The famous game-theoretic couple, Alice & Bob, live in the set-theoretic universe, $V$, a model of $ZFC$. Just like many other couples they sometimes argue over a statement, $\sigma$, expressible in the language of set theory. (One may think of $\sigma$ as a family condition/decision in the real life, say having kids or living in a certain city, etc.) 


Alice wants $\sigma$ to be true in the world that they live but Bob doesn't. In such cases, each of them tries to manipulate the sequence of the events in such a way that makes their desired condition true in the ultimate situation. Consequently, a game of forcing iteration emerges between them as follows:


Alice starts by forcing over $V$, leading the family to the possible world $V[G]$. Then Bob forces over $V[G]$ leading both to another possible world in which Alice responds by forcing over it and so on. Formally, during their turn, Alice and Bob are choosing the even and odd-indexed names for forcing notions, $\dot{\mathbb{Q}}\_{0}$, $\dot{\mathbb{Q}}\_{1}$, $\dot{\mathbb{Q}}\_{2}$, $\cdots$, in a forcing iteration of length $\omega$, $\mathbb{P}=\langle\langle\mathbb{P}\_{\alpha}: \alpha\leq\omega\rangle, \langle\dot{\mathbb{Q}}\_{\alpha}: \alpha<\omega\rangle\rangle$, where the ultimate $\mathbb{P}$ is made of the direct/inverse limit of its predecessors (depending on the version of the game). Alice wins if $\sigma$ holds in $V^{\mathbb{P}}$, the *ultimate future*. Otherwise, Bob is the winner. 



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> **Question 1.** Is there any characterization of the statements $\sigma$ for which Alice has a winning strategy in (the direct/inverse limit version of) the described game? How much does it depend on the starting model $V$?
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Clearly, Alice has a winning strategy if $\sigma$ is a consequence of $ZFC$, a rule of nature which Bob can't change no matter how tirelessly he tries and what the initial world, $V$, is! However, if we think in terms of *buttons* and *switches* in [Hamkins' forcing multiverse](https://arxiv.org/abs/math/0509616), the category of the statements for which Alice has a winning strategy seems much larger than merely the consequences of $ZFC$. 


I am also curious to know how big the difference between the direct and inverse limit versions of the described game is:



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> **Question 2.** What are examples of the statements like $\sigma$, for which Alice and Bob have winning strategies in the direct and inverse limit versions of the described game respectively? 
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**Update.** Following [Mohammad's comment](https://mathoverflow.net/questions/304973/forcing-and-family-contentions-who-wins-the-disputes#comment760522_304976), it seems a variant of this game in which Alice and Bob are restricted to choose certain types of forcing notions (e.g. c.c.c. or proper) might be of interest as well. So the following version of the question 1 arises:



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> **Question 3.** Is there any characterization of the statements $\sigma$ for which Alice has a winning strategy in the described game restricted to forcing notions from the class $\Gamma$ where $\Gamma$ is the class of all c.c.c./proper/... forcings?
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Forcing and Family Contentions: Who wins the disputes?
I like this question a lot. It provides an interesting way of talking about
some of the ideas connected with the maximality principle and the
modal logic of forcing.


Let me make several observations.


First, Alice can clearly win, in one move, with any forceably
necessary statement $\sigma$, which is a statement for which
$\newcommand\possible{\Diamond}\newcommand\necessary{\Box}\possible\necessary\sigma$
holds in the modal logic of forcing, meaning that one can force so
as to make $\sigma$ remain true in all further forcing extensions.
She should simply force to make $\necessary\sigma$ true, and then
Bob cannot prevent $\sigma$ in any further extension, including the
limit model. Under the [maximality
principle](http://jdh.hamkins.org/maximalityprinciple/), all such
statements are already true.


Let me point out that there are some subtle issues about
formalization in the question. For example, the strategies here
would be proper class sized objects, and so one must stipulate
whether one is working in ZFC with only definable classes or
whether one has GBC or KM or whatever and whether global choice
holds. Another difficulty concerns the determinacy of the game,
since even [open determinacy for class
games](http://jdh.hamkins.org/open-determinacy-for-class-games/) is
not a theorem of GBC.


