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](https://mathoverflow.net/questions/257613/getting-measures-on-omega-2-from-potential-clubs).




---


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.



"Potentially club" filters on $\omega_2$
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.


