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Let $M$ be an inner model (of height $\mathsf{Ord}$) containing all the reals. I am wondering about the consistency strength of the statement "Every game in $M$ is determined in $V$."

MOTIVATION

For the statement of AD, games are restricted to be played on natural numbers, terminating after $\omega$ plays. In that case payoffs correspond to subsets of $\omega^\omega$.

There are various ways of generalizing this; for example, given $X$ and $A \subset X^\omega$, one can define the game $G_X(A)$ where each player takes turns choosing from $X$, and where the payoff is given by $A$.

With that definition of "game," it is not possible that every game in $M$ is determined in $M$. Indeed, suppose $AD^M$ holds: I define a nondetermined game played on $\omega_1$ (following Kanamori).

On the first move, player I picks some $\alpha < \omega_1$. On the subsequent moves, player II picks elements of $\{0, 1\}$. Player II wins iff his/her sequence of bits encodes $\alpha$.

Letting $A$ be the payoff corresponding to the above, then $G_{\omega_1}(A)$ cannot be determined in $M$. But it is certainly determined in $V$!

QUESTION

So, to reiterate, I am curious about the hypothesis "Every game in $M$ is determined in $V$." Note that this encompasses $AD^M$ since $M$ contains every real. There are a couple subquestions:

(a) Is this hypothesis outright inconsistent?

(b) Is this hypothesis equivalent to $AD^M$, or to, say, $AD_{R}^M$?

(c) Does this hypothesis depend on the precise generalization of games I use? (Another generalization would be to allow ordinal-length games, for example.)

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  • $\begingroup$ Your statement "every game in $M$ is determined in $V$" has the unspecified parameter $M$. When you ask for the consistency strength of the statement, do you have a specific $M$ in mind? Or perhaps you are actually asking instead about the consistency strength of the statement "$\exists M$ such that every game in $M$ is determined in $V$"? $\endgroup$ – Joel David Hamkins Apr 3 '13 at 3:06
  • $\begingroup$ I was thinking of it as a separate question for each inner model $M$; is it possible to formalize the latter? $\endgroup$ – Douglas Ulrich Apr 3 '13 at 11:20
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Two observations.

First, I note that your hypothesis implies projective determinacy, since $M$ and $V$ have exactly the same projective sets, and if they have winning strategies in $V$ then those strategies are also winning in $M$.

Second, I observe that for any set $X$ for which $V$ has an $\omega$ sequence from $X$ that is not in $M$, then player II can in $V$ win any game on $X$ whose payoff set for player I is in $M$. The idea is simply that player II can play that sequence on his or her own moves, and since this sequence is not in $M$, the resulting play will not even be in $M$ and consequently will not be in the payoff set; so player II will win the play. It follows that any game in $M$ on any set larger than $X$ will also be determined in $V$ for the same reason.

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    $\begingroup$ It is an interesting question. But since my answer isn't really a full answer, perhaps you've accepted it prematurely? We could wait and see if someone comes in with a fuller account... $\endgroup$ – Joel David Hamkins Apr 3 '13 at 12:40
  • $\begingroup$ I confused myself very badly. I am going to think for a bit. . . $\endgroup$ – Douglas Ulrich Apr 5 '13 at 4:38
  • $\begingroup$ I hope you do not get an email for each of my comments. (This is my 8th or so.) You almost answered the question: Suppose $AD^M$. Let $X \in M$. If $X^\omega \subset M$ then every game on $X$ is determined in $V$ via a coding of the winning strategy; else the same holds by your argument. So $AD^M$ is equivalent. $\endgroup$ – Douglas Ulrich Apr 5 '13 at 5:14
  • $\begingroup$ No, I don't get email comment notifications. In your comment, you seem to be assuming more than just $\text{AD}^M$, since you want games played on $X$ to be determined. When $X$ is uncountable, then $\text{AD}_X$ can be much stronger than mere $\text{AD}$. For example, set theorists often study $\text{AD}_{\mathbb{R}}$, where the players play an entire real number on each turn. So I'm not quite sure about a full answer to the question... $\endgroup$ – Joel David Hamkins Apr 5 '13 at 19:10
  • $\begingroup$ Right. I'm finding it tricky to keep everything straight, sorry. $\endgroup$ – Douglas Ulrich Apr 5 '13 at 20:50

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