I have some trouble with hereditarily countable sets.
Let $H_\kappa=\{x:|trcl(x)|< \kappa\}.$ Prove the following.
a) For every infinite cardinal $\kappa$, $H_\kappa$ is transitive.
b) For every infinite cardinal $\kappa$, $H_\kappa \cap ORD=\kappa$, where $ORD=\{x:x$ is an ordinal$\}$
c) For every infinite cardinal $\kappa$, $H_\kappa \subseteq V_\kappa$, where $V_\alpha$ is defined as follows :$\alpha$ is an ordinal $V_0=\emptyset, V_\beta = \mathcal P(V_\alpha)$, and if $\gamma$ is limit ordinal, $V_\gamma=\cup\{V_\alpha:\alpha<\gamma\}$.
d) For every regular uncountable cardinal $\kappa$, $H_\kappa = V_\kappa$ iff $\kappa$ is strongly inacessible.
This is my working so far.
For a). Let $z \in x$. Note that $|trcl(z)|< |trcl(x)| <\kappa$. Thus $z \in H_\kappa$ and hence $x \subseteq H_ \kappa$. Thus $H_\kappa$ is transitive. (Is this correct?)
For b). $ORD$ is transitive and since $x \subseteq ORD$ then $ trcl(x) \subseteq ORD$. Then I am stuck here.
I am stuck for c) and d) too. Help would be greatly appreciated.
(a) is correct except that it should be $|\operatorname{trcl}(z)|\le |\operatorname{trcl}(x)|.$
(b) If $\alpha$ is an ordinal then $\alpha$ is transitive, so $\operatorname{trcl}(\alpha)=\alpha.$
(c) Prove that for any $\alpha < \operatorname{rank}(x)$ there is a $y\in \operatorname{trcl}(x)$ with $\operatorname{rank}(y)=\alpha.$ It follows that $|\operatorname{rank}(x)|\le|\operatorname{trcl}(x)|.$
(d) If $\kappa$ is not a strong limit, it's easy to find an set in $V_\kappa\setminus H_\kappa.$ For the converse, show that if $\kappa$ is strongly inaccessible, $|V_\alpha|<\kappa$ for every $\alpha<\kappa.$