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 $|tc(x)|\lt\kappa\leftrightarrow\exists\lambda\lt\kappa(|tc(x)|=\lambda)$)



Does $H\vDash AC$
No, yes, and not sure.


$H$ (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 $H$.


