I've asked [this question](https://math.stackexchange.com/questions/3228725/can-power-set-axiom-be-proved-in-a-class-theory-of-well-ordered-hereditarily-acc) 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 [$\text{MK}$](https://en.wikipedia.org/wiki/Morse%E2%80%93Kelley_set_theory#MK_axioms_and_ontology)


**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](https://math.stackexchange.com/questions/3228725/can-power-set-axiom-be-proved-in-a-class-theory-of-well-ordered-hereditarily-acc) 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.



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


