# Is V, the Universe of Sets, a fixed object?

When I first read Set Theory by Jech, I came under the impression that the Universe of Sets, $V$ was a fixed, well defined object like $\pi$ or the Klein four group. However as I have read on, I am beginning to have my doubts. We define

\begin{align} V_0 & :=\emptyset. \\[10pt] V_{\beta +1} & :={\mathcal {P}}(V_\beta). \\[10pt] V_\lambda & :=\bigcup_{\beta <\lambda} V_\beta \text{ for any limit ordinal } \lambda, \end{align}

and finish by saying $$V:=\bigcup_{\alpha \in \operatorname{Ord}} V_\alpha.$$

However this definition seems to create many problems. I can see at least two immediately: First of all, the Power Set operation is not absolute, that is it varies between models of ZFC. Secondly (and more importantly) this definition seems to be completely circular as we do not know a priori what the ordinals actually are. For instance, if we assume some two, mutually inaccessible large cardinals $\kappa , \kappa'$ to exist, and model ZFC as $V_\kappa, V_{\kappa'}$ respectively, then we get two completely different sets of ordinals! So we seem to be at an impasse:

In order to define the Universe of Sets we must begin with a concept of ordinals, but in order to define the ordinals we need to have a concept of the Universe of Sets!

So my question is to ask: Is this definition circular? The only solution I can think of is that when we define $V$, we implicitly assume a model of ZFC to begin with. Then after constructing the ordinals in this model, we construct $V$ off of them, so to speak. Is this what is being assumed here?

• The way I'd like to think about these things is: we always work in a model of ZFC to start with (called V). Then you can develop ordinals. After that you can define $V_\beta$. The assertion that $V=\bigcup_{\beta\in ORD} V_\beta$ should really be thought of as a theorem. It is an easy consequence of the axiom of foundation. What I mean here is if we can out the axiom of foundation from ZFC, it can be shown that $V=\bigcup_{\beta\in ORD} V_\beta$ is equivalent to the axiom of foundation. – Jing Zhang Feb 28 '18 at 23:54
• Wars have been started for less. – Asaf Karagila Mar 1 '18 at 0:00
• Also, what exactly is $\Bbb N$? – Asaf Karagila Mar 1 '18 at 0:02
• I always liked $\mathbb{V}=\{x:x=x\}$ for my universe, lowercase symbols denoting sets; no need for any mention of ordinals or a hierarchy, just the first order language of sets and the class building axiom schemata. This is used, for example, in Donald Monk's MK class theory book. – Alec Rhea Mar 1 '18 at 3:21
• Please don't use $...$ for emphasising text when you write, that's what markdown is for (or \emph in LaTeX). – David Roberts Mar 1 '18 at 5:45

As you noticed, the iterative conception of sets requires a pre-existing universe of sets, and ordinals with which we can label the stages. So if you work within ZFC itself, in other words within an existing model of ZFC, you can perform that iterative construction to obtain $V$. Like Asaf Karagila says here, you cannot get nothing from nothing. Typically, in set theory you work in ZFC, where you have ordinals, and construct $V_k$ for each ordinal $k$. Note that $V$ is not a set, but the entire universe (the one you are working in).

I still think your question is mainly philosophical, your comment to Nik Weaver notwithstanding. After all, you ask:

First of all, the Power Set operation is not absolute, that is it varies between models of ZFC.

Of course, since ZFC proves that $V$ is the whole universe, different models of ZFC will have different $V$. If ZFC is consistent, then it has a countable model $V$. That should not be surprising; one can never 'pin down' the set-theoretic universe, not to say using $V$. The same issue shows up with the natural numbers, as Asaf alluded to in a comment; second-order PA with full semantics does not 'pin them down' because our meta-system MS must always be computable, and hence even if MS has (proves existence of) a full semantics model of second-order PA, there are models of MS whose interpretation of the naturals are not isomorphic (if there are any models at all).

In order to define the Universe of Sets we must begin with a concept of ordinals, but in order to define the ordinals we need to have a concept of the Universe of Sets! So my question is to ask: Is this definition circular?

We cannot define anything, much less whatever is meant by a "universe of sets", without already working under some assumptions. It is not necessary that you work in ZFC, but what other alternative meta-system do you have in mind? Remember that to construct $V$ you need all those set-theoretic operations that you are using, plus the collection of ordinals, and any meta-system that supports all these is going to look very much like ZF or some extension of it.

Boolos noted the same philosophical circularity in this paper (page 15) (which I rephrased to the language used here and emphasized some points):

There is an extension of the stage theory from which the axioms of replacement could have been derived. We could have taken as axioms all instances of a principle which may be put, 'If each set is correlated with at least one stage (no matter how), then for any set $z$ there is a stage $s$ such that for each member $w$ of $z$, $s$ is later than some stage with which $w$ is correlated'.

This bounding or cofinality principle is an attractive further thought about the interrelation of sets and stages, but it does seem to us to be a further thought, and not one that can be said to have been meant in the rough description of the iterative conception. For that there are exactly $ω_1$ stages does not seem to be excluded by anything said in the rough description; it would seem that $V_{ω_1}$ (see below) is a model for any statement that can (fairly) be said to have been implied by the rough description, and not all of the axioms of replacement hold in $V_{ω_1}$. (*) Thus the axioms of replacement do not seem to us to follow from the iterative conception.

(*) Worse yet, $V_{δ_1}$ would also seem to be such a model. ($δ_1$ is the first uncomputable ordinal.)

To put it more cogently, if you take for granted the power-set operation as a primitive, and start with the empty-set, and also take for granted the ability to consolidate into a single operation the union of any iteratively generated sequence of sets, which may itself be used iteratively to generate more sets, then what you can generate appears to be entirely contained within $V_{δ_1}$. And if you additionally allow taking union of arbitrary (countable) and potentially indescribable sequences, then what you can generate is still contained within $V_{ω_1}$.