Meanwhile, here is something positive to say. I shall consider only the direct-limit version of the game.


**Theorem.** Suppose there is a $\omega$-closed unbounded class of cardinals
$\kappa$ such that the statement $\sigma$ is forced by the collapse
forcing of $\kappa$ to $\omega$. Then Alice has a winning strategy
in the $\sigma$ game.


**Proof.** Let $C$ be the class $\omega$-club of such $\kappa$, and let
Alice simply play always to collapse the next element of $C$ above
the size of the previous forcing played by Bob. It follows that the
limit forcing $\mathbb{P}$ will collapse all the cardinals up to an
element $\kappa\in C$, and since $\kappa$ will have cofinality
$\omega$, it will also collapse $\kappa$ itself. Since the forcing
will also have size $\kappa$, in the direct limit case, it follows
that the forcing is isomorphic to $\text{Coll}(\omega,\kappa)$, and
so $\sigma$ holds in the model $V[G]$, so Alice has won. $\Box$


For example, if the GCH holds, then CH will be such a statement
$\sigma$, even though this is a switch, because the collapse
forcing will collapse $\kappa$ and the CH will hold in $V[G]$, as a
residue of the GCH in $V$. Thus, the GCH implies that Alice can win
the CH game.


A dual analysis is:


**Theorem.** If the class of cardinals $\kappa$ of countable
cofinality for which the collapse forcing
$\text{Coll}(\omega,\kappa)$ forces $\sigma$ is stationary, then
Alice can defeat any strategy of Bob in the $\sigma$ game.


**Proof.** For any strategy for Bob, there is a club of cardinals
$\theta$ such that $V\_\theta$ is closed under the strategy, in the
sense that if Alice plays a poset in $V\_\theta$ then Bob's strategy
will reply with a strategy in $V\_\theta$. So by the stationary
assumption of the theorem, there is a $\kappa$ of countable
cofinality that is closed under the strategy. Alice can now play so
as to collapse more and more of $\kappa$, and Bob will always reply
with a poset below $\kappa$. So the limit forcing will again be the
collapse of $\kappa$, which forces $\sigma$. So Alice can defeat
this strategy.
$\Box$


For example, if the GCH fails on a $\omega$-closed unbounded class of
cardinals (this contradicts SCH), then
Alice can win with $\neg$CH. And if it fails on a stationary class
of such cardinals with countable cofinality, then Bob cannot win the $\neg$CH game.


**Theorem.** From suitable consistency assumptions, it is
consistent with GBC that the CH game is not determined with respect
to class strategies.


**Proof.** Using the Foreman-Woodin theorem that it is relatively
consistent that GCH fails everywhere, we can perform additional
forcing by first adding a generic class of cardinals, and then
forcing certain instances of GCH by collapsing cardinals. The result will be a model of GBC where the class of
cardinals $\kappa$ of cofinality $\omega$ at which the GCH holds is
both stationary and co-stationary. By the theorem above, considered
from either Alice's or Bob's perspective, either player can defeat
any strategy of the other player. So the game is not determined.
$\Box$


I guess the argument isn't just about CH, but rather any statement $\sigma$ such that there is a stationary/co-stationary class of $\kappa$ of countable cofinality such that $\sigma$ holds after the collapse of $\kappa$. In such a case, neither player can have a winning strategy.


The ideas appear to culminate in answer to question 1. 


**Theorem.** The following are equivalent for any statement $\sigma$. 


1. Alice has a winning strategy in the $\sigma$ game.
2. The class of cardinals $\kappa$ of countable cofinality, such that the collapse forcing of $\kappa$ to $\omega$ forces $\sigma$, contains a class $\omega$-club.


**Proof.** Statement 2 implies statement 1 by the first theorem above. Conversely, if statement 2 fails, then there is a stationary class of such $\kappa$ where $\sigma$ fails in the collapse extension. In this case, Bob can defeat any strategy for Alice, by Bob's analogue of the second theorem.
$\Box$ 


