Today it is known that $AD$ (the axiom of determinacy of games played with integers) is true in $L(\mathbb{R})$. Has it been proven that this is the only model in which $AD$ is true? Have other models been identified in which $AD$ is true? Of course, I am asking about genuine models, since we can force over $L(\mathbb{R})$ and still keep enough $AD$. A related question is the following: how different from $L(\mathbb{R})$ is the universe $V$? Thx.
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asked Nov 20, 2011 at 21:22
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1$\begingroup$ There are many models where AD holds, but of course this depends on what background assumptions you allow. You can get models of the form $L(\Gamma,{\mathbb R})$ where $\Gamma$ is a collection of sets of reals, for example, and the larger $\Gamma$ is, the more interesting the model you obtain. $\endgroup$– Andrés E. CaicedoCommented Nov 20, 2011 at 22:02
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4$\begingroup$ I have to say that I am a bit puzzled by the statement "today it is known that AD is true in $L(\mathbb R)$". I understand what you mean, under large cardinal assumptions AD is true in $L(\mathbb R)$, but most people, no matter how close to the Berkeley school of set theory, usually feel the need to mention that large cardinal assumptions are necessary here. $\endgroup$– Stefan GeschkeCommented Nov 21, 2011 at 13:41
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2 Answers
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I'm not sure what you mean by "genuine models", but let me comment on how different $L(\mathbb R)$ is from $V$. They look very different to me. Partly this is because the axiom of choice holds in $V$ and fails rather spectacularly in $L(\mathbb R)$. For example, AD implies that $\aleph_n$ is singular whenever $3\leq n\leq\omega$, so the cardinal structure of $L(\mathbb R)$ looks very different from that of $V$. Even where they agree, for example at $\aleph_1$ (which is the same in $L(\mathbb R)$ as in $V$), there's a big difference as to what subsets are present. AD implies that the club filter on $\aleph_1$ is an ultrafilter, so all of $V$'s stationary co-stationary subsets of $\aleph_1$ are missing from $L(\mathbb R)$.
A more philosophical (by which I mean imprecise and not mathematical) reason to think $L(\mathbb R)$ differs greatly from $V$ is that it seems entirely implausible to me that the whole universe should be constructible from any single set. I expect to see more and more complexity the higher up I go in the cumulative hierarchy --- and not just complexity of ordinals.
answered Nov 20, 2011 at 21:35
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$\begingroup$ Thx for your answer. I should have made my second question more precise, sorry about that. What I wanted to mean is what difference is there in terms of truth. For instance do they agree for $\Sigma_3$ or $\Pi_2$ statements? But still I am more interested in knowing if there is some other model in which $AD$ is true. By genuine I mean, not a model constructed by forcing over $L(\mathbb{R})$ while keeping $AD$. $\endgroup$ Commented Nov 20, 2011 at 21:49
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To answer the first question, like Andres mentioned, larger models $L(\Gamma,\mathbb{R})$ of AD can behave quite differently from $L(\mathbb{R})$. For example they can satisfy AD$_{\mathbb{R}}$, the axiom of determinacy for Gale-Stewart games played on $\mathbb{R}$, which fails in $L(\mathbb{R})$. (This is because it implies the Axiom of Uniformization, i.e., that every binary relation on $\mathbb{R}$ contains a function with the same domain, whereas in a model such as $L(\mathbb{R})$ where every set is ordinal-definable from a real, the set of pairs $(x,y)$ such that $y$ is not ordinal-definable from $x$ cannot be uniformized.)
To add to Andreas's answer to the second question, there is a $\Sigma_1$ statement in the parameter $\mathbb{R}$ that is true under ZFC but false under AD, namely the existence of an injection $\omega_1 \to \mathbb{R}$. (This is easily seen to be inconsistent with a countably complete nonprincipal ultrafilter on $\omega_1$.)
