# Can power set axiom be proved in a class theory of well ordered hereditarily accessible sets?

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 would be subnumerous to pure accessible ordinals via $$\text{GCH}$$. However I'm not so sure.