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, 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 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 $\sigma$.
- Alice has a winning strategy in the $\sigma$ game.
- 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.
$\Box$
(Working in ZFC.)
No (re $\mathcal{L}_{\omega_1,\omega}$). Suppose it is. Consider the signature $\Sigma$ with just one binary relation symbol $<$. Let $\Sigma',\eta$ witness SED-ness for $\Sigma$.
Let $\pi:M\to V_\theta$ be elementary,with $\theta$ some sufficiently large limit ordinal, $M$ transitive, $M$ of cardinality $\kappa=2^{\aleph_0}$, with $\mathbb{R},2^{\aleph_0}\subseteq M$. So $\mathrm{crit}(\pi)=\kappa^{+M}<\kappa^+$ and $\pi(\kappa^{+M})=\kappa^+$. Let $\mathfrak{A}=(\kappa^{+M},{\in}\upharpoonright\kappa^{+M})$ and $\mathfrak{B}=(\kappa^+,{\in}\upharpoonright\kappa^+)$. Note that $\mathfrak{A}\equiv_{\omega_1,\omega}\mathfrak{B}$, since $\pi$ is elementary and all the sentences in $\mathcal{L}_{\omega_1,\omega}$ are in $M$, since $\mathbb{R}\subseteq M$. So by hypothesis, there is some $\Sigma'$-structure $\mathfrak{G}$ such that $\mathfrak{G}\models\eta$, $A^{\mathfrak{G}}\upharpoonright\Sigma\cong\mathfrak{A}$ and $B^{\mathfrak{G}}\upharpoonright\Sigma\cong\mathfrak{B}$.
Now let $G$ be $V$-generic for $\mathrm{Coll}(\omega,\max(\kappa^{+},|\mathfrak{G}|))$.
In $V[G]$, there is an $\mathcal{L}_{\omega_1,\omega}$ sentence $\varphi$ in the signature $\Sigma$ asserting that the model is (isomorphic to) $\kappa^{+M}$. So in $V[G]$, $\mathfrak{A}\models\varphi$ but $\mathfrak{B}\models\neg\varphi$. So $V[G]$ models the statement "there are countable structures $\mathfrak{A}_0,\mathfrak{B}_0$ in the signature $\Sigma$ and an $\mathcal{L}_{\omega_1,\omega}$-sentence $\varphi_0$ such that $\mathfrak{A}_0\models\varphi_0$ and $\mathfrak{B}_0\models\neg\varphi_0$
and there is a countable structure $\mathfrak{G}_0$ in the signature $\Sigma'$ such that $\mathfrak{G}_0\models\eta$ and $A^{\mathfrak{G}}\upharpoonright\Sigma\cong\mathfrak{A}_0$ and $B^{\mathfrak{G}}\upharpoonright\Sigma\cong\mathfrak{B}_0$". But this statement is $\Sigma^1_2$ in a real coding $\eta$. (I don't see that it is $\Sigma^1_1$ in such a real, because to assert that $\varphi_0$ is really a sentence of $\mathcal{L}_{\omega_1,\omega}$ involves saying that it is built along a real ordinal.) Since $\eta\in V$ and by Shoenfield absoluteness, $V$ models the same statement. But this contradicts our assumptions.
Best Answer
These games have many uses. I think they're a lot of fun, but proofs that Duplicator has a winning strategy tend to get tedious quickly. Several applications are given (either in the body of the text or as exercises) in Elements of Finite Model Theory by Libkin. One application of E-F games that I like: to show that in first-order logic with equality and a single unary relation symbol R there is no sentence such that for all structures A the sentence holds in A iff the cardinality of the interpretation of R in A is even. Lots of other "cardinality properties" can be similarly shown with E-F games to be undefinable in FOL, such as FOL's inability to capture that the cardinality of the interpretation of two unary relations R and S are equal (in the finite case, without regard for the infinite case). In my Ph.D. dissertation I applied E-F games to a fun little example in the same spirit: given a first-order signature with three unary relation symbols V, E, and F, one cannot give a formula that captures the class of structures A for which V^A - E^A + F^A = 2. That is to say, in a suitable sense Euler's polyhedron formula cannot be captured in FOL.