$, like $Col(\omega,\omega_1)$ (call such forcings relevant). Unfortunately, the associated filters are never ultrafilters, at least in ZFC: this is because they are always $\omega_2$-closed (since $\omega_1$ is made countable, and $V^\mathbb{P}\models$ the intersection of countably many clubs is countable).

That said, if we drop choice, then such filters can be ultrafilters! Let $V$ be a model of $ZF+V=L(\mathbb{R})$ + "The theory of $L(\mathbb{R})$ is absolute" - formally, this last condition is a scheme asserting that for every sentence $\varphi$ in the language of set theory with real parameters and every set forcing $\mathbb{P}$, $V\models\varphi\iff L(\mathbb{R})^{V[G]}\models\varphi$. (Informally, this should suggest that $V$ is the $L(\mathbb{R})$ of some choice model with a proper class of Woodins.)

Now consider the "labelled Levy collapse" $\mathbb{Q}=Col_{\mathbb{R}}(\omega,\omega_1)$ - a condition in this forcing is a finite sequence of reals coding well-orderings, and conditions are ordered by extension. Like the Levy collapse, this forcing makes $\omega_1$ countable, but - importantly - does so by adding a single real, $G$; in particular, the generic extension $V[G]$ satisfies "$V=L(\mathbb{R})$."

This gives us lots of absoluteness between $V$ and $V[G]$. In particular, the sentence "The club filter on $\omega_1$ is an ultrafilter" is true in $V$, and hence in $V[G]$; and in particular, this means that in $V$ the filter $\mathcal{F}_{Col_\mathbb{R}(\omega, \omega_1)}$ is an ultrafilter on $\omega_2^V$. Indeed, this filter witnesses that $\omega_2^V$ is measurable in $V$.

EDIT: The above argument relies on the assumption that $\omega_1$ in the new $L(\mathbb{R})$ is the old $L(\mathbb{R})$'s $\omega_2$; this seemed obvious to me at first, but now that I think about it (prompted by Asaf's answer below) I see it is wildly unjustified.

Hopefully the examples above motivate the following questions.

First, we can consider specific forcings in the ZFC context:

What is $\mathcal{F}_{Col(\omega,\omega_1)}$?

Second, we can consider individual sets in the ZFC context:

Is there a forcing $\mathbb{P}$ such that $\mathcal{F}_\mathbb{P}$ contains $B$ (the set of $x\in\omega_2$ of uncountable $V$-cofinality)?

I suspect the answer is no, but I don't immediately see how to prove it.

Third, we may consider the natural ideal which this notion induces:

Let $\mathcal{I}$ denote the set of subsets of $\omega_2$ which are not in any $\mathcal{F}_\mathbb{P}$, and $\mathcal{I}_{rel}$ the set of subsets of $\omega_2$ which are not in any $\mathcal{F}_\mathbb{P}$ with $\mathbb{P}$ relevant. What can we say about $\mathcal{I}$ and $\mathcal{I}_{rel}$?

I suspect $\mathcal{I}=\mathcal{I}_{rel}$ - that is, if we can force $x\subseteq\omega_2$ to contain a club, we can do so with a forcing which collapses $\omega_1$ and preserves $\omega_2$. In terms of understanding the ideals, one thing I'm interested in is the associated forcing notions: force with "positive" subsets of $\omega_2$ modulo "null" sets.

Finally, we can ask determinacy-flavored questions. The one which seems most interesting to me is:

Supposing ZF+AD+ whatever else, what are the measures on $\omega_2$ of the form $\mathcal{F}_{\mathbb{P}}$ for some (or some relevant) $\mathbb{P}$?

I suspect the answer is "all of them," but I don't see how to prove it.

EDIT: The first three questions were answered by Monroe Eskew below; the fourth, however, seems like a fundamentally different question. I've accepted Monroe's answer, and have asked the fourth question separately here.

And, of course, the standard: what are some good sources on this sort of thing? My problem here is that I don't know how to google this effectively - too many unrelated hits (e.g. about $\omega_2$ satisfying generic versions of large cardinal properties) keep coming up. So I'm probably missing some well-known material.

  • $\begingroup$ What do you mean "or some relevant $\Bbb P$"? in that choiceless question? $\endgroup$
    – Asaf Karagila
    Dec 12, 2016 at 6:42
  • $\begingroup$ (Also, if you're assuming ZFC+AD, then the answer is both yes and no... :-)) $\endgroup$
    – Asaf Karagila
    Dec 12, 2016 at 7:51
  • $\begingroup$ Re: ZFC, whoops. :P And "relevant" means "collapses $\omega_1$ but preserves $\omega_2$," see the third paragraph below the second line (beginning "Of particular interest . . ."). $\endgroup$ Dec 12, 2016 at 10:09

2 Answers 2


First, your notation is nonstandard; when we write $\mathrm{Col}(\kappa,\lambda)$, this typically means the set of partial functions $p : \kappa \to \lambda$ of size $<\kappa$, i.e. reverse of yours.

