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It is a well-known result by Woodin that the Hartogs number $h(\mathbb{R})$ (more commonly known as $\Theta$) is a Woodin cardinal (in HOD) assuming ZF + AD + DC. This is equivalent to $h(\mathcal{P}(\aleph_0))$. However, I was wondering the simple question of this very generalization; specifically the following question:

$$\exists\kappa\exists\alpha\in\mathrm{Ord}(\aleph_\alpha\not\leq^*|\mathcal{P}(\kappa)|)?$$

If $\kappa$ satisfies the inner quantification of this formula, one could say $\kappa$ satisfies the Hartogs property (or $\mathrm{H}(\kappa)$).

In, ZF + AD + DC what properties does $\mathrm{H}(\aleph_1)$ have? What about even larger ordinals? What about in HOD, or even assuming $V=\mathrm{HOD}?$ (The original question was quickly shown true)

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    $\begingroup$ (And the answer is of course yes. ) $\endgroup$ Commented Oct 17, 2017 at 1:04
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    $\begingroup$ $|A|\le|B| $ means that $A $ injects into $B $. $|A|\le^*|B|$ means that either $A $ is empty or else $B $ surjects onto $A $, but this notation is less common than the one for injections so you should probably remind the reader of its meaning the first time it shows up. $\endgroup$ Commented Oct 17, 2017 at 1:08
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    $\begingroup$ ZF+AD+AC implies that the empty set is not empty. $\endgroup$
    – Asaf Karagila
    Commented Oct 17, 2017 at 12:33
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    $\begingroup$ More generally, AD seems like a red herring to this entire question. And it seems to me that the question is made by poorly understanding the structure of the universe of set theory without choice. $\endgroup$
    – Asaf Karagila
    Commented Oct 17, 2017 at 12:34
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    $\begingroup$ The notation of this question is a total mess. You're overloading $H(\kappa)$, which is often used for sets whose transitive closure has size $<\kappa$; you're still using the name Hartogs property, even though it is not the right name; and what does it even mean in the first displayed formula that $\kappa$ has an existential quantifier? Then you move to HOD which is a model of choice, making the whole thing even more confused. $\endgroup$
    – Asaf Karagila
    Commented Oct 18, 2017 at 6:42

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This is a bit confused, but I think what you're asking is:

Let $S(\alpha)$ be the supremum of all the ordinals onto which $\mathcal{P}(\alpha)$ surjects. What can we say about e.g. $S(\omega_1)$?

(And your original question was whether $S(\alpha)$ always exists, which it does.) If this is your question, I think it's a good one (if it's not then you need to clarify what your question is), and the key fact is Moschovakis' coding lemma (see around page 397 in Kanamori's book):

(ZF+AD) Suppose there is a surjection from $\mathcal{P}(\omega)$ to $\alpha$. Then there is a surejction from $\mathcal{P}(\omega)$ to $\mathcal{P}(\alpha)$.

(There are also non-AD versions of the coding lemma.) This gives us:

Corollary (ZF+AD): Suppose $S(\omega)>\alpha$ (so $\mathcal{P}(\omega)$ surjects onto $\alpha$). Then $S(\omega)=S(\alpha)$.

That is: for $\alpha<\Theta$, $S(\alpha)=\Theta$. In particular, $S(\omega_1)=\Theta$ and $\Theta$ is a limit ordinal, since the powerset of a cardinal always surjects onto that cardinal's successor.

Proof: Since $\mathcal{P}(\omega)$ surjects onto $\alpha$, by the coding lemma $\mathcal{P}(\omega)$ surjects onto $\mathcal{P}(\alpha)$. But if $\beta<S(\alpha)$ this means that $\mathcal{P}(\alpha)$ surjects onto $\beta$, so by composing surjections we get that $\mathcal{P}(\omega)$ surjects onto $\beta$, so $\beta<S(\omega)$. (And conversely clearly $S(\omega)\le S(\alpha)$.)


Having calculated $S(\alpha)$ up to $\Theta$, it's worth saying a bit at this point about $\Theta$ itself. The coding lemma shows that $\Theta$ is a limit, but this isn't the end of the story - lots more can be said. For example, Solovay showed that $\Theta=\omega_\Theta$ and that $\Theta$ has uncountable cofinality. Additional hypotheses give you further "largeness" properties of $\Theta$, and in particular "$\Theta$ is regular" plays an important role in inner model theory (see e.g. this paper of Sargsyan).

Note that it is quite possible that $\Theta$ is not regular - the regularity of $\Theta$ is a strong assumption, especially in conjunction with stronger determinacy principles (e.g. AD$_\mathbb{R}$). In fact, $\Theta$ can have cofinality $\omega$. Interestingly, this cannot occur if $V=L(\mathbb{R})$ - Solovay showed that in $ZF+AD+V=L(\mathbb{R})$, $\Theta$ is regular.

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    $\begingroup$ Noah, you should indicate that the cofinality of $\Theta $ can be countable, just not in $L (\mathbb R) $. $\endgroup$ Commented Oct 18, 2017 at 16:05
  • $\begingroup$ @AndrésE.Caicedo Sure, I'll add a bit at the end. $\endgroup$ Commented Oct 18, 2017 at 16:06

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