(Why cumulative hierarchy of Sets is not model of ZF) As seen from this post, it is proven that V is a structure that satisfies ZFC and is model of it. As we know an inaccessible cardinal k implies Vk (a segment of V) meaning that inaccessible cardinals are apart of the cumulative hierarchy (In what sense are inaccessible cardinals inaccessible?). This is where the problem comes in. It has been accepted that k is inconsistent with ZFC (meaning in the ZFC universe there is no stage Vk) but if V is a model of ZFC (and contains k within it), does that not mean k exists within ZFC? I've only started researching models fairly recently, so the answer to this question may seem 'obvious' to some and would like some clarification. Any answers would be appreciated!

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    $\begingroup$ "It has been accepted that k is inconsistent with ZFC" here's the mistake. ZFC cannot prove the consistency of ZFC+there is an inaccessible, but that doesn't mean that an inaccessible is inconsistent with ZFC $\endgroup$ Sep 21, 2022 at 5:05
  • $\begingroup$ @AlessandroCodenotti Thanks for the clarification, but doesn't the problem still arise? V (which is a model of ZFC) contains k, so paradoxically doesn't k still exist within ZFC even if it can't be proven? $\endgroup$
    – SI J
    Sep 21, 2022 at 5:31
  • $\begingroup$ Do you mean "as we know, existence of inaccessible $k$ implies that $V_k$ exists"? $\endgroup$
    – C7X
    Sep 21, 2022 at 22:52

2 Answers 2


An explanation for the comment thread, some of this is based on Carl Mummert's answer here.

I argue that any discussion about "the universe of ZFC" is actually about $V$. When people say "the universe for [a theory]", they most likely are talking about the intended model of that theory, because that's what we imagine variables ranging over when we discuss it. For example if we talk about "the universe for Peano arithmetic", the set we're thinking while saying this is N, i.e. we imagine the variables ranging over naturals. This is the case even though there are uncountable models of Peano arithmetic, if you mention "universe of PA" those aren't the intended sets that leap to mind. Kleene says the reason for this is because formalization comes after our intuitive idea of what a "set" is: (Kleene, Mathematical Logic (1967), page 200)

Since a formal system (usually) results in formalizing portions of existing informal or semiformal mathematics, its symbols, formulas, etc. will have meaning or interpretations in terms of that informal or semiformal mathematics. These meanings together we call the (intended or usual or standard) interpretation or interpretations of the formal system.

Not only that, $V$ isn't a true "completed" object, because if we try looking at it as the set of all "collections", Russell's paradox occurs. So it's difficult to talk about the intended model of set theory as if it's a "true" object, which makes it hard to connect models of ZFC to $V$. (I credit this MO question for this point.)

So $V$ is what we make of it, and has the properties that we assert it does. If I were to claim "I believe V is the collection of all sets constructed from $\omega$ and the ZFC operations", this would be me saying that sets are only the things constructed in this way and no more, in particular I would claim an inaccessible cardinal does not exist. If I said "V is a collection that happens to contain all sets constructed from $\omega$ and the ZFC operations", I have only said the ZFC axioms are correct, but have left open questions about the sets not constructed this way, in particular the question of an inaccessible existing. The latter is the ZFC-user's perspective on inaccessibles, we can't prove there exists a cardinal which is inaccessible, but it's also consistent that they do exist. $V$ is tameable through the construction of the levels $V_\alpha$, which is still an example of $V$ being what we make of it - this construction works only if we assert "$V$ satisfies the axiom of foundation".

We can still try asking "how are models of ZFC connected to V", by asking what do models of ZFC have in common with the post-Russell view of V? Like in the open-ended perspective from the last paragraph (seeing the ZFC operations as giving a not-necessarily-exhaustive enumeration of all sets that exist,) we can construct models with or without inaccessibles, with even larger cardinals that have many inaccessibles below them, and even a model that we as observers know is countable, but the model proves "there exists an uncountable set". As observers working using all the tools of ZFC available to us, it's describable (but I would avoid describing it) as "looking down from $V$ onto the countable model."