The crux is that you cannot generate $V_{ω_1}$ without essentially having $ω_1$. And this corresponds to two logical facts: that there is no countable sequence of ordinals before $ω_1$ whose union is $ω_1$, and that $V_{ω_1}$ is a model of ZF with replacement restricted to countable sequences.

More philosophically, if you envision the stages as being generated rather than pre-existing, then necessarily you cannot generate stage $ω_1$ until you have generated all the stages corresponding to countable ordinals. But there is no way to generate all countable stages without having a generation process that already has length at least $ω_1$. And since $ω_1$ does not appear in any stage up to $V_{ω_1}$, you have no choice but to assume the ability to 'run a generation process' of length $ω_1$ if you want to obtain $V_{ω_1}$ and further stages, which implies that the iterative conception cannot give ontological justification for the existence of $ω_1$.

Just to add, it is true that uncountable well-orderings do appear much earlier than $V_{ω_1}$, but the very fact that $ω_1$ does not appear even at stage $ω_1$ (union of all prior stages) should be a warning that one should not consider all well-orderings of the same length to be on equal footing. In particular, to have a well-ordering as a binary relation on a set that makes it totally ordered with no strictly descending sequence is not the same as being able to iterate along it.

Perhaps someone may find a non-circular way to justify ZFC philosophically, but the iterative conception seems to get us no further than countable replacement.

• Yeah, I'm also sure of that. Sometime after my death, someone will scour the data dumps from SE and compile these into something along the lines of Cantor's letters to Dedekind or whatever. (Okay, seriously now, I hope this never happens.) – Asaf Karagila Mar 1 '18 at 10:09
• @AsafKaragila: For you, the popular quote might be something like "I can suggest that you take comfort that in mathematics the intuition you have initially (or hoping to develop "immediately") is almost always wrong. − Asaf 2015". – user21820 Mar 1 '18 at 10:11
• Ha! I didn't even remember writing that. Good quote. – Asaf Karagila Mar 1 '18 at 10:26
• Is this $\delta_1$ another notation for $\omega_1^{CK}$? – Pedro Sánchez Terraf Mar 2 '18 at 11:07
• @PedroSánchezTerraf: $δ_1$ is just the notation that Boolos used in the cited paper, and is indeed $ω_1^{CK}$ in modern notation. – user21820 Mar 2 '18 at 13:19

This seems like more of a philosophy of math question than a proper math question. However, in the past Mathoverflow has often been tolerant of such questions.

The basic concern is that the universe of all things there are surely cannot itself be a separate thing. Various responses have been given. On the iterative conception of sets, as it is usually expressed, there is no "completed" $V$, but rather an unending series of stages which are built up iteratively in a process which can never be completed. The obvious objection is that an abstract platonic object is not something which can be "built", nor can it appear in "stages" if it is timeless. One may then be told that the language about building in stages is merely metaphorical, which is not so satisfying.

Russell considered $V$ to be "self-reproductive" in the sense that "we can never collect all of the terms having the said property into a whole, because, whenever we hope we have them all, the collection which we have immediately proceeds to generate a new term also having the said property". Of course the language about collecting into a whole and generation are still dubious, but Dummett reformulated the idea in terms of "indefinitely extensible" concepts where it is one's conception of $V$, not $V$ itself, which keeps changing. However, if you look at it carefully you discover that in order to make sense of the idea of indefinite extensibility you already have to understand the difference between sets and proper classes, which is what the idea was supposed to explain.

I don't think there are settled answers to these questions. Joel Hamkins' work on the "set-theoretic multiverse" is a provocative recent approach which might appeal to you. My view on these matters is given in the last chapter of my book Truth & Assertibility --- in brief, my position is that there is a well-defined class of all sets, but there is no class of all classes because classes cannot be explicitly listed out, as sets (in principle) can. In other words, classes can only be presented indirectly by linguistic expressions, which creates the possibility of there being expressions whose status as representing a class cannot be decided. This means that reasoning about proper classes demands the use of intuitionistic logic and renders the concept "class of all classes" illegitimate. This view is worked out in detail in my book mentioned above.

• See also Hamkins' recent work on set-theoretic potentialism jdh.hamkins.org/set-theoretic-potentialism which takes the ideas in your second and third paragraphs very seriously. – David Roberts Mar 1 '18 at 5:47
• Thanks Nik. I think you misunderstood the questions slightly (I have edited it to make it more clear). I am not primarily asking a philosophical question. The question I had was: In order to define $V$ we have to use the class of ordinals. But the class of ordinals depends on what Universe of Sets we are working in. Therefore we seem to have a circular definition. – Elie Ben-Shlomo Mar 1 '18 at 9:15
• @ElieBen-Shlomo: What, in Jech's text, have you read about Godel's constructible universe $L$? – Thomas Benjamin Mar 1 '18 at 9:43
• I have read his chapter on $L$ showing the consistency of AC and GCH relative to ZF. Why do you ask? – Elie Ben-Shlomo Mar 1 '18 at 11:52
• @ElieBen-Shlomo: As you are probably aware, every model $M$ of $ZF$ (and of course $ZFC$) will have $L$ as its smallest inner model (so $L$ for $M$ will contain all the ordinals of $M$). Since $L$ for $M$ is a proper class relative to $M$, this proper class $L$ can be generated by the Godel operations (in fact, the ordinals of $M$ can be generated by the Godel operations as well). I would argue that $L$ and the ordinals generated by the Godel operations are, in fact, the pre-existing universe of sets and ordinals spoken of in user21820's answer to you. – Thomas Benjamin Mar 1 '18 at 15:41

First of all, I think that part of the confusion stems from talking about "models of ZFC." My recommendation, if you want to sort out what's going on, is to start by forgetting what a model of ZFC is. That way madness lies.

