Skip to main content
MathOverflow

Return to Answer

Added a small remark about why we assume large cardinals
Source Link
Andrés E. Caicedo
Andrés E. Caicedo
  • 32.5k
  • 5
  • 133
  • 240

I didn't address this above, but I should probably say something:

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.

Under weaker assumptions, the partial ordering may vary greatly. For example, $L({\mathbb R})\ne L$ does not suffice to preclude choice in $L({\mathbb R})$.

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.)

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.

I didn't address this above, but I should probably say something:

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.

Under weaker assumptions, the partial ordering may vary greatly. For example, $L({\mathbb R})\ne L$ does not suffice to preclude choice in $L({\mathbb R})$.

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.)

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.

Source Link
Andrés E. Caicedo
Andrés E. Caicedo
  • 32.5k
  • 5
  • 133
  • 240

Ricky:

I assume you mean to ask your question in $L({\mathbb R})$.

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.

Assuming enough large cardinals (in $V$ or, equivalently, determinacy in $L({\mathbb R})$), the ordering of cardinals in $L({\mathbb R})$ is much more complicated than you suggest.

To give you an idea of how little we know: We do not know yet whether there are infinitely many successors of $|{\mathbb R}|$.

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:

$|2^{\mathbb N}/E_0|$ 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.

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 countable (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.

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.

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 is 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.