Does ZFC prove existence of proper class many $H_\kappa$ sets with $|H_\kappa|=\kappa$?

Question

This question is a follow up to a prior question

Let $H_\kappa$ be the set of all sets hereditarily strictly smaller than $\kappa$, where $\kappa$ is cardinal.

Is it a theorem of $\sf ZFC$ that we have a proper class of sets $H_\kappa$ such that $|H_\kappa|=\kappa$?

Answer 1

Suppose that $H_\kappa$ has size $\kappa$. First, note that $\kappa$ is regular, otherwise $[\kappa]^{\operatorname{cf}\kappa}$ is a subset of $H_\kappa$, and by König's lemma that would have a strictly larger size.

Next, note that if $\kappa=\mu^+$, then this means that $2^\mu=\kappa$.

So, either $\sf GCH$ holds on a proper class or there is a proper class of inaccessible cardinals.

Since $\sf ZFC$ does not prove either claim, the answer is no.

Answer 2

Under the Axiom of Choice, one can also explicitly compute the size of $H(\kappa)$ to be $2^{<\kappa}$, so that this question amounts to asking whether there is a proper class of cardinals with $\kappa=2^{<\kappa}$ (which Asaf answered already).

$H(\kappa)$ has size at least $2^{<\kappa}$ because the power set of each $\lambda<\kappa$ has transitive closure of size $<\kappa$, so each $\mathcal{P}(\lambda)$ is in $H(\kappa)$.

To see that is has size at most $2^{<\kappa}$, recall that each set can be uniquely coded as a relation on its cardinal (AC used here), and by Gödel's pairing, as a subset of its cardinal. So if $x\in H(\kappa)$, then one can find an ordinal $\lambda<\kappa$ such that $x$ is coded as a subset of $\lambda$. This injects $H(\kappa)$ into $2^{<\kappa}$.