Two things:

  • Some details on working in $V_k$: $k$ is not an actual object we have access to when working within $V_k$. In particular, how would we formalize "$k$ exists"? We could try doing this by seeing if $V_k$ satisfies $``k=k"$, in order to do this we could set $\phi(x)\iff \vDash_{V_k}\ulcorner x=x\urcorner$ and try setting $x=k$, the result is a false statement. Then we might try $\vDash_{V_k}\exists x(x=k)$, we get the same false result... when working only within the model $V_k$, since $k\notin V_k$, $k$ is no longer in the domain of discourse. Variables are "prohibited" from taking on the value $k$. In the formalization of $\vDash_{V_k}$, this is done by restricting the ranges of variable assignments - functions that map variable symbols to their members of $V_k$. Since no $\textrm{varSymbol}\to V_k$ function will have output $k$ on any value, any direct formalization of "there is a set equal to $k$" will be a false statement.
  • There are plenty of cases where restricting our axioms prevents us from proving $\exists x\phi(x)$, even when in our model there really is such a set with property $\phi$. For example, assuming ZFC is sound, there is a non-measurable subset of the reals in any model of ZFC. Working within one, we may restrict our proofs to only assume ZF, since these axioms are still satisfied by the model. But then, the non-measurable set is still there, we're just no longer able to prove the true statement "there exists a set of reals which is non-measurable".

More info about models of ZFC:

  • Let $\kappa$ be the least inaccessible. $V_\kappa$ is a model of ZFC, and $V_\kappa$ is closed under powerset. $V_{\kappa+1}$ is not a model of ZFC, and it satisfies the negation of the powerset axiom (e.g., it satisfies "$L_\kappa$ has no powerset".)
  • However, being closed under powerset is not sufficient to be a model of ZFC: $V_{\omega2}$ is not a model of ZFC.
  • Also, being closed under powerset is not even necessary to be a model of ZFC: by the Lowenheim-Skolem theorem, since ZFC has an infinite model (for example $V_\kappa$), there must be a countable model of ZFC. There are some technically difficult constructions of them, and they are able to contain $\omega$ but not $\mathcal P(\omega)$.
  • $\begingroup$ "Even if we pass to V", you do mean if V becomes the model? and to sum up you said that Vk (and above, e.g. Vk + 1) do satisfy ZFC's axioms but none the less still cannot be proven in ZFC due to restrictions? Just to confirm. $\endgroup$
    – SI J
    Sep 22, 2022 at 9:34
  • $\begingroup$ Also my conception of models is that in effect, models show the 'height' (how far it extends in the Von Neumann universe) of the universe of a theory. For example the model of a purely finite set theory would be Vω and that's also how far the universe for this theory goes. I am confused about how (standard) ZFC's universe is less than Vk (as it can only prove sets constructed via powerset) and how it's universe is also V (which includes all sets)? This is in part why I made this question, to understand the relation between these two 'universes'. Could you please clarify this for me? $\endgroup$
    – SI J
    Sep 22, 2022 at 9:50
  • $\begingroup$ For your first comment, I will add a new section to my answer. For your second, I think you will like this answer: math.stackexchange.com/a/121131, in particular, a set is just an object that acts in the way the axioms say it does, this is substrate-independent and V is then just "the collection of all things which have behavior described by ZFC". Working over a theory for finite sets, V is supposed to consist of only finite sets. (But there are models of our finite set theory containing infinite sets, so V can't be thought of as a concrete object in this way.) $\endgroup$
    – C7X
    Sep 23, 2022 at 7:00
  • $\begingroup$ Also, why we can't say sets are just the objects that may contain other objects, and V is the collection of them all: "naive set theory" is the point of view of V as some collection that just is, without being generated from some axioms, and it leads to Russell's paradox as there aren't any restrictions on what a set is, or any criteria to determine if some object is in V or not. $\endgroup$
    – C7X
    Sep 23, 2022 at 7:06
  • $\begingroup$ @SIJ About "even if we pass to V", sorry about that, it's now removed as it's not that important and may make it more confusing. I used it in order to make it seem like in the example we weren't assuming ZFC holds in V then cryptically dropping choice and becoming unable to prove the true statement "there exists a nonmeasurable set", instead I wanted to do it in a model so it would sound more clear to say "this is a model of ZF" instead of "let's drop the axiom of choice which we just assumed was true" $\endgroup$
    – C7X
    Sep 23, 2022 at 7:19

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