Having cleared our minds of madness, I will agree with you that there is something subtle going on with the very last step $V := \bigcup_{\alpha\in Ord} V_\alpha$. The subtlety derives from the fact that $V$ is not a set. Therefore if you want to conceptualize it as a "fixed, well-defined object like $\pi$" then you should adopt something like NBG or Morse–Kelley, which allows you to talk about proper classes as entities in their own right. The class existence axioms of these systems are what let you define things like the class of all sets without any circularity.

If we insist on using ZFC, then the way to understand the definitions you cited is as follows. You can define ordinals (though not the class of all ordinals) using the set existence axioms of ZFC. That is, you can give a mathematically precise definition of "$\alpha$ is an ordinal" but you cannot define the set of all ordinals (and in ZFC, the only things your axioms tell you exist are sets). Similarly, each individual $V_\alpha$ makes sense. However, the very last step $V := \bigcup_{\alpha\in Ord} V_\alpha$ cannot be conceived of as taking a set-theoretic union to form a new set. You must either think of it as an informal, non-rigorous "definition" or come up with some formal way to handle "proper classes." In Kunen's book on set theory, he adopts the subterfuge of "stepping outside the system" and defining proper classes as formulas. I won't say more because I don't want to go mad; suffice it to say that Kunen's approach lets you stick with ZFC, but at the cost of declaring that $V$ is not a "well-defined object like $\pi$."

There is a "philosophical" way to interpret your question, which is how ZFC can say that there exist sets without first saying that there is a universe of sets and picking things inside of that universe, but I think that this is not really your question.

• I am just curious: why do you characterize Kunen's approach as "madness"? I'm only a student in this area, and haven't read much beyond Kunen's book, but this approach seemed sensible to me. – Nate Eldredge Mar 2 '18 at 5:28
• @Nate I would actually insist that talking about models of ZFC is the best way to keep oneself sane (i.e. the opposite of madness). Though this might sidestep the philosophical issue at hand (by declaring classes to be only formulas in disguise). – Pedro Sánchez Terraf Mar 2 '18 at 11:22
• NBG (at least with AC rather than global choice) is no different from treating proper classes as formulae, because each class is defined by a formula over ZFC. In some sense NBG is simply ZFC but where you can talk about proper-classes inside rather than outside the system, so I don't see a conceptual difference. As for MK, it seems more madness to me, because if we can't even imagine a model of ZFC without going mad, then it seems a worse madness to imagine quantifying over all classes... No? =) – user21820 Mar 2 '18 at 13:22
• @TimothyChow: Fair enough. For me personally, I felt that the approach of saying that classes are formulas in disguise was helpful in resolving my confusion, rather than adding to it. – Nate Eldredge Mar 2 '18 at 17:04
• @TimothyChow: Well, I would say that $V$ is the formula $\varphi(x)$ which says "there exists an ordinal $\alpha$ such that $x \in V_\alpha$." Then it's a theorem of ZFC that $\forall x \varphi(x)$. In other words, it's not inherently $x=x$, but it's a consequence of some axioms that the two formulas are equivalent. And in some weaker set theory without foundation, say $ZFC^-$, we can say $V \vDash \mathsf{ZFC}$. This is the metatheorem that every axiom of ZFC, relativized to $V$, is a theorem of $ZFC^-$. – Nate Eldredge Mar 3 '18 at 0:01

For completeness, it should be mentioned that yet another answer could have been: yet, in some foundational approaches, $V$ literally is an object (albeit not a "fixed" one, whatever that means), namely, an 'initial object' (in the usual sense) of the category of 'ZF-algebras' (with ZF-algebra-morphisms). You can find more on this here and in the book

Laura Crosilla, Peter Schuster, From Sets and Types to Topology and Analysis: Towards Practicable Foundations for Constructive Mathematics Oxford Logic Guides, Volume 48, Oxford University Press, 2005 ISBN 9780198566519

Basic remarks to make are that $V$ is the ZF-algebra for the endofunctor with object-classfunction $x\mapsto \{x\}$, and the class $\mathrm{Ord}$ is the ZF-algebra for the endofunctor with object-classfunction $\beta\mapsto\{\beta,\{\beta\}\}$, and that the class-function $\mathrm{Ord}\to V$, $\beta\mapsto V_\beta$ is the object-function of a functor which is the right-adjoint of the unique functor $V\to\mathrm{Ord}$.

I believe the answer to your question revolves around correcting a subtle confusion between classes and sets in the Cumulative Hierarchy. This can be shown by reference to Samuel Coskey's Senior Thesis, "Partial Universes and the Axioms of Set Theory", found under title on the Web on SemanticScholar. This thesis shows which $$ZFC$$ axioms hold at each stage of the Cumulative Hierarchy. Here is a short synopsis of his results:

Axioms that always hold (at each stage of the hierarchy): Extensionality, Foundation, Union, Axiom Schema of Separation, Choice.

Axioms that holds in $$V_{\alpha}$$ iff $$\alpha$$ $$\gt$$ 0: Empty Set.

Axioms that hold in $$V_{\alpha}$$ iff $$\alpha$$ $$\gt$$ $$\omega$$: Infinity.

Axioms that hold at limit ordinals: Power Set, Pairing.

Axioms that hold at inaccessible cardinals: Replacement.

(Note that although the existence of the empty set can be treated as an axiom, it can also be derived from Separation as follows:

{$$y$$ $$\in$$ $$x$$ | $$y$$ $$\ne$$ $$y$$})

Coskey shows, in his thesis:

Theorem 5.8. The Empty Set axiom holds in $$V_{\alpha}$$ iff $$\alpha$$ $$\gt$$ 0.

Proof. $$V_0$$ = $$\emptyset$$ (= { }), so nothing postulating the existence of a set holds. On the other hand, $$\emptyset$$ $$\in$$ $$\mathcal P$$($$\emptyset$$) = $$V_1$$.

