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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.

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