Can somebody explain more in detail on why cumulative hierarchy of Sets (or Von Neumann universe) is not model of ZF (according to 1st answer in this post https://mathoverflow.net/questions/13609/models-of-zfc-set-theory-getting-started)?

  • 2
    $\begingroup$ Usually, people do think that the cumulative hierarchy is a model of ZF. Could you please expand the question to explain why you think it is not? Moreover, the proof that the constructible universe $L$ relative to the cumulative hierarchy $V$ is a model of ZF assumes that $V$ already satisfies ZF. $\endgroup$ Commented Oct 29, 2017 at 19:52
  • 3
    $\begingroup$ Thank you for the link to MO. I didn't easily find something there about the cumulative hierarchy not being a model of ZFC. Could you tell me which part, in particular, to look at? I did notice this answer: mathoverflow.net/a/13775/5442 Similarly, in Jech's book, does he explicitly say that the cumulative hierarchy is not a model of ZF, or have you extrapolated that from him not saying that the cumulative hierarchy is a model? Do you mean Jech's book "Set Theory" published by Springer? $\endgroup$ Commented Oct 29, 2017 at 19:59
  • $\begingroup$ Theorem.(ZF). $\forall x\; \exists a\in On\;(x\in V_a).$ Every set belongs to the cumulative hierarchy. $\endgroup$ Commented Oct 29, 2017 at 20:12
  • 3
    $\begingroup$ @User1006: Perlmutter says that $V$ is a class model of ZFC. This cannot be proved formally in ZFC - but neither can the fact that $L$ is a class model of ZFC. As I pointed out in the answer I just wrote, it is unlikely for an advanced set theory book to focus much energy on whether $V$ is a model of ZFC - from their viewpoint this is an obvious fact. $\endgroup$ Commented Oct 29, 2017 at 20:22

3 Answers 3


The other answers are quite correct (and Carl's seems particularly relevant to your question). But let me add a bit of a spin on the picture:

There are two ways to approach the cumulative hierarchy. Phrased a bit Platonistically:

  • We start with the cumulative hierarchy, and then argue that it satisfies ZFC. (It sounds like this is how you're thinking about it.)

  • We start with a "universe of sets" which we assume to satisfy ZFC, and then show that it is in fact a cumulative hierarchy.

In many ways the latter is superior to the former - for example, while the former tends to involve some philosophical commitment, it is easy to give the latter a completely formalist interpretation:

ZFC proves the statement "For all $x$ there is some ordinal $\alpha$ such that $x\in V_\alpha$"

(where "$V_\alpha$" is defined in the language of ZFC in the usual way). And in between the "there is a genuine universe of all sets" approach and pure formalism, we have the "model-y" interpretation:

Given any model $M$ of ZFC, we have $M=V^M$.

Here "$V^M$" denotes the definable subclass of $M$ consisting of all sets with ordinal rank. (There is a really unfortunate overloading of "$V$" in set theory.)

It is this latter approach which Jech is taking and which others tend to take. And I think the following might help clear things up:

The formalist version of the statement "$V\models ZFC$" is

For each axiom $\varphi$ of ZFC, ZFC proves $\varphi$.

This is completely trivial. The formalist version of the statement "$L\models ZFC$" is

For each axiom $\varphi$ of ZFC, ZFC proves $\varphi^L$

where $\varphi^L$ is intuitively the sentence expressing "$\varphi$ is true in $L$." This is very much nontrivial, and in particular even figuring out what "$\varphi^L$" is takes serious work.

The model-y versions of each of these principles are

For $M\models ZFC$, we have $M\models ZFC$.


For $M\models ZFC$, we have $L^M\models ZFC$.

As in the formalist interpretation, the "$V$-version" is trivial while the "$L$-version" is not. This is why you don't see something like "$V\models ZFC$" proved in Jech: he begins by assuming that $V$ is a model of ZFC.


When we talk about $V$ and $L$ being models of ZFC, as Asaf Karagila's answer points out we first need to remember that these are class models, not set models in the usual sense. Also, for this purpose there is no real difference in looking at ZF or ZFC, except that $L$ will satisfy ZFC if $V$ satisfies ZF.

So, moving on from those beginning issues, there are two things we might mean by "$V$ is a model of ZFC" or "$L$ is a model of ZFC".

First option - start with a model of ZFC and define $V$ relative to the model

The first option is that we begin with a model $M$ of ZFC (which could be a class model or a set model). Inside $M$, we identify $V$ with the cumulative hierarchy. We define this "cumulative hierarchy" as follows: $V = \bigcup_{\lambda \in \text{Ord}}V_\lambda$, where $$ V_{\lambda} = P\left ( \bigcup_{\beta < \lambda} V_\beta \right ). $$ These definitions can be phrased entirely in the language of ZFC. We can then show that $M$ satisfies $(\forall x) [x \in V]$, so $M$ believes this $V$ to be the class of all sets.