Before I continue further, let me recall Elie's conundrum as stated in his question:

In order to define the Universe of Sets we must begin with a concept of ordinals, but in order to define the ordinals we need to have a concept of the Universe of Sets!...So my question is to ask: Is this definition circular?

Coskey's Theorem 5.8 seems (at least to me) to suggest that the conundrum is real because (to repeat)

$$V_0$$ = $$\emptyset$$, so nothing postulating the existence of a set holds. On the other hand, $$\emptyset$$ $$\in$$ $$\mathcal P$$($$\emptyset$$) = $$V_1$$.

yet $$V_0$$ = $$\emptyset$$ seemingly postulates the existence of the 'empty set' outright. Since one can derive the Empty Set Axiom from Separation, it is instructive to look at Coskey's proof that Separation always holds at each stage of the cumulative hierarchy:

Axiom 6 (Separation). If $$x$$ is a set and $$y$$ is a class such that $$y$$ $$\subset$$ $$x$$, then $$y$$ is a set.

Theorem 5.6. The Separation axiom schema holds in all $$V_{\alpha}$$

Proof. Suppose $$x$$ is an element of $$V_{\alpha}$$. If $$y$$ is any class at all which is a subset of $$x$$, then we have $$y$$ $$\subset$$ $$x$$ $$\in$$ $$V_{\alpha}$$ an so by Corollary 4.6 [If $$y$$ $$\subset$$ $$x$$ and $$x$$ $$\in$$ $$V_{\alpha}$$, then $$y$$ $$\in$$ $$V_{\alpha}$$], $$y$$ $$\in$$ $$V_{\alpha}$$ as well.

But now consider the proof at $$V_0$$ = $$\emptyset$$. By Coskey's definition of $$\subset$$, $$\emptyset$$ $$\subset$$ $$\emptyset$$ so that $$\emptyset$$ $$\in$$ $$\emptyset$$, which seems to defy Foundation. With this in mind, I now consider Coskey's rendition and proof that Foundation holds for all $$V_{\alpha}$$:

Axiom 10 (Foundation). If $$x$$ is a set, then there is a $$z$$ $$\in$$ $$x$$ such that $$z$$ $$\cap$$ $$x$$ = $$\emptyset$$

Theorem 5.2. The Axiom of Foundation holds for all $$V_{\alpha}$$.

Proof. If $$x$$ $$\in$$ $$V_{\alpha}$$, then there is a $$z$$ $$\in$$ $$x$$ such that $$z$$ $$\cap$$ $$x$$ = $$\emptyset$$. Since $$V_{\alpha}$$ is transitive, in fact $$z$$ $$\in$$ $$V_{\alpha}$$ as well.

Again, consider the proof of Theorem 5.2 for $$V_0$$. Since $$V_0$$ = $$\emptyset$$, obviously by definition of $$\emptyset$$ ($$\existsy\forallz$$($$z$$ $$\notin$$ $$y$$)), the proof of foundation fails for $$V_0$$ but the empty set axiom ensures that $$\emptyset$$ $$\notin$$ $$\emptyset$$. In fact, if one were to assume that $$\emptyset$$ $$\in$$ $$\emptyset$$, this would contradict Coskey's Proposition 4.11

Proposition 4.11. For every $$\alpha$$, we have $$\alpha$$ $$\notin$$ $$V_{\alpha}$$, but $$\alpha$$ $$\in$$ $$V_{\alpha+1}$$. In other words, $$rank$$($$\alpha$$)= $$\alpha$$.

What Proposition 4.11 shows is that (if one views the cumulative hierarchy as being generated rather than preexisting) the $$V_{\alpha}$$ (where $$\alpha$$ is either a successor or limit ordinal) which has been currently generated should be deemed a proper class (since no class $$V_{\alpha+1}$$ for which $$V_{\alpha}$$ $$\in$$ $$V_{\alpha+1}$$ has yet been generated) until $$V_{\alpha+1}$$ has been generated, and only then should be deemed a set (since when $$V_{\alpha+1}$$ has been generated, $$V_{\alpha}$$ $$\in$$ $$V_{\alpha+1}$$). Propositon 4.11 also shows that Extensionality, Foundation, Union, Separation schema, and choice should be classified as axioms that hold in $$V_{\alpha}$$ iff $$\alpha$$ $$\gt$$ 0 (since as Coskey rightly points out for the Empty Set axiom at $$V_0$$, "nothing postulating the existence of a set holds", and at $$V_0$$, $$\emptyset$$ = $$V_0$$ is a proper class). The reason that the $$V_{\alpha}$$ should be deemed proper classes at the point of generation is because of the following definition (courtesy of Olivier Esser from his paper, "On the Consistency of a Positive Theory", Mathematical Logic Quarterly, Vol. 45, No. 1 (1999), pp. 105-116):

"$$a$$ is a set" $$\Leftrightarrow$$ $$\existsy$$( "$$y$$ is a class" $$\land$$ $$a$$ $$\in$$ $$y$$)

