# "Potentially club" filters on $\omega_2$

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.

• What do you mean "or some relevant $\Bbb P$"? in that choiceless question? Dec 12, 2016 at 6:42
• (Also, if you're assuming ZFC+AD, then the answer is both yes and no... :-)) Dec 12, 2016 at 7:51
• 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 . . ."). Dec 12, 2016 at 10:09

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.

 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 $\omega_1$.

• 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? Dec 14, 2016 at 17:24
• That is correct. It was just a mention of some "possibly relevant forcing" from some models of AD which collapses $\omega_1$ but preserves $\omega_2$. Dec 14, 2016 at 17:29
• 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/…. Dec 19, 2016 at 23:35