Does this consequence of measurability in terms of games of length $\omega+1$ imply measurability?

For any two structures $\mathcal{M}$ and $\mathcal{N}$ in the same first-order language $\mathcal{L}$ and any ordinal $\theta$, let $G_\theta(\mathcal{M},\mathcal{N})$ be the two-player game of perfect information of length $\theta$ such that in round $\alpha$ player I plays an element $x_\alpha\in \mathcal{M}$ and then player II plays an element $y_\alpha \in \mathcal{N}$, and player II wins if and only if the augmented structures $(\mathcal{M}, x_\alpha)_{\alpha < \theta}$ and $(\mathcal{N}, y_\alpha)_{\alpha < \theta}$ have the same theory in the language obtained from $\mathcal{L}$ by adding $\theta$ constant symbols.

Note that if there is an elementary embedding $j$ of $\mathcal{M}$ into $\mathcal{N}$ then player II has winning strategies in $G_\theta(\mathcal{M},\mathcal{N})$ for all $\theta$ obtained by letting $y_\alpha = j(x_\alpha)$. Conversely, if player II has a winning strategy in $G_\theta(\mathcal{M},\mathcal{N})$ where $\theta$ is the cardinality of $\mathcal{M}$ then we may obtain an elementary embedding of $\mathcal{M}$ into $\mathcal{N}$ by letting $j(x_\alpha) = y_\alpha$ where $(x_\alpha:\alpha < \theta)$ is an enumeration of $\mathcal{M}$. The existence of a winning strategy for player II in the case $\theta = \omega$ is weaker: it is equivalent to the existence of an elementary embedding of $\mathcal{M}$ into $\mathcal{N}$ in every generic extension of $V$ by the poset $\text{Col}(\omega,\mathcal{M})$.

Let us define a cardinal $\kappa$ to be $\theta$-strategically measurable if there is an ordinal $\kappa'$ and a transitive set $N$ such that $\kappa < \kappa' \in N$ and letting $\mathcal{M} = (H_{\kappa^+}; \mathord{\in},\kappa, \xi)_{\xi < \kappa}$ and $\mathcal{N} = (N; \mathord{\in}, \kappa', \xi)_{\xi < \kappa}$, player II has a winning strategy in the game $G_\theta(\mathcal{M},\mathcal{N})$.

Remarks:

1. $2^\kappa$-strategic measurability is equivalent to measurability.

2. If $0^\sharp$ exists then every Silver indiscernible is $\omega$-strategically measurable in $L$. (Use $j:L \to L$ and the absoluteness of existence of winning strategies for closed games of length $\omega$).

3. Every $(\omega+1)$-strategically measurable cardinal is a Ramsey cardinal and a limit of Ramsey cardinals. (Proof given below.)

Questions:

1. Is ($\omega+1$)-strategic measurability equivalent to measurability?
2. If not, what is its consistency strength?
3. Have these games been studied before for $\theta > \omega$? Do they have a name?

Proof of remark 3:

We use a winning strategy for player II in the game $G_{\omega+1}(\mathcal{M},\mathcal{N})$ (or just the nonexistence of a winning strategy for player I) to build increasing sequences $(M_n, n<\omega)$ and $(\mu_n, n<\omega)$ such that $M_n \prec H_{\kappa^+}$, $\left| M_n\right| = \kappa$, $\mu_n$ is an $M_n$-normal ultrafilter on $\mathcal{P}(\kappa)\cap M_n$, and $M_n, \mu_n \in M_{n+1}$. In round $n$ player I plays an enumeration of $\mathcal{P}(\kappa)\cap M_n$ in order type $\kappa$ and uses player II's response to define $\mu_n$. Then letting $M_\omega = \bigcup_{n<\omega}M_n$ and $\mu_\omega = \bigcup_{n<\omega} \mu_n$, we see that $\mu_\omega$ is a weakly amenable $M_\omega$-normal ultrafilter on $\mathcal{P}(\kappa)\cap M_\omega$. Moreover it is countably complete (in the sense of nonempty intersection) because otherwise player I can win by playing a counterexample to countable completeness in round $\omega$. Since we can take $M_0$ to contain any given subset of $\kappa$, it follows that $\kappa$ is a Ramsey cardinal. Because $M_\omega \prec H_{\kappa^+}$ we can reflect Ramseyness below $\kappa$.

(See Gitman, Ramsey-like cardinals for the relationship between Ramsey cardinals and weakly amenable countably complete ultrafilters.)

Further remarks:

1. The proof of Ramseyness outlined above resembles the "filter games" introduced by Holy and Schlicht and further studied by Nielsen and Welch.
2. Defining ($\omega+1$)-strategic strongness in an analogous way, I think I can prove that it is equiconsistent with strongness. (And again the argument only requires the nonexistence of a winning strategy for player I. The previous sentence is wrong; I will add more details later.) However, the proof goes through ${\bf\Sigma}^1_4$ generic absoluteness and does not seem to generalize to other strategic large cardinals.
• Being $\omega+1$-strategically measurable implies that the "nonempty" player has a winning strategy in the Mycielski game (cut and choose). Thus, it is at least equiconsistency to a measurable cardinal (see "The evolution of large cardinals axiom in set theory" by Kanamori and Magidor, section 27). – Yair Hayut Aug 16 '18 at 8:56
• I see! That essentially answers half of the questions. I wonder if one can show it is not equivalent to measurability by a forcing argument. – Trevor Wilson Aug 16 '18 at 9:35
• @YairHayut Do you think you could write out the argument with the Mycielski game as an answer? – Victoria Gitman Aug 16 '18 at 13:18

This answer addresses only the consistency strength of $\omega+1$-strategically measurable.

