What does the partially ordered class of cardinals look like in L(R)? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T01:06:54Z http://mathoverflow.net/feeds/question/46863 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/46863/what-does-the-partially-ordered-class-of-cardinals-look-like-in-lr What does the partially ordered class of cardinals look like in L(R)? Ricky Demer 2010-11-21T21:18:43Z 2010-11-21T22:20:58Z <p>Assuming the existence of enough large cardinals (I'm not sure whether I mean in the original V or in L(R), do whatever is standard), is the partially ordered class of cardinals order-isomorphic to something simpler? <br> If so, what is the weakest large cardinal assumption that gives this result? Is L(R) ≠ L sufficient?</p> <p>My best guess would be that it is order-isomorphic to $\omega \cup (\operatorname{Ord} \times \operatorname{Ord})$, with all elements of $\omega$ being smaller than all elements of $\operatorname{Ord} \times \operatorname{Ord}$, and $\operatorname{Ord} \times \operatorname{Ord}$ having the product partial order.</p> http://mathoverflow.net/questions/46863/what-does-the-partially-ordered-class-of-cardinals-look-like-in-lr/46868#46868 Answer by Andres Caicedo for What does the partially ordered class of cardinals look like in L(R)? Andres Caicedo 2010-11-21T22:02:12Z 2010-11-21T22:20:58Z <p>Ricky:</p> <p>I assume you mean to ask your question in $L({\mathbb R})$. </p> <p>In general, without choice, the ordering of cardinals tends to be rather pathological, although we do not yet know by how much. Here is an example: It is open whether ZF proves that, if there is no infinite set all of whose members are pairwise incomparable (cardinality-wise) then choice holds. On the other hand, if choice fails, then for every finite $n$ there are $n$ pairwise incomparable sets.</p> <p>Assuming enough large cardinals (in $V$ or, equivalently, determinacy in $L({\mathbb R})$), the ordering of cardinals in $L({\mathbb R})$ is <em>much more complicated</em> than you suggest. </p> <p>To give you an idea of how little we know: We do not know yet whether there are infinitely many successors of $|{\mathbb R}|$.</p> <p>For an example of the immense complexity present in the ordering, recall that $E_0$ is the equivalence relation on $2^{\mathbb N}$ given by $x E_0 y$ iff $\exists n\forall m\ge n (x(m)=y(m))$. Then: </p> <blockquote> <p>$|2^{\mathbb N}/E_0|$ is a successor of $|{\mathbb R}|$, and above it but still below $|{\mathcal P}({\mathbb R})|$, you can embed the partial order of Borel subsets of ${\mathbb R}$ under containment. </p> </blockquote> <p>In fact, you can realize these cardinals by taking suitable quotients of the reals by Borel equivalence relations. All these relations can be taken to be <em>countable</em> (i.e., each class has countably many members) and their definitions trace back to free measure preserving actions of $F_2$ (the free group in two generators) on Polish spaces.</p> <p>I take this as a strong indication that there is no sense in which we may have a "reasonable" description of the whole partial ordering. But really, we are far from being able to say much. Ketchersid and I have some recent results at the very bottom of the ordering, see "A trichotomy theorem in natural models of $AD^+$," to appear in the Proceedings of Boise Extravaganza in Set Theory, and also our forthcoming paper on "$G_0$-dichotomies in natural models of $AD^+$." You may also want to take a look at a paper by Woodin that deals with a slightly more demanding version of determinacy, "The cardinals below $|[\omega_1]^{\lt\omega_1}|$," Annals of Pure and Applied Logic 140 (2006) 161–232.</p> <p>Even if we restrict ourselves to better understood classes (specific collections of Borel sets), the ordering is rather elaborate and far from well understood. For a nice subclass for which there <em>is</em> a very decent picture, see Alex Andretta, Greg Hjorth, and Itay Neeman, "Effective cardinals of boldface pointclasses," J. of Mathematical Logic, vol. 7 (2007) 35–82.</p> <hr> <p>I didn't address this above, but I should probably say something: </p> <p>The assumption that there are enough large cardinals ensures that the theory of $L({\mathbb R})$ is invariant under forcing, so it is in a sense as canonical as we can hope. </p> <p>Under weaker assumptions, the partial ordering may vary greatly. For example, $L({\mathbb R})\ne L$ does not suffice to preclude choice in $L({\mathbb R})$. </p> <p>Assuming you do not have choice, it is open (even for $L({\mathbb R})$) whether well-foundedness of the partial ordering of cardinalities must fail. This is believed to be the case, and it is certainly so in all reasonable cases I have checked. It is easy to give examples by forcing over $L$ where the $L({\mathbb R})$ of one extension and the $L({\mathbb R})$ of another have non-equivalent partial orderings of cardinalities. (For example, by replicating or excluding the behavior mentioned above of quotients by free actions.) </p> <p>Sometimes we have some form of "control", for example, if $L({\mathbb R})$ is a kind of Solovay model. But it already takes effort to show that in "nice" situations there are no infinite Dedekind finite subsets of ${\mathbb R}$ in $L({\mathbb R})$. In my view, the "right" version of these questions is under large cardinals, so we have canonicity, but already without it there are many difficulties and possibilities that may be interesting to explore. </p>