Strongly inaccessible cardinals and certain kinds of universes Write $\mathrm{tc}(*)$ for transitive closure.
For all cardinals $\kappa$, let $H(\kappa)$ denote the collection of all sets $X$ such that


*

*$|X|<\kappa$, and

*For all $x \in \mathrm{tc}(X)$, it holds that $|x| < \kappa$.


Consider the class $K$ of all cardinals $\kappa$ such that $H(\kappa)$ is closed under powersets and unions. i.e. the class of all $\kappa$ such that


*

*For all $X \in H(\kappa)$ it holds that $\mathcal{P}(X) \in H(\kappa)$

*For all $X \in H(\kappa)$ it holds that $\bigcup X \in H(\kappa)$.


Is $K$ precisely the class of all strongly inaccessible cardinals? (include $0$ and $\omega$ in the definition of strongly inaccessible cardinals)
 A: Fact: $H(\kappa)$ is closed under unions iff $\kappa$ is regular.
Proof: If $\kappa$ is not regular then any cofinal sequence of ordinals in $\kappa$ witnessing its singularity also witnesses that $H(\kappa)$ is not closed under unions.  Conversely, suppose $\kappa$ is regular, and let $X \in H(\kappa)$.  Since $\mathrm{tc}\left(\bigcup X\right) \subseteq \mathrm{tc}(X)$, it'll suffice to show that $\left|\bigcup X\right| < \kappa$.  But this follows from the fact that $\kappa$ is regular, and that $\bigcup X$ is a union of $< \kappa$ many elements each of size $< \kappa$.
Fact: $H(\kappa)$ is closed under power sets iff $\kappa$ is strong limit.
Proof: If $H(\kappa)$ is closed under power sets, then $\kappa$ is clearly strong limit.  Conversely, suppose $H(\kappa)$ were not closed under power sets, and suppose $X \in H(\kappa)$ witnesses this.  Since every element of the transitive closure of $\mathcal{P}(X)$ is either a subset of $X$ or an element of $\mathrm{tc}(X)$, the only way we can have $\mathcal{P}(X) \notin H(\kappa)$ is if $|\mathcal{P}(X)| \geq \kappa$.  But then we have $|X| < \kappa$; $|\mathcal{P}(X)| = 2^{|X|} \geq \kappa$.  And so $\kappa$ is not strong limit.
Strictly speaking, a cardinal must be uncountable to be considered inaccessible, so the two properties you listed aren't equivalent to inaccessibility.  But your two properties, together with uncountability, are equivalent to inaccessibility.
EDIT: Just noticed you say you want to consider $0$ and $\omega$ to be strongly inaccessible.  If you do this, then yes, your two conditions are equivalent to strong inaccessibility.  But why would you do this?