Note that by setting Extensionality, Foundation, Union, Axiom Schema of Separation, and Choice to hold at $$V_{\alpha}$$ iff $$\alpha$$ $$\gt$$ 0, one has the Cumulative Hierarchy definable in ZFC (the Wikipedia article, "Von Neumann universe", claims that Judith Roitman, in her book, Introduction to Modern Set Theory [2011 editon,pg 79], states without reference that the realization that the Axiom of Foundation is equivalent to the equality of the universe of $$ZF$$ sets to the Cumulative Hierarchy is due to Von Neumann). Note also, however, that the existence of the Cumulative Hierarchy does not imply what ordinals and cardinals actually exist. For example, if one assumes the existence of a strongly inaccessible cardinal $$\kappa$$, it can be shown that $$V_{\kappa}$$ is a model of ZFC. However, by Theorem 4.11, $$\kappa$$ $$\notin$$ $$V_{\kappa}$$ (but by Theorem 4.11, $$\kappa$$ $$\in$$ $$V_{\kappa + 1}$$ and the aforementioned Wikipedia entry states that $$($$ $$V_{\kappa}$$, $$\in$$, $$V_{\kappa + 1}$$ $$)$$ when $$\kappa$$ is strongly inaccessible is a model of $$MK$$ set theory). This shows that $$\kappa$$ is a proper class relative to $$V_{\kappa}$$ and a set relative to $$V_{\kappa+1}$$. However, $$V_{\kappa + 1}$$ is now a proper class and one needs to justify the move from $$V_{\kappa}$$ to $$\mathcal P$$($$V_{\kappa}$$) = $$V_{\kappa + 1}$$ since $$V_{\kappa}$$ is a model of $$ZFC$$ (if one sets $$\kappa$$ for $$V_{\kappa}$$ to be the least inaccessible, $$V_{\kappa}$$ is a model of $$ZFC$$ + "There is no inaccessible cardinal" and such a $$V_{\kappa}$$ can be considered to be the universe of all sets $$V$$) and $$\mathcal P$$ is only defined relative to $$\kappa$$ (in fact, since $$\kappa$$ is a strong limit, one has that for every $$\beta$$ $$\lt$$ $$\kappa$$, $$2^{\beta}$$ $$\lt$$ $$\kappa$$ so that the power set operation definable in $$V_{\kappa}$$ is not the power set operation needed to define $$\mathcal P$$($$V_{\kappa}$$) = $$V_{\kappa + 1}$$). As has been mentined before, if one sets $$\kappa$$ for $$V_{\kappa}$$ to be the least inaccessible cardinal (call it $$\kappa_0$$ for $$V_{\kappa_0}$$), $$V_{\kappa_0}$$ is a class model of $$ZFC$$ + "There are no inaccessible cardinals" and can be rightly called $$V$$.

But what of $$V$$, the proper class of all sets? By the Burali-Forti paradox, $$V$$ must always be a proper class (which means that one cuts off the cumulative hierarchy at $$\kappa_0$$ or at some inaccessible $$\kappa$$). However, one could possibly imagine $$\mathcal P$$($$V$$) (the class of all subclasses of $$V$$) so that $$V$$ $$\in$$ $$\mathcal P$$($$V$$). In order to do this, however, one must have a set/class theory like Ackermann set theory. Fotunately, one has the Levy theory (or the Levy-Vaught interpretation of Ackermann set theory, as it is called) which is $$ZFC$$ + $$V_{\kappa}$$ $$\prec$$ $$V$$ + $$\kappa$$ inaccessible. Joel David Hamkins, in his answer to my mathoverflow question, " Forcing in Ackermann set theory", has this to say regarding the Levy theory:

Ackermann set theory is a version of set theory where one views the set-theoretic hierarchy as continuing far past the the construction of sets, into the construction of classes, classes of classes and so on. In the Ackermann theory, it is as though one is building the full $$V_{\alpha}$$ hierarchy, but then part-way through one finds a particularly robust $$V_{\delta}$$ and declares its elements to be real "sets", with everything above $$\delta$$ declared "classes". (Critics would say that Ackermann's sets are only some of the sets, since his classesbehave fundamentally like sets.)

As Francois [Dorais--my comment] points out in his comments, however, the Ackermann theory seems to provide less than what one may want in the realm of classes, a weakness in the theory that is addressed by its natural strengthenings to various set theories in a more $$ZFC$$-like context. Namely, the Levy theory is $$ZFC$$ + $$V_{\delta}$$ $$\prec$$ $$V$$ + $$\delta$$ is inaccessible, where $$V_{\delta}$$ $$\prec$$ $$V$$ is the scheme asserting $$\forallx$$ $$\inV_{\delta}$$ ($$\varphi$$($$x$$) $$\Leftrightarrow$$ $$\varphi({x})^{V_{\delta}}$$), which is expressible in the language of set theory augmented with the constant symbol $$\delta$$. The set $$V_{\delta}$$ here plays the part of $$V$$ in Ackermann's theory, and so every model of the Levy scheme is a model of Ackermann set theory, if one regards the elements of $$V_{\delta}$$ as the official "sets" and the sets above $$V_{\delta}$$ as the "classes". But the Levy theory asserts more than Ackermann, because not only is the collection of sets existing as an object in the theory, but also it is an elementary substructure of the full universe. In addition, the levy theory has a fuller treatment of classes, making them more set-like, in that the larger universeabove $$\delta$$, which correspond to the classes of the Ackermann theory, actually satisfy $$ZFC$$.

For my part (at least), I would like to set $$\delta$$ for $$V_{\delta}$$ as equal to $$\kappa_0$$, the least inaccessible cardinal. That way, the natural models $$V_{\kappa}$$ of $$ZFC$$ + "There are no inaccessible cardinals, $$ZFC$$ + "There exists an inaccessible cardinal", $$ZFC$$ +There are two inaccessible cardinals", etc., all correspond to proper classes (and by Godel's Second Incompleteness Theorem are all separate, distinct theories). As can be easily seen (I think, since Esser's criterion for distinguishing sets from classes holds for the Levy theory), using the Levy theory as a metatheory (or, perhaps better, a schema for generating models for increasingly more comprehensive metatheories) dissolves Elie's conundrum.