So, sitting outside $M$, we see that this $V$ is a model of ZFC, in the same sense that we already saw that $M$ was a model of ZFC. In this sense, it is a triviality that $V$ satisfies ZFC. This is the reason why few advanced books are likely to dwell long on that point. The other reason is that many set theory texts avoid almost all discussions about the metatheory they are working with when they study class models of set theory.

The reason that we need to go outside the model $M$ to see that $V$ is a model of ZFC is that, with $V$ as stated above, the language of ZFC is not strong enough to even express that "$V$ is a model of ZFC" (and the same goes for $L$). However, for each axiom $\phi$ of ZFC, $M$ does prove that $\phi$ holds as a statement about $V$ (and also holds as a statement about $L$ - there's no difference between $V$ and $L$ related to this issue). This issue is mentioned briefly in Jech's Set Theory (2003) in the section "Models of Set Theory and Relativization" on pp. 161-162.

Second option - take $V$ from a pre-existing concept of sets, and argue that it satisfies ZFC

The second option is to start with some pre-existing concept about "sets" (not ZFC), to let $V$ be the class of all "sets" according to the previous concept, and then to argue that $V$ satisfies ZFC using those concepts. One concrete example of this kind of argument is in Shoenfield's article in the Handbook of Mathematical Logic.

This kind of argument is philosophically relevant because it is used to argue that the ZFC axioms are a reasonable set of axioms for set theory, in the sense that they match well with our pre-existing concept of "set". In this second sense it is not as much of a triviality that $V$ in this sense satisfies ZFC - it depends on our pre-existing concept of "set" whether $V$ will satisfy ZFC at all.

In either case, option 1 or option 2, once we have $V$ we may formally define a subclass $L$ of $V$. We can then argue that $L$ will satisfy ZFC in the same sense that we already knew that $V$ satisfied ZFC. If tried to form $L$ from a hierarchy that did not model ZFC (e.g. if we only formed $L$ through stage $\omega$), there would be no natural way to prove this $L$ satisfies ZFC.


The question is what do you mean by "a model of $\sf ZF$".

Models are sets coupled with some relations, constants, and functions which interpret a given language and satisfy a particular theory.

But $V$, the set theoretic universe, is not a set. $L$ is not a set either. So $L$ is not a model of set theory.

What Gödel actually proved in his construction of $L$, is that assuming $V$ satisfies $\sf ZF$, then for every given axiom in $\sf ZF$, as well as $\sf AC$ and $\sf GCH$ (and the axiom $V=L$ which can be formulated in FOL), the relativization of this axiom to $L$ holds. Namely, for every $\varphi$ in $\sf ZF$ etc., $\sf ZF\vdash\varphi^L$.

But this is a meta-theorem. It is a schema. Every axiom has a different proof. Sure, all the proofs look-kind-of-the-same, but they are not.

Moreover, we start by assuming that $\sf ZF$ holds in $V$ to begin with. This means that you can re-dress the whole thing and say "Given a model of $\sf ZF$, $V$, we can find another model $L$ such that $L\subseteq V$, has the same ordinals, and $L$ satisfies $\sf ZFC+\rm V=L$.", but now we assume that we start with a model.

To sum up, $V$ is not a model of set theory because it is not a set, and $L$ is not a model of set theory either because it is also not a set. It is often an abuse of the language to say that they are, and when one is sufficiently familiar with these type of subtle points, one can make these abuses without getting confused all that often.

  • $\begingroup$ In Jech's set theory book, he does prove constructible universe a model of ZF. So I think it is related to constructible of sets. In Godel's case, he assumed that every set is constructible, which could mean that every class is set. $\endgroup$ Commented Oct 29, 2017 at 20:28
  • 2
    $\begingroup$ @User1006: I looked up Jech's theorem, which is 13.3 on page 176, and you are correct that it only says "$L$ is a model of ZF". However, his proof goes on to only show that, given an axiom $\sigma$ of ZFC, then ZFC proves the relativized axiom $\phi^L$. This way of writing is, for better or worse, common in set theory books. In any case Jech does not prove $L$ is a set model of ZFC, nor even that ZFC proves $L \vDash \text{ZFC}$, because the latter is not expressible within the language of ZFC. To be fair, Jech's book is intended for readers who already know quite a bit of set theory. $\endgroup$ Commented Oct 29, 2017 at 20:34
  • 1
    $\begingroup$ I now see that Jech clarified this on p. 162: "We shall often ... say that the model $(M,E)$ satisfies $\phi$. We point out however that while this is a legitimate statement in every particular case of $\phi$, the general satisfaction relation is formally undefinable in ZF.". $\endgroup$ Commented Oct 29, 2017 at 20:40
  • $\begingroup$ In theorem 13.16 on p187, Jech proves that $L$ is the smallest inner model of ZF. $\endgroup$ Commented Oct 29, 2017 at 20:44
  • 3
    $\begingroup$ @User1006 In set theory, an "inner model" is usually a class model, rather than a set model. Note that $V$ is an inner model of itself, and $L$ is also an inner model of $V$. So $V$ and $L$ are again models in the same sense. $\endgroup$ Commented Oct 29, 2017 at 20:46

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .