Can power set axiom be proved in a class theory of well ordered hereditarily accessible sets? I've asked this question at MathStackExchange, only to receive no answer.
I'll repost this question here:
Working in a pure class theory, where sets are defined as elements of classes. That is: 
Define: $set(x) \iff \exists y (x \in y)$
Let's have the following known three axioms from $\text{MK}$
Extensionality: $\forall x\forall y [\forall z (z \in x \leftrightarrow z \in y) \to x=y]$
Class Comprehension: if $\varphi$ is a formula in which the symbol $``x"$ is not free, then $(\exists x \forall y (y \in x \leftrightarrow set(y) \wedge \varphi))$ is an axiom.
Define: $x=\{y|\varphi\} \iff \forall y (y \in x \iff set(y) \wedge \varphi )$ 
Pairing: $\forall a,b [set(a) \wedge set(b) \to set(\{a,b\})]$
Define purely accessible ordinal as any ordinal that does not have a subclass of it that is an uncountable regular [weak] limit cardinal; i.e., no inaccessible cardinal is a subclass of it.
Now if we add an axiom stating that any class is a set if and only if it is hereditarily subnumerous to a purely accessible ordinal. Formally this is:
Accessibility: $\forall x [set(x) \leftrightarrow \exists \alpha (\alpha \text{ is purely accessible ordinal } \wedge \\ \forall y (y \in TC(x) \lor y=x \to \exists f (f:y \rightarrowtail \alpha)))]$
Where $\text TC(x)$ means the transitive closure of $x$ defined in the usual manner as the intersectional class of all transitive super-classes of $x$.

Would this theory prove the power set axiom for sets? that is:
$\forall x (set(x) \to \exists y (y=\{z|z \subseteq x\}  \wedge set(y)))$

Note: its clear that if we drop the requirement of $y=x$ in Accessibility axiom, then we can get the power set axiom. But here $x$ itself must be also subnumerous to some purely accessible ordinal. Can for example $P(\aleph_0)$ be equinumerous to the proper class $ORD$ of all set ordinals? We note that this theory is as strong as $\text{ZFC}$ as regards proving existence of ordinals, i.e. every ordinal provable to exist in $\text{ZFC}$ is also provable to exist here as a set ordinal, also all axioms of $\text {ZFC}$ except power and regularity are provable here. But apparently the power set axiom is not provable here? The reason why I say that is because the cardinality of the continuum is not controllable, it is consistent for it to even be of inaccessible cardinality. I can see how to interpret $\text {ZFC - Power}$ in this theory, i.e. interpret adding Regularity, but I don't know how to interpret the power set axiom? This theory must be able to do that, i.e. interpret power set axiom, although I think it is not able to prove that axiom. I suspect interpreting power can be done through constructible sets, i.e. through $L$, since powers in $L$ would be subnumerous to pure accessible ordinals via $\text{GCH}$. However I'm not so sure.
 A: Your system indeed couldn't prove even that $\mathcal{P}(\omega)$ is a set.
Let $M$ be a countable transitive of $\mathsf{ZFC}+\mathsf{GCH}+\mbox{there exists an inaccessible}$. Let $\kappa\in M$ be the first inaccessible in $M$. Let $M[G]$ be the forcing extension of $M$ by $\kappa$-many Cohen reals. Note that by the standard facts about Cohen forcing, $\kappa$ is the first weakly inaccessible in $M[G]$ and $M[G]\models (\kappa=2^{\aleph_0})$. Let $K$ be the transitive model $(\mathcal{P}(H\kappa))^{M[G]}$; we will treat $K$ as a model of class theory, e.g. $K$-sets will be precisely elements of $(H\kappa)^{M[G]}$. Clearly, $K$ is a model of your theory and in $K$ there exists a (class) bijection between the classes $\mathcal{P}(\omega)$ and $On$ (the latter is due to the fact that in $M[G]$ there is a bijection between $\kappa$ and $(\mathcal{P}(\omega))^{M[G]}$).
However I am not certain whether your theory shows that $L$ is an interpretation of $\mathsf{ZFC}$. It would be very plausible that it is the case if it would be possible to prove the axiom of collection for sets (naturally formalized as a single sentence in the setting with classes) in your system, but I don't know whether it is indeed possible.
