Von Neumann hierarchy has a critical role in set theory. It is well-known that $V_\alpha$ is a model of $\mathsf{ZC}$ if $\alpha$ is a limit ordinal. Furthermore, $V_\alpha$ satisfies the cumulative hierarchy axiom ($\mathsf{CHA}$), an axiom claiming the existence of sequence $\langle V_\eta\mid \eta<\xi\rangle$ for each ordinal $\xi$ with that every set is contained in some $V_\eta$, which is apparently not a theorem of $\mathsf{ZC}$. (Caution. $\mathsf{CHA}$ is not equivalent to the claim that $V_\xi$ exists for all ordinal $\xi$. See Hamkins' answer.)

However, $V_\alpha$ does not satisfy Replacement even for $\Delta_1$-formulas.

Theorem. $V_{\omega+\omega}$ does not satisfy Replacement for $\Delta_1$-formulas.

Proof. Consider

\begin{align} \phi(x,y) :\equiv \exists n<\omega\exists f [x<n+1\land f\colon (n+1)\to V \land f(0)=\omega \\ \land \forall m<n [f(m+1) = f(m)+1] ]\land f(x)=y. \end{align}

We can see that the above formula is equivalent to \begin{align} \forall f \forall n<\omega [x<n+1\land f\colon (n+1)\to V \land f(0)=\omega \\ \land \forall m<n [f(m+1) = f(m)+1] ]\to f(x)=y. \end{align}

Furthermore, $\forall x\in\omega\exists! y\phi(x,y)$, but Replacement for $\phi$ results in $\{\omega+n\mid n<\omega\}$ that is not a member of $V_{\omega+\omega}$.

(I previously claimed that $\Delta_0$-Replacement is invalid over $V_{\omega+\omega}$, but I found there is a gap in my proof.)

Then we can ask the following questions:

  1. Is there any non-trivial consequences of Replacement that are valid over any $V_{\alpha}$ for limit $\alpha>\omega$? (According to Mathias's The Strength of Mac Lane Set Theory, $\mathsf{ZC}$ proves Replacement for stratified formulas.)

  2. Can we characterize the theory of $V_\alpha$ for limit $\alpha>\omega$?

The second question needs some clarification. Assume that $\mathsf{ZFC}$ is consistent. My question is we can characterize the theory $T$ defined by

A sentence $\sigma$ is a member of $T$ if and only if, for any models $M\models \mathsf{ZFC}$ and $\alpha\in M$ such that $M\models \alpha>\omega\text{ is a limit ordinal}$, $M\models (V_\alpha\models \sigma)$.

Obviously $T\supseteq \mathsf{ZC+CHA}$. Furthermore, if we assume $\mathsf{ZFC}$ + there is a worldly cardinal is consistent, then

  • $T\nvdash\lnot\sigma$ for axioms $\sigma$ of $\mathsf{ZFC}$: If $M$ is a model of $\mathsf{ZFC}$ and $M\models \kappa \text{ is worldly}$, then $V^M_\kappa\models \mathsf{ZFC}$.
  • If $\mathsf{ZFC}$ proves we can force $\sigma$, then $T\nvdash\lnot\sigma$. Again, let $M$ be a countable model of $\mathsf{ZFC}$ and $M\models \kappa \text{ is worldly}$. By the assumption, we can find a forcing poset $\mathbb{P}\in V^M_\kappa$ such that $V^M_\kappa$ thinks $\mathbb{P}$ forces $\sigma$. Now consider the $\mathbb{P}$-generic extension $M[G]$ of $M$. Then $V^{M[G]}_\kappa = V^M_\kappa[G]$ and $V^M_\kappa[G]\models \mathsf{ZFC}+\sigma$.

These results provide some upper bound for $T$. Could we find a better upper bound for $T$?