How? Well, consider that the metatheories set the context in which a universe of sets can be defined; in fact, the metatheories define the models, a.k.a the "pre-existing universe of sets and ordinals" (as user21820 states in his answer) which are necessary for the cumulative hierarchy to exist and function, but only in a virtual sense (for example, the metatheory defines the power set operation and the union operation by which the cumulative hierarchy can be defined, and the initial set, the empty set, on which the power set operation can operate). It is, however, the cumulative iterative construction which actually forms the model $$V_{\kappa}$$. However, what it forms is an internal cumulative hierarchy cut off at the inaccessible cardinal of your choice (e.g., $$V_{\kappa_0}$$ without the existence of $$\kappa_0$$, which is, of course, the natural model of $$ZFC$$ + "There are no inaccessible cardinals"). In this aforementioned example, $$V_{\kappa_0}$$ = $$V$$, so $$V_{\kappa_0}$$ is a proper class subject to the Burali-Forti paradox. Note, though, that I have kept (for reference purposes only) the subscript $$\kappa_0$$ for $$V_{\kappa_0}$$ (though in actuality I should properly refer to $$V_{\kappa_0}$$ in this case as $$V$$). By keeping $$\kappa_0$$ as an index for $$V_{\kappa_0}$$, one implicitly assumes that there exists a model $$V_{\kappa_1}$$ where $$\kappa_0$$ $$\lt$$ $$\kappa_1$$ and $$\kappa_1$$ is also inaccessible in which $$\kappa_0$$ is a set (and therefore a set model of $$ZFC$$ + "There is no inaccessible cardinal"), or (in the Levy theory) a proper class. In fact, one has the following rule:

$$R1$$. If one assumes the existence of an inaccessible $$\kappa_{i}$$ where $$i$$ $$\in$$ $$Ord$$, one implicitly assumes the existence of the inaccessible $$\kappa_{i + 1}$$.

This follows from the following theorems found in Jeroen Hekking's Bachelor's Thesis, "Natural Models, Second-order Logic & Categoricity in Set Theory":

Theorem 3.1.2. A cardinal $$\kappa$$ is inaccessible iff $$\mathcal M_{\kappa}$$ $$\vDash$$ $$ZFC^2$$.

Proposition 3.1.1. For all Henkin models $$<$$ $$\mathcal M$$, $$\mathcal G$$ $$>$$ in $$\sigma$$ satisfying $$ZFC^2$$ we have $$\mathcal M$$ $$\vDash$$ $$ZFC$$ [where $$\sigma$$ is the signature consisting of the symbol $$\in$$--my comment. A Henkin model is a pair $$<$$ $$\mathcal M$$, $$\mathcal G$$ $$>$$ with $$\mathcal M$$ a first-order model and $$\mathcal G$$ a collection of relations and functions satisfying second-order choice and second-order comprehension. The latter determines the range of our second-order variables and contains, by comprehension, all second-order definable functions and relations on $$\mathcal M$$. If we take all relations and functions on $$\mathcal M$$ we get a full second-order model, which will be denoted simply by $$\mathcal M$$ (this is a direct quote of Hekking's definition of Henkin model, and his second-order deductive system consists of Second-order Choice, Second-order Comprehension, and "some straight-forward rules for manipulating quantifiers and logical connectives as given in Shapiro's Foundations without foundationalism; A case for second-order logic, (1991), pg. 66"--my comment also). Note also that full models are Henkin models as well].

Theorem 3.2.2. Let $$\mathcal M$$ be a full model in $$\sigma$$ of $$ZFC^2$$. Then $$\mathcal M$$ is uniquely isomorphic to the natural model $$\mathcal M_{\kappa}$$ with $$\kappa$$ $$\cong$$ $$O^{\mathcal M}$$ [where $$O^{\mathcal M}$$ is the class of ordinals of $$\mathcal M$$. $$\kappa$$ is defined as the ordinal height of $$\mathcal M$$--my comment].

Corollary 3.2.3 (External Semi-categoricity). The theory $$ZFC^2$$ is semi-categorical with respect to full models. That is, for any two models $$\mathcal M$$, $$\mathcal N$$ $$\vDash$$ $$ZFC^2$$ in $$\sigma$$ we can uniquely embed $$\mathcal M$$ as an initial segment into $$\mathcal N$$ [that is, if the ordinal height of $$\mathcal M$$ is less than the ordinal height of $$\mathcal N$$--my comment], or the other way around.

(Here is the proof: By 3.2.2. and 3.1.2., $$O^{\mathcal M}$$ $$\cong$$ $$\kappa$$ so that $$O^{\mathcal M}$$ is inaccessible (so for the least inaccessible cardinal $$O^{\mathcal M_0}$$ $$\cong$$ $$\kappa_0$$, where $$\mathcal M_0$$ $$\vDash$$ $$ZFC$$ + "There are no inaccessible cardinals, $$\kappa_0$$ must exist as a set or as a proper class in the Levy theory in order that Replacement holds in $$\mathcal M_0$$). Since $$rank$$($$O^{\mathcal M}$$)= $$\kappa$$, $$O^{\mathcal M}$$ $$\in$$ $$V_{\kappa + 1}$$, so for $$V_{\kappa + 1}$$ = $$\mathcal P$$($$V_ {\kappa}$$), one of necessity needs to assume (because $$\mathcal P$$($$x$$) can only be defined at limit ordinals) that there exists a larger inaccessible cardinal $$\kappa^{'}$$ (i.e. $$\kappa$$ $$\lt$$ $$\kappa^{'}$$ so that by Corollary 3.2.3, $$\mathcal M_{\kappa}$$ is an initial segment of $$\mathcal M_{\kappa^{'}}$$, i.e., $$V_{\kappa}$$ is an initial segment of $$V_{\kappa^{'}}$$ and Replacement will hold for $$V_{\kappa^{'}}$$). But then by the previous argument, there must of necessity exist a larger inaccessible cardinal $$\kappa^{''}$$, etc. in order to keep the power set operator defined for the entire cumulative hierarchy (since the $$\kappa$$ for the $$V_{\kappa}$$ for $$\kappa$$ inaccessible are ordinals, they themselves can be well-ordered and can be indexed by ordinals,hence the theorem is proved--note again that for the least inaccessible cardinal $$\kappa_0$$, it must exist for the simple reason that if $$\kappa_0$$ is inaccessible then Replacement holds for $$V_{\kappa_0}$$, which then is a model of (by Proposition 3.1.1) $$ZFC$$ + "There are no inaccessible cardinals"--note also that if one sets $$V_{\kappa_0}$$ to be the class of all sets $$V$$ and any larger inaccessible $$V_{\kappa^{'}}$$ to be 'proper classes', then one has a model of the Levy theory since Corollary 3.2.3 shows that $$V_{\kappa_0}$$ $$\prec$$ $$V$$). Since it is known that the largest of the large cardinals can be expresed in terms of elementary embeddings, it behooves me to find out what elementary embeddings represent strongly inaccessible cardinals.