First note the following fact:

(1) If $\kappa$ is regular and $\mathbb P$ is $\kappa$-c.c., then every club subset of $\kappa$ in $V^\mathbb{P}$ contains a club from $V$.

To answer, "What is $\mathcal F_{\mathrm{Col}(\omega,\omega_1)}$?": It is just the club filter on $\omega_2$. Since this forcing has size $<\omega_2$, every new club contains a ground model club, so the new club filter is generated by the old.

For the next question, we use the following well-known result due to Harvey Friedman (generalized by Stavi and Abraham-Shelah):

(2) If $S \subseteq \omega_1$ is stationary, then the forcing $\mathbb C(S)$ consisting of closed bounded subsets of $S$ ordered by end extension adds a club through $S$ without adding reals.

Let $S = S^{\omega_2}_{\omega_1}$, i.e. the ordinals of uncountable cofinality below $\omega_2$. Then $\mathrm{Col}(\omega,\omega_1) * \dot{\mathbb C}(\check S)$ forces $S$ is in the new club filter on $\omega_2^V$. By (1), $S$ remains stationary, so we may apply (2).

For the next question, the above argument shows that your ideal $\mathcal I$ is simply the nonstationary ideal on $\omega_2$.

I don't know much about forcing without choice, so I'll leave the rest to someone else.


Without choice, the Levy collapse, being $\omega_1^{<\omega}$ ordered by reverse inclusion, is well-orderable so every set of ordinals has a canonical name, and the usual choice-arguments continue to hold.

Namely, every club of $\omega_2$ in the extension, contains a club from the ground model, by repeating the same proof as before. Or, if you would like, you could consider the proof inside $L[A,\dot C]$ with $\dot C$ being a name for a club of $\omega_2^V$ and $A$ being some maximal antichain (or a dense set, if one so desires) which forces sufficient relevant information on $\dot C$. Then we can find a club of $\omega_2^V$ in $L[A,\dot C]$ which is forced to be inside $\dot C$, and since $\omega_2^V$ is still regular in $V$, this club is still a club.

In other words: a well-orderable forcing behaves nicely with respect to sets of ordinals; so if we have a $\kappa$-c.c. well-orderable forcing, every measure on $\kappa$ extends to a measure in the generic extension.

With the labeled collapse $\Bbb Q$, this argument is not going to work directly, since the forcing is not well-ordered nor it lies in some model of choice. But something bothers me here: if the theory of $L(\Bbb R)$ is absolute, and the generic extension satisfies $V=L(\Bbb R)$, then $\sf AD$ holds after the extension. But this means that we collapsed all the $\omega_n$'s for $n>2$ because the new $\omega_2$ has to be measurable, and in particular regular. So I suspect that your labeled collapse—under assumptions of absoluteness—might not be relevant here.

However, do note that under $V=L(\Bbb R)+\sf DC+AD$, we can take the product of the Levy collapse of $\omega_1$ to $\omega$ and a Prikry forcing on $\omega_2$ with the measure of $\operatorname{cf}(\alpha)=\omega$. This should be1 a model where $\omega_2^V$ is the new $\omega_1$ and singular. There, as you noted, things trivialize, but it's still an option.

[1] You can think about this as an iteration, first do the Prirky forcing, you didn't collapse $\omega_2$, then do the Levy collapse which is a "nice enough" forcing that it won't collapse any cardinal other than $\omega_1^V$, regardless of its cofinality. So $\omega_2^V$ is the new and singular $\omega_1$.

  • $\begingroup$ Thanks! Just so I'm clear, though - the third paragraph doesn't directly address the question, right? Since it isn't looking at measures in the ground which are the "pullback" of the club filter in some extension? $\endgroup$ Dec 14, 2016 at 17:24
  • $\begingroup$ That is correct. It was just a mention of some "possibly relevant forcing" from some models of AD which collapses $\omega_1$ but preserves $\omega_2$. $\endgroup$
    – Asaf Karagila
    Dec 14, 2016 at 17:29
  • $\begingroup$ So it turns out that $\omega_2$ is (consistently) preserved, since it can be $\delta^1_2$ and $L(\mathbb{R})$ computes $\delta^1_2$ correctly: see Juan's answer to my question mathoverflow.net/questions/257614/…. $\endgroup$ Dec 19, 2016 at 23:35

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