# Does $H\vDash AC$

The set $$H_\kappa$$ of sets hereditarily of cardinality less than $$\kappa$$ is defined as $$H_\kappa=\{x||tc(x)|\lt\kappa\}$$. What if we define the set $$H=H_{Ord}$$ of sets hereditarily of cardinality less than $$Ord$$; $$H$$ is the class of sets with some ordinal number as there cardinality. Equivalently, $$H$$ is the class of hereditarily well-ordered sets.

It is immediately obvious $$V=H$$ is equivalent to $$AC$$. My questions are as follows:

1. Is it true that $$ZF\vdash H\vDash ZF$$? Is it true that $$ZF\vdash H\vDash ZFC$$?

2. More generally, does $$ZF\vdash H_\kappa\vDash AC$$?

3. What is the consistency strength of the existence of some $$j: H\prec H$$? Is the existence of such an embedding first order expressible?

(I am obviously working in the context of $$ZF$$ without choice. I define $$|tc(x)|\lt\kappa\leftrightarrow\exists\lambda\lt\kappa(|tc(x)|=\lambda)$$)

$$H$$ (and $$H_\kappa$$ in general) always satisfies choice, because any family of nonempty sets in $$H$$ has a well orderable transitive closure, from whence we can define a choice function, which is easily hereditarily well orderable as well.
However, $$H\models\sf Power$$ if and only if choice holds on $$V$$. Otherwise, let $$\alpha$$ be the least ordinal with a power set that cannot be well ordered, then $$\mathcal P(\alpha)\subseteq H$$ but it is not an element of $$H$$.
• Do you have an idea on the consistency strength of some $j: H\prec H$? Commented Dec 17, 2020 at 19:28
• It's also worth mentioning replacement can fail in $H_{\kappa}.$ For example, in the Feferman-Levy extension of $L,$ $H_{\omega_1}$ fails replacement since the $\omega_n^L$'s are definable in $H_{\omega_1}.$ Commented Dec 17, 2020 at 20:18
• Even if it is determined $HOD$ is definable in $H$, what is the consistency strength of the existence of some $j: HOD\prec HOD$? Commented Dec 17, 2020 at 20:48