  • 1
    $\begingroup$ Re: (1), Borel determinacy famously requires quite a lot of replacement but is already "detected" by $V_{\omega+1}$, hence by $V_\alpha$ for every limit $\alpha>\omega$. $\endgroup$ Jan 17, 2022 at 2:46
  • $\begingroup$ @Noah It would complicate determining $T$ (or $\mathsf{ZC+CHA}$ proves Borel determinacy.) $\endgroup$
    – Hanul Jeon
    Jan 17, 2022 at 4:27

3 Answers 3


The sets $V_\lambda$ for limit ordinals $\lambda>\omega$ are precisely the models of second-order Zermelo set theory (including foundation scheme, infinity, and choice) plus the cumulative hierarchy axiom CHA, which asserts that every set $x$ is in some $V_\beta$, where this is a set for which there is well-ordered sequence $\langle V_\alpha\mid\alpha\leq\beta\rangle$ obeying the recursive definition of the cumulative hierarchy.

It is easy to see that every $V_\lambda$ satisfies that theory, and conversely, if a model satisfies this theory, then because we have the second-order separation axiom in second-order Zermelo, it will be correct about power sets, and so the cumulative hierarchy that it builds will be correct. So the model will be (isomorphic to) $V_\lambda$ for some limit ordinal $\lambda>\omega$.

This theorem was observed by an undergraduate student of mine Donghui Jia in Oxford last term, who wrote on it (under my supervision) emphasizing the analogy with Zermelo's quasi-categoricity theorem, in which Zermelo proved that the models of second-order ZFC are precisely $V_\kappa$ for inaccessible cardinals $\kappa$. The theorem I mentioned above is the corresponding result for the weaker Zermelo theory, when augmented with the cumulative hierarchy axiom.

Meanwhile, you had asked for consequences of replacement that hold in every $V_\lambda$, and the cumulative hierarchy axiom is one. Mathias constructed supertransitive models of Zermelo set theory in which $V_\omega$ does not exist. Modifications of the Mathias slim-model technique allow one to construct models of Zermelo set theory where the first omitted $V_\lambda$ is for any desired limit ordinal $\lambda$, even though the model has order-types much exceeding $\lambda$ and sets of rank higher than $\lambda$. I believe also that there are models of Zermelo set theory that do not satisfy transitive containment---that is, not every set has a transitive closure; this holds in every $V_\lambda$, but one uses replacement to prove it. And Mathias has other crazy models of Zermelo set theory, where forcing extensions can have new larger ordinals not in the ground model, but this never happens over any $V_\lambda$.

Warning. To get the CHA, it isn't quite enough to say that for every ordinal $\beta$ there is a set $V_\beta$ arising from a sequence $\langle V_\alpha\mid\alpha\leq\beta\rangle$ obeying the recursive definition of the cumulative hierarchy. The reason is that the ordinals might run out before you expect, and perhaps you have $V_\alpha$ for all the ordinals $\alpha$ that exist, but there are still more sets of higher rank, whose ordinals do not exist.

  • $\begingroup$ Thank you for your interesting answer. I have two questions about your answer: first, I thought that $\mathsf{ZC+CHA}$ proves there is a rank function (equivalently, $\langle V_\alpha \mid \alpha\in\mathrm{Ord}\rangle$ form a cofinal sequence) when I wrote my question. It is now like there is no obvious way to prove this. Do you have any idea about it? Second, is there Donghui Jia's thesis or report? It looks interesting, so I want to look at whether there is any further detail or results there. $\endgroup$
    – Hanul Jeon
    Jan 17, 2022 at 4:13
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    $\begingroup$ The point of my warning is that it is a little finicky to say that the cumulative hierarchy exists---if you just say $\forall\beta\exists V_\beta$, it isn't good enough, and I believe Mathias has models of ZC with sets having no ordinal rank, and I think we can have them satisfy $\forall\beta\exists V_\beta$, simply because there aren't many $\beta$. But meanwhile, if you have CHA as I stated it, then you get the ordinals of every rank inside those $V_\beta$, and you can define rank $r(x)=$least $\beta$ such that $x\in V_{\beta+1}$. This doesn't need replacement. $\endgroup$ Jan 17, 2022 at 4:18
  • 1
    $\begingroup$ Donghui's thesis is not yet completed, but I have alerted him to this question, and perhaps he will post something when it is ready (in the coming term). $\endgroup$ Jan 17, 2022 at 4:19
  • 1
    $\begingroup$ I missed that the statement $\mathsf{CHA}$ includes every set is contained in some $V_\beta$. By the way, I am looking forward to Donghui's thesis. $\endgroup$
    – Hanul Jeon
    Jan 17, 2022 at 4:23
  • 5
    $\begingroup$ The result discovered by Donghui was published earlier in section 5 of the following paper by Gabriel Uzquiano: Models of Second-Order Zermelo Set Theory, The Bulletin of Symbolic Logic, Vol. 5, No. 3 (Sep., 1999), pp. 289-302. $\endgroup$
    – Ali Enayat
    Jan 17, 2022 at 17:30

