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


1 Answer 1


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.

  • $\begingroup$ I think this theory can prove replacement for sets, and so collection for sets as well. $\endgroup$ May 31, 2019 at 18:09
  • $\begingroup$ I think the lemma is to prove that the union of every well orderable set of well orderable sets, is well orderable also. If that is provable then its easy to prove both Replacement and Collection! $\endgroup$ May 31, 2019 at 19:46
  • $\begingroup$ I think the proof of this Lemma is easy, now if $x$ is bijective to some ordinal $\alpha$ via function $f$, then code each element $m$ of $x$ after that bijection, i.e. by $f(m)$, now let's have a function $r$ that replaces each element $m$ of $x$ by $r(m)$, now $r(m)$ itself is bijective to an ordinal by function $h$, now send each element $m$ of $x$ to the set $\{\langle f(m), h(n) \rangle| n \in r(m)\}$, now the set union of all those sets is clearly well orderable. The rest of proof of replacement (and collection) over sets is simple. $\endgroup$ May 31, 2019 at 21:26

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