In the Cantor's Attic entry, "Elementary Embedding", one finds the following under the subheading "Use in Large Cardinal Axioms":

There are two ways of making the critical point as large as possible:

1. Making $$\mathcal M$$ as large as possible, much larger than $$\mathcal N$$ (meaning that a "large" class can be embedded into a smaller class)
2. Making $$\mathcal M$$ and $$\mathcal N$$ more similar(for example, $$\mathcal M$$ = $$\mathcal N$$ yet $$j$$ is nontrivial)

Using the first method, one can simply take $$\mathcal M$$ = $$V$$ (the universe of all sets), and the resulting critical point is always a measurable cardinal, a very strong type of cardinal, e.g. the first measurable is larger than infinitely many weakly compact cardinals (and much more).

Using the second method, one can take, say, $$\mathcal M$$ = $$\mathcal N$$= $$L$$, i.e. create [a non-trivial--my comment] embedding $$j$$: $$L$$ $$\rightarrow$$ $$L$$, whose existence has very important consequences, such as the existence of $$0^{\sharp}$$ (and thus $$V$$ $$\neq$$ $$L$$) and implies that every or dinal that is an uncountable cardinal in $$V$$ is strongly inaccessible in $$L$$. By taking $$\mathcal M$$ = $$\mathcal N$$ = $$V_{\lambda}$$, i.e. a rank of the cumulative hierarchy, one obtains he very powerful rank-into-rank axioms, which sit near the very top of the large cardinal hierarchy. However, this second method has its limits, as shown by Kunen, as he showed that $$\mathcal M$$ = $$\mathcal N$$ = $$V$$ leads to an inconsistency with the axiom of choice, a theorem now known as the Kunen inconsistency. He also showed that a natural strengthening of the rank-into-rank axioms, $$\mathcal M$$ = $$\mathcal N$$ = $$V_{\lambda+2}$$ for some $$\lambda$$ $$\in$$ $$Ord$$, was inconsistent with the $$AC$$.

Most large cardinal axioms in between measurables and rank-into-rank axioms are obtained by mixing those two methods: one usually sets $$\mathcal M$$ = $$V$$ then requires $$\mathcal N$$ to satisfy strong closure properties to make it "larger", i.e. closer to $$V$$ (that is, to $$\mathcal M$$). For example, $$j$$: $$V$$ $$\rightarrow$$ $$\mathcal N$$ is nontrivial with critical point $$\kappa$$ and the cumulative hierarchy rank $$V_{j(\kappa)}$$ is a subset of $$\mathcal N$$ then $$\kappa$$ is superstrong; if $$\mathcal N$$ contains all sequences of elements of $$\mathcal N$$ of length $$\lambda$$ for some $$\lambda$$ $$\gt$$ $$\kappa$$ then $$\kappa$$ is $$\lambda$$-supercompact, and so on.

The existence of a nontrivial elementary embedding $$j$$: $$\mathcal M$$ $$\rightarrow$$ $$\mathcal N$$ that is definable in $$\mathcal M$$ implies that the critical point $$\kappa$$ of $$j$$ is measurable in $$\mathcal M$$ (not necessarily in $$V$$). Every measurable ordinal is weakly compact and (strongly) inaccessible therefore its existence in any model is beyond $$ZFC$$, meaning that $$ZFC$$ cannot prove that such a cardinal exists [Note that at least according to Noah Schweber, "...my impression is that when we say '$$\kappa$$ is ... $$I_0$$' we mean '$$\kappa$$ is the critical point of an $$I_0$$ embedding,' and this is always inaccessible (and measurable, and etc.)...Regardless, even if you refer to the rank level of the embedding_, the property 'is the critical point of...' is equiconsistent and does define an inaccessible...." so (if Noah is correct) that even the largest known large cardinal axiom fits into the pattern $$j$$: $$V$$ $$\rightarrow$$ $$M$$ and also into the pattern $$\mathcal Z$$ = {$$V_{\kappa}$$ | $$\kappa$$ inaccessible} satisfying the Universe Axiom for $$V_{\kappa}$$ $$\vDash$$ $$ZFC^2$$ needing, for $$V_{\kappa}$$ to be a set (or a proper class according to the Levy theory), even larger inaccesible $$\kappa$$'s above it. This suggests, by the theorems for $$ZFC^2$$ listed above, that $$\mathcal Z$$ = {$$V_{\kappa}$$| $$\kappa$$ inaccessible} is the 'universe $$V$$' of $$ZFC^2$$, which, by the above argument, can never be completed. (Note also that if Choice fails above $$I0$$ in such fashion that above $$I0$$, $$ZF$$ + "There exists a Reinhardt cardinal" holds, then it can be shown in $$ZF$$ that the Reinhardt cardinal $$\kappa_{Reinhardt}$$ is inaccessible [according to Prof. Hamkins in his answer to Tim Campion's mathoverflow question, "Does $$Con$$($$ZF$$ + Reinhardt) really imply $$Con$$($$ZFC$$ + $$I0$$)?"] so that the 'choiceless cardinals' can be elements of $$\mathcal Z$$ = $$V^2$$ $$\vDash$$ $$ZF^2$$ (and by Proposition 3.1.1 of Hekking, of $$V$$ $$\vDash$$ $$ZF$$) as well, lending credence to the view that $$V^2$$ (and therefore, by Proposition 3.1.1, $$V$$) is not fixed according to height.).

