Let $T$ be the theory consisting of Zermelo's original set theoretic axioms (extensionality, empty set, pairing, union, powerset, infinity, separation, choice) together with foundation. Put more succinctly, $T$ consists of ${\rm ZFC}$ axioms without the replacement axiom scheme. The theory $T$ is too weak for most set theoretic purposes because, for instance, it cannot even prove the existence of transitive closures. We consider strengthening $T$ in the following two ways. Let $A$ be the axiom asserting that $V\_\alpha$ exists for every $\alpha$ and let $A^\*$ be the axiom asserting that every $x$ is an element of a $V\_\alpha$. For example, $V\_{\omega+\omega}$ is a model of $T+A^\*$. Does $T$ together with $A$ imply $A^\*$? Or is it possible that $V\_\alpha$ exists for every ordinal $\alpha$, but there is a set without an ordinal rank?



Does the existence of the von Neumann hierarchy in models of Zermelo set theory with foundation imply that every set has ordinal rank?
Take the Zermelo ordinals to be defined by $Z(0) = 0$, $Z(\alpha+1) = \{Z(\alpha)\}$, and $Z(\lambda) = \{Z(\alpha): \alpha<\lambda\}$ (where $\alpha, \lambda$ are von Neumann ordinals). Then if we add $Z(\omega+ \omega)$ to $V\_{\omega +\omega}$ and close under pairing, union, subsets, and powersets, we get a model of $T$ - Choice + $A$ which is not a model of $A^\*$. 


More precisely, let $D\_0 = V\_{\omega+\omega} \cup\{Z(\omega+\omega)\}$ and $D\_{n+1}$ be the result of adding pairs, unions, subsets, and powersets of element of $D\_n$ to $D\_n$. Clearly, $M = \bigcup\_{n <\omega}D\_n$ is transitive and models $T$ - Choice. A simple induction shows that:


For every $x\in D\_n$ there is an $\alpha<\omega+\omega$ such that the (von Neumann) ordinals in $tc(x)$ are less than $\alpha$. 


Since $V\_{\omega+\omega}\subseteq M$, it follows that the von Neumann ordinals in $M$ are just those in $\omega+\omega$. So it models $A$. Since the rank of $Z(\omega+\omega)$ is $\omega+\omega$, it doesn't model $A^\*$.


(To get Choice in the form ``every set is well-orderable" we just throw in $x \times x$ at $D\_{n+1}$ for $x\in D\_n$).