Claim: If there is a $\omega+1$-strategically measurable cardinal then there is an inner model with a measurable cardinal.

Proof: A winning strategy for $G_{\omega+1}(H(\kappa^{+}), \mathcal{N})$ for some transitive $\mathcal{N}$ as in the definition of $\omega+1$-measurability implies that there is a winning strategy for the Mycielski game (the "cut and choose" game) on $\kappa$.

In the Mycielski game, a sequence of subsets of $\kappa$ $A_0, A_1, \dots, A_n, \dots$ in constructed in the following way. Let us denote $A_{-1} = \kappa$. At step $n < \omega$, player I picks a partition of $A_{n-1}$ into two sets $B_n, C_n$ (so $B_n \cup C_n = A_{n-1}$) and player II chooses one of them to be $A_n$. Player II wins if $|\bigcap_{n < \omega} A_n| > 1$.

Given a winning strategy for the game $G_{\omega+1}(\mathcal{M},\mathcal{N})$, where $\mathcal{M}, \mathcal{N}$ are as in the question, we can construct a winning strategy for the Mycielski game as follows. Player II will maintain an auxiliary play in the game $G_{\omega+1}(\mathcal{M},\mathcal{N})$.

At step $n < \omega$, let $(B_n, C_n)$ be the partition that player I played. Player II define $x_n = (B_n, C_n)$ and use the winning strategy of $G_{\omega+1}(\mathcal{M},\mathcal{N})$ in order to produce some $y_n \in \mathcal{N}$.

Clearly, $y_n = (B_n^\star, C_n^\star)$. Now, player II will pick $A_n$ to be $B_n$ iff $\kappa \in B_n^\star$ (and then $A_n^\star = B_n^\star$. Otherwise, $A_n^\star = C_n^\star$). We want to make sure that player II wins.

Indeed, in the constructed play in the game $G_{\omega+1}(\mathcal{M},\mathcal{N})$, we obtained the sequence of steps $\langle x_n, y_n \mid n < \omega\rangle$. Player II can ask what the winning strategy will answer to $x_{\omega} = \langle A_n \mid n < \omega\rangle$. Let $y_\omega$ be the answer. Then, by elementarity $y_n = \langle A_n^\star \mid n < \omega\rangle$. Now, since $\kappa \in \bigcap A_n^\star$, $\bigcap A_n$ cannot be bounded in $\kappa$.

It is known that the existence of a winning strategy for player II in the Mycielski game is equiconsistent with a measurable cardinal (see "The evolution of large cardinals axiom in set theory" by Kanamori and Magidor, section 27. The argument is attributed there to Silver and Solovay). QED

Let me remark that it is consistent to have a winning strategy for the Mycielski game at accessible cardinals. For example, it is consistent there is a winning strategy for the Mycielski game on $\omega_2$.

• Great, Yair! Thank you. – Victoria Gitman Aug 16 '18 at 14:04
• Yair, there's a detail in the proof of the Silver-Solovay fact you mentioned that I don't quite understand. Do you know why the ultrapower in their proof is wellfounded, as they're only considering functions which are elements of their set $A$? And if their ultrapower is really just $\text{Ult}(A,U)$ then I don't see why that should give a generic embedding $j:V\to M$. This is on page 249 in the book you referenced. – Dan Saattrup Nielsen Sep 17 '18 at 12:57
• Assume that the ultrapower is ill-founded. Let $\dot{f} = \langle \dot{f}_n \mid n < \omega\rangle$ be a name for an infinite decreasing sequence of ordinals. Let $M$ be a countable elementary sub-model of $H(\theta)$ for large $\theta$ containing the name $\dot{f}$. Let $H$ be an $M$-generic filter and let $U$ be the corresponding $M$-ultrafilter. Note that $B_n = \{\alpha \in A\mid f_{n+1}(\alpha) \in f_n(\alpha)\} \in U$ and by the definition of $U$, $B = \bigcap B_n \neq \emptyset$. Let $x\in B$. Then $f_n(x)$ is an infinite decreasing sequence of ordinals - a contradiction. – Yair Hayut Sep 17 '18 at 19:27
• I wrote this in a slightly confusing way, sorry. The $B_n$ are defined inside $M[H]$ and not $V[G]$ (the sequence of $f_n$ at this point of the argument is the interpretation of $M[H]$ to the name $\dot{f}$, and not the sequence of functions in $V[G]$). In particular, since $M, H\in V$ also the sequence $\langle B_n \mid n < \omega\rangle \in V$ and thus has non empty intersection in $V$. – Yair Hayut Sep 20 '18 at 7:11
• Yes. I think that they mean that they construct the ultrapower using all functions $f \colon A \to V$ from $V$ and not $V[G]$. – Yair Hayut Sep 20 '18 at 11:00