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Ackermann in 1956 proposed an axiomatic set theory.

Reinhardt proved that Ackermann's set theory equals ZF

It's clear that Zermelo set theory can be interpreted in Ackermann's set theory minus class comprehension.

Is Ackermann's set theory minus class comprehension also equal to ZF?

In other words: did Reinhardt's proof made an essential use of the class comprehension schema in proving the equivalence between Ackermann's set theory and ZF?

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  • $\begingroup$ Did you ever find out the answer? I also wonder whether extensionality and regularity are necessary to interpret ZFC. Can we keep just the axiom schema of set comprehension, the axiom of elements, and the axiom of subsets? $\endgroup$
    – user76284
    Commented Mar 25, 2020 at 20:00
  • $\begingroup$ I don't know the answer yet, although the below presented answer seem to say that it is equivalent, so no need for class comprehension, but yet I'm not sure, regularity is generally not needed for the interpretation of ZFC, and I think extensionality not needed as well because the class of all co-extensional sets to some set, is a set. If class comprhension is not needed, then we only really need two axioms one of completness (any element or subset of a set ,is a set) and set comprehension. But I'm not sure really. $\endgroup$
    – Zuhair Al-Johar
    Commented Mar 26, 2020 at 9:51
  • $\begingroup$ IF class comprehension is not needed, then this would mean that we only need two axiom schemata that of "class comprehension" and "set construction" presented in linked question below: mathoverflow.net/questions/334208/… $\endgroup$
    – Zuhair Al-Johar
    Commented Mar 26, 2020 at 10:15

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Let $V$ be the $V$ from $A$. It is clear $V$ satisfies exstensionality, seperation, and regularity. Pairing follows from reflection ($x=a\lor x=b$). Inductively, we can prove each $n\in V$, and so we have $\omega\in V$. If $X\in V$, then $\cup X\subseteq V$. Then define $\phi(x)\leftrightarrow \exists y(y\in X\land x\in y)$. By reflection $\cup X=\{x|\phi(x)\}$, and so $\cup X\in V$. A similar argument goes for powerset. For replacement, let $F=\{(x,y)|\phi(x,y)\}$ be a function. Then $F(X)\subseteq V$ (By definition). Then we can find some $V_\alpha$ that reflects $\phi(x,y)$ relative to $V$, and $x\in F(X)\leftrightarrow \exists y(y\in X\land\phi(x,y)^{V_\alpha})$, and so $F(X)\in V$. Therefore $V\vDash \phi$ for each axiom $\phi$ of $ZFC$. Now suppose an axiom of $ZFC\,\phi$, satisfies $\phi\vdash\psi$. Then every model of $\phi$ satisfies $\psi$, and so $V\vDash\psi$.

Proof the $V$ satisfies the reflection theorem: Note that the statement $x\in V_\alpha$ is $\Delta_1$, and so $V_\alpha^V=V_\alpha$. In addition, the mapping $\phi(x,\alpha)\leftrightarrow rank(x)<\alpha$ is $\Delta_1$, and so $\phi(x,\alpha)^V\leftrightarrow\phi(x,\alpha)$. Then for any $\alpha\in V$, $V_\alpha=\{x|rank(x)<\alpha\}$, and so $V\vDash Replacement\,for\,\alpha\mapsto V_\alpha$. As a consequence, for any $C\subseteq V$, the set $\hat C=\{x\in C|\forall y\in C(rank(x)\leq rank(y))\}$ is in $V$.

Let $H(u_0...u_n)=\hat C$ and $C=\{x|\phi^V(x_0...x_n)\}$. Then, for any set $M_0\in V$, define a sequence starting at $M_0$ with $M_{i+1}=M_i\cup(\cup\{H(u_0...u_n)|u_0...u_n\in M_i\})$. The function $F^V:{i\mapsto V_{ran(M_i)+1}}$ can be defined inductively, and it is easy to see that if $F(i)\in V$, $F(i+1)\in V$. Then $D=\cup \{H(u_0...u_n)|u_0...u_n\in\cup F(\omega)\}$ is in $V$. Then $x\in F(i)\leftrightarrow (i=0\land x\in M_0)\lor (\exists j(j+1=i\land x\in F(i)\cup ((D'\restriction F(i))(F(i))))$, where $D'$ is the union of the class of all sets such that each $x\in D'$ is $\hat C$ for some $C$ and $\cup D'=D$. What I mean by $(D'\restriction F(i))(F(i))$, is really $(G\restriction F(i))(F(i))$, where $G=\{(x,u_0...u_n)|x=H(u_0...u_n)\land u_0...u_n\in F(i)\}$. And so $V_\alpha=\cup F(\omega)$ is in $V$ and $V_\alpha\vDash\phi\leftrightarrow V\vDash\phi$ (By induction on formula complexity).

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  • $\begingroup$ I didn't get the argument for Replacement. for example your $F(X)$ seems to be a subclass of the domain of $F$ it consists of elements of $dom(F)$ that have their images among their elements, I don't see why this $F(X)$ class should be a subset of $V$ for an arbitrary function $F$? Its not that clear. $\endgroup$
    – Zuhair Al-Johar
    Commented Jun 9, 2019 at 4:26
  • $\begingroup$ I think I made a typo on the second to last line. Is it clearer now? $\endgroup$
    – Master
    Commented Jun 9, 2019 at 4:31
  • $\begingroup$ what do you mean by $F(X) \subseteq V$ (by definition), you have an arbitrary class $X$ and $F(X) $is the image of $X$ under function $F$, why you think that images must be subsets of $V$? for example let $X$ be $\{V\}$ and let $F$ be the equality function, clearly $F(X)=X=\{V\}$ which is not a subset of $V$ and it is not an element of $V$. $\endgroup$
    – Zuhair Al-Johar
    Commented Jun 9, 2019 at 11:01
  • $\begingroup$ What we are trying to prove is first that $V\vDash ZFC$. To prove this we have to show that every definable function $F: V\rightarrow V$ has its image in $V$. By the definition of a function, its image is a subset of $V$ (You can't map using replacement, say $n\mapsto \kappa_n$, where $\kappa_n$ is the $n$th worldly cardinal, and therefore prove $Con(ZFC)$). $\endgroup$
    – Master
    Commented Jun 9, 2019 at 16:05
  • $\begingroup$ If you show that for every $X \in V$, any definable function $F:X \to V$ implies that $F(X) \in V$, then yes this would definitely interpret Replacement over $V$! BUT the problem is how to prove that? Ackermann minus comprehension can only prove that statement for the kind of functions that are definable after formulae that do not have $V$ occurring in them and only when all their parameters are elements of $V$ also. It cannot show it for all kinds of formulas, in particular it cannot show it for bounded by $V$ formulas, which are the wanted formulas for a proof of replacement! $\endgroup$
    – Zuhair Al-Johar
    Commented Jun 9, 2019 at 18:22

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