Models of $AD$ different from $L(\mathbb{R})$ - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T17:13:11Z http://mathoverflow.net/feeds/question/81455 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/81455/models-of-ad-different-from-l-mathbbr Models of $AD$ different from $L(\mathbb{R})$ alephomega 2011-11-20T21:22:14Z 2011-12-02T07:18:33Z <p>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.</p> http://mathoverflow.net/questions/81455/models-of-ad-different-from-l-mathbbr/81457#81457 Answer by Andreas Blass for Models of $AD$ different from $L(\mathbb{R})$ Andreas Blass 2011-11-20T21:35:38Z 2011-11-20T21:35:38Z <p>I'm not sure what you mean by "genuine models", but let me comment on how different <code>$L(\mathbb R)$</code> 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 <code>$L(\mathbb R)$</code>. For example, AD implies that <code>$\aleph_n$</code> is singular whenever $3\leq n\leq\omega$, so the cardinal structure of <code>$L(\mathbb R)$</code> looks very different from that of $V$. Even where they agree, for example at <code>$\aleph_1$</code> (which is the same in <code>$L(\mathbb R)$</code> as in $V$), there's a big difference as to what subsets are present. AD implies that the club filter on <code>$\aleph_1$</code> is an ultrafilter, so all of $V$'s stationary co-stationary subsets of <code>$\aleph_1$</code> are missing from <code>$L(\mathbb R)$</code>.</p> <p>A more philosophical (by which I mean imprecise and not mathematical) reason to think <code>$L(\mathbb R)$</code> 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.</p> http://mathoverflow.net/questions/81455/models-of-ad-different-from-l-mathbbr/82434#82434 Answer by Trevor Wilson for Models of $AD$ different from $L(\mathbb{R})$ Trevor Wilson 2011-12-02T07:18:33Z 2011-12-02T07:18:33Z <p>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.)</p> <p>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 $\omega_1$.)</p>