# $\Theta$ and the Hartogs of $2^\mathbb R$

Let $$a,b$$ be sets, we write $$a\leq^\ast b$$ if either $$a=\varnothing$$ or there exists a surjection $$f\colon b\to a$$. With the axiom of choice this is a linear ordering equivalent to the usual ordering of cardinals $$a\leq b$$ if there is an injection from $$a$$ into $$b$$).

Without the axiom of choice this order need not be linear, or even a partial order. Indeed there are models where $$a < b$$ (so trivially $$a\leq^\ast b$$) but also $$b\leq^\ast a$$.

Let $$\Theta(a)$$ denote the least ordinal $$\alpha$$ such that $$\alpha\nleq^\ast a$$. This is a dual notion to $$\aleph(a)$$ which is the least ordinal $$\alpha$$ such that $$\alpha\nleq a$$. It is not very hard to see that both these ordinals are in fact cardinals, and that they exist for every set in $$\mathrm{ZF}$$. In the case where $$a=\mathbb R$$ we simply write $$\Theta$$ for $$\Theta(\mathbb R)$$.

It is consistent that for some set $$a$$ it holds $$\aleph(a)<\Theta(a)$$. For example if $$V$$ is a Solovay model then $$\aleph(\mathbb R)=\aleph_1\leq^\ast\mathbb R$$.

We also know that if $$a\leq^\ast b$$ then $$a\leq 2^b$$, where the injection is simply the sending a point in $$a$$ to its fiber under a fixed surjection from $$b$$. It follows, if so, that $$\Theta(a)\leq\aleph(2^a)$$. Even in $$\mathrm{ZFC}$$ one can see that it is consistent that there is an equality, and it is consistent that there is no equality (e.g. $$a=\omega$$ and take a model in which $$\frak c=\aleph_1$$ and another where it is is $$\aleph_2$$).

Question: Suppose that $$V$$ is a model of $$\mathrm{ZF+AD}$$, we know that if $$\alpha\leq^\ast\mathbb R$$ then $$2^\alpha\leq^\ast\mathbb R$$, and therefore $$\alpha^+\leq^\ast\mathbb R$$. Is it consistent that $$\Theta<\aleph(2^\mathbb R)$$?

Equivalently, suppose $$\alpha<2^\mathbb R$$, can we find a surjection from $$\mathbb R$$ onto $$\alpha$$?

## 1 Answer

Yes, it is consistent with AD. In $L(\mathbb{R})$, if AD holds then $\Theta$ injects into $\mathcal{P}(\mathbb{R})$, or equivalently, into $2^{\mathbb{R}}$. The OD sets of reals are Wadge-cofinal and $\Theta$ is regular, so we can define an injection $F : \Theta \to \mathcal{P}(\mathbb{R})$ by induction. Let $F(\alpha)$ be the $<_{OD}$-least OD set of reals with Wadge rank greater than the Wadge ranks of all the sets of reals $F(\gamma)$, $\gamma < \alpha$.

Going further, we can determine the Hartogs number $\aleph(2^{\mathbb{R}})$ in all models of $AD + V=L(\mathcal{P}(\mathbb{R}))$. The proof splits into two cases corresponding to the length of the Solovay sequence being a successor ordinal or a limit ordinal.

If the length of the Solovay sequence is a successor ordinal, this means that there is a set of reals $A$ such that every set of reals is OD from $A$ and a real. In this case $\aleph(2^{\mathbb{R}})$ has the largest possible value, namely $\Theta(2^{\mathbb{R}})$. Given a surjection $F : \mathcal{P}(\mathbb{R}) \to Z$ for any set $Z$ we can construct an injection $Z \to \mathcal{P}(\mathbb{R})$. Let $G(z)$ be the set of pairs $(x,y)$ of reals such that $y$ is in the least $OD_{A,x}$ set of reals $B$ with $F(B) = z$ (if it exists.)

On the other hand, if the length of the Solovay sequence is a limit ordinal then $\aleph(2^{\mathbb{R}})$ is equal to $\Theta$; that is, to $\Theta(\mathbb{R})$. We have $\aleph(2^{\mathbb{R}}) \ge \Theta$ on general grounds like you said in the question. But any function $F : \Theta \to \mathcal{P}(\mathbb{R})$ is OD from a set $A$ of reals because $V=L(\mathcal{P}(\mathbb{R}))$, so the range of $F$ consists of $OD_A$ sets of reals. The Solovay sequence has limit length, so the range of $F$ cannot be all of $\mathcal{P}(\mathbb{R})$.

• Hmm. Thanks. I was hoping for such answer. Is there a known bound on $\aleph(2^\mathbb R)$? Also, this answer is for models of regular $\Theta$, what happens when we have a singular $\Theta$? – Asaf Karagila Jul 25 '12 at 20:15
• I don't know about bounds on $\aleph(2^{\mathbb{R}})$, but the updated answer now shows that if $\Theta$ injects into $\mathcal{P}(\mathbb{R})$ and $V = L(\mathcal{P}(\mathbb{R}))$ then $\Theta$ is regular. – Trevor Wilson Jul 25 '12 at 20:27
• So the atypical case is actually $\Theta=\aleph(2^\mathbb R)$? I have to admit that I am somewhat surprised... – Asaf Karagila Jul 25 '12 at 20:48
• I wouldn't necessarily say the case where the Solovay sequence has limit length is atypical. By the way, I've updated the answer to include a calcuation of the Hartogs number for all cases except when the Solovay sequence has limit length and $V \ne L(\mathcal{P}(\mathbb{R}))$. – Trevor Wilson Jul 25 '12 at 21:24