• Why the downvote? – Thomas Benjamin Oct 24 '18 at 6:07

I come a little late, just to clearly state and sum up something that was began to be outlined by some here. Namely, my syntacticalist view that nothing in math or formal sciences in general is a "fixed object" unless its's an actual syntactic object; otherwise it's just a Concept in one(s)'s mind(s), possibly formalized into actual syntax and/or conceived from actual syntax. And the Concepts themselves are subjectively as diverse as conceivers, only given a same name if those agree enough about their intuitions. So here, the Concepts of π, the Klein Four-Group or the Von Neumann Universe aren't more fixed than by traditional common shared intuition. But, where the question becomes interesting (to me) and your answer quite true (as your intuition had guessed), is once rephrased given all the above. Sort of: "Has the Concept of Von Neumann Universe any concrete syntactical counterpart, an actual avatar, such as π or the Klein Four-Group?" And the answer indeed is: not really. Yes π has any algorithmic Recursive Enumeration of its digits in any fixed base. Yes Klein's Group has his Cayley Table. Though V only has, as said Nate Eldredge in a comment, its defining Predicate - as a Unary Relation Symbol for e.g. ZFC, as a Constant Symbol of Class for e.g. NB. However as this phrase with a hyphen shows, as a defining Formula it really is itself well-"defined" only relatively to some fixed Formal System, for which it actually defines something local. Finally, it doesn't alone have an actual independent existence that could unequivocally manifest all its features; only as a String of Symbols devoid of most of the answers to the questions one can ask about V as a Concept, not at all intrinsically tied to this intuition of ours, enough so to embody it into an actual reality.

I might argue that it is indeed a fixed object, but in a different sense. Let me preface by saying that I (am not an expert in set theory, but also) have been saying to others for years something to the effect of: we may be better served by removing the Axiom of Regularity/Foundation from ZFC. In order to explain why, let me first take a small detour into the realm of Turing machines.

One remarkable concept that any CS major probably sees (disclaimer: I wasn't one) is that there exist Turing machines that can output their own descriptions. I will skip the details of returning descriptions of inner functions (subroutines) and assert that such a machine should essentially have code of the form $C+\text{str}(C)$ where $+\text{str}(C)$ means add string of the code to the code. The idea here is that the instructions $C$ should include something to the effect of

output the code of a machine whose code is defined by

stripping $\text{str}(C)$ of delimiters (i.e. quotes, etc) (thus giving you the raw $C$)

appending $\text{str}(C)$ to the end of it as some variable

Clearly the output then describes a Turing machine that is identical (up to isomorphism) to the Turing machine that created it. (Technically I should say that the output is a string and not a code, and that some other Turing machine (possibly the one that created it) wants to simulate it as a subroutine, say, for iterating, and can then pick up that string, strip the quotes (which then makes it code), and then execute.)

Such a Turing machine may be construed as a fixed point with respect to an implementation function: $I(C+\text{str}(C))=C+\text{str}(C).$ And also note there is no circular problem here; we simply write the code twice, but put quotes around it the second time (and assign it to some variable). So it's not like $\text{str}(C)$ has some infinite length to it or anything (at least not literally, but semantically it will).

Now you may notice this form $C+\text{str}(C)$ is eerily similar to the form of a successor set in set theory: $s(x)=x\cup\{x\}.$ If we abandoned the Axiom of Regularity and considered a cyclic set (containing itself) of the form $A=\{\text{elements of }A,A\},$ then it is readily seen to be a fixed point for the successor function $s.$ In particular, the set $N:=\{N,0,1,2,...\}$ would also satisfy the Axiom of Infinity (in addition to the ordinal $\omega$) but would also be a successor fixed point (whereas $\omega\neq\omega+1$ is not a fixed point). Of course, ZFC$^-$ (ZFC minus Regularity) cannot prove the existence of cyclic sets (without being inconsistent), but they are consistent relative to ZFC$^-$.

The universe though should also be closed under the power set and union operations. Indeed if $V=\{V,\textsf{all other sets}\}$ (so that, in effect, a "set of all sets" could be discussed) then it would be closed under successor, power set, and union operations (as it would have anyway with Regularity). But it now also self-referentially resolves the paradox of needing set theory to generate set theory: set theory becomes precisely the thing needed to describe itself. In particular, if I think of a choice function on $V$ as something analogous to a Turing machine choosing a subroutine to implement, then $c(V)=V$ becomes analogous to a Turing machine simulating itself as a subroutine (or, say, set theory being used to rigorously construct set theory---after which point we call the first one a meta'' one, simply to distinguish the two identical copies). In this interpretation, one can "step outside the system" without having to step outside the system, which is a win-win.

• "I might argue that it is indeed a fixed object" It doesn't seem that you do in this answer, though, unless I'm missing something. So while this is interesting, I don't think it answers the question asked here. – Noah Schweber Jul 5 '18 at 13:39
• If $V$ is the "set of all sets" (so we'd have to be in $ZFC^-$), then for any set-theoretic operation $O$ we're interested in (power set, successor, etc), we'd have $O(V)=V.$ That is, it is a fixed object in the sense of being a fixed point. – Andrew Aug 10 '18 at 19:32
• $ZFC^-$ does not permit a set-of-all-sets either. If $a$ is any set, separation tells us that $\{b\in a: b\not\in b\}$ is a set, but it clearly can't be in $a$; so it is separation, not powerset, which kills the possibility of a universal set. Note that New Foundations with urelements has powerset and a universal set but not separation, and is known to be consistent (relative to a tiny fragment of ZFC). (And I still don't see how this argues for a "single $V$;" what prevents there from being two distinct things with the properties you describe?) – Noah Schweber Aug 10 '18 at 19:44