This answer concerns the second question (which asks whether there is a characterization of the theory of models of the form $V_{\alpha}$, where $\alpha$ ranges over limit ordinals), and its elaboration which asks whether there is a characterization of the theory $T$ defined by:

A sentence $\sigma$ is a member of $T$ if and only if, for any models $M\models \mathsf{ZFC}$ and $\alpha\in M$ such that $M\models \alpha>\omega\text{ is a limit ordinal}$, $M\models (V_\alpha\models \sigma)$.

Let $T^{+}$ be the theory of models of the form $V_{\alpha}$, where $\alpha$ ranges over limit ordinals. The observations below make it clear that $T \neq T^{+}$.

Observation 1. $T$ coincides with the collection of set-theoretical sentences $\sigma$ such that the implication "For all $\alpha$, if $\alpha$ is a limit ordinal greater than $\omega$, then $\sigma^{V_{\alpha}}$" is provable in $\mathsf{ZFC}$.

Proof: this is a simple consequence of the completeness theorem for first order logic.

Observation 2. $T$ is recursively enumerable.

Proof: This is an immediate corollary of Observation 1.

Observation 3. $T^{+}$ is not arithmetically definable (and indeed, $T^{+}$ is not definable in $n$-order arithmetic for any $n \in \omega$).

Proof: Given an arithmetical sentence $\varphi$, $\mathbb{N}\models \varphi$ iff the set-theoretical sentence expressing [if $\omega$ exists, then $\mathbb{N}\models \varphi$] is a member of $T^{+}$. So by Tarski's undefinability of truth theorem, $T^{+}$ is not arithmetically definable (the parenthetical clause of Observation 3 has a similar proof).

Finally, looking at the definition of $T^{+}$, $T^{+}$ is definable by a $\Pi_2$-formula (in the Levy hierarchy). Based on a "back-of-the-envelope-proof", this calibration is optimal. The proof is based on the "folklore" characterization of $\Sigma_2$-statements, as in Lemma 2 (page 4) of the paper The universal finite set of Hamkins and Woodin (note that $\sigma \in T^{+}$ iff the implication "if there is no last ordinal, then $\sigma$" is true in all structures of the form $V_{\alpha}$).


I don’t have a comprehensive answer, but a couple of observations:

  • $T$ is included in ZFC unconditionally (i.e., not assuming the consistency of worldly cardinals).

    Indeed, for any sentence $\sigma$, ZFC proves $\neg\sigma\to\exists\alpha>\omega\text{ limit }V_\alpha\nvDash\sigma$ by Lévy’s reflection principle. Thus, if ZFC proves $\forall\alpha>\omega\text{ limit }V_\alpha\vDash\sigma$, then ZFC proves $\sigma$.

  • $T$ is strictly stronger than ZC + CHA, and in particular, it is not included in ZC + $\Sigma_n$-replacement for any fixed $n$ (assuming ZFC is consistent).

    The reason is that ZFC proves the consistency of ZC + $\Sigma_n$-replacement (using Lévy’s reflection principle again, and universal $\Sigma_n$-formulas), which is a $\Pi^0_1$ statement; thus so does $T$, by absoluteness of $\omega$.

    More generally, $T$ includes the $\Pi_1^P$-fragment of ZFC, i.e., the $\Pi_1^P$-reflection principle for ZC + $\Sigma_n$-replacement for each fixed $n$, where a formula is $\Pi_1^P$ if it consists of a block of universal quantifiers followed by a bounded formula that may use the powerset function (I don’t know if there is a standard notation for this).


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