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Is $V$, the union of the von Neumann hierarchy, necessarily a proper class? Or is it only a proper class after you assume that it contains every set? (In that case, $V$ can't be a "set of all sets" because the foundation axiom implies a "set of all sets" can't exist.)

Restated as an analogy: Is it actually more like $V_\omega$, whose existence can't be proved or disproved until you assume the infinity axiom?

Or am I misunderstanding the meaning of "proper class" or something more fundamental?

(I have similar questions about the collection of all ordinals.)

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How do you propose to construct $V$ internally? Unions can only be taken over collections indexed by a set, and $V$ is a "union" over a collection indexed by the ordinals, which do not form a set (the Burali-Forti paradox: en.wikipedia.org/wiki/Burali-Forti_paradox). –  Qiaochu Yuan Apr 22 '11 at 21:49
    
@Qiaochu: While you are essentially correct, due to the well foundedness of the $\in$ relation one can approximate "as high rank as you want" internally, sort of how you can approximate an unbounded sequence by finite numbers high as you want. Furthermore, I think there might be something to say about that in the context of $V=L$, but I'm not completely sure and would rather to wait until Andres or Joel (or someone else who is more knowledgeable on the topic) find their way around here. –  Asaf Karagila Apr 22 '11 at 22:01
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@Asaf: yes, I am aware of this. The question is about all of $V$, not $V_{\alpha}$ for any particular ordinal $\alpha$. –  Qiaochu Yuan Apr 22 '11 at 22:04
    
@Qiaochu: I did not doubt that for a single moment, I did mention that because other readers might not be aware of this. –  Asaf Karagila Apr 22 '11 at 22:10
    
@Qiaochu: Can't I postulate its existence using a first-order statement? And in that case, wouldn't I discover that the formula can be neither proved nor disproved in ZFC? –  Neil Toronto Apr 22 '11 at 23:32

2 Answers 2

up vote 6 down vote accepted

Asaf has already given a good answer to your question but I'll add a second perspective.

You don't need to assume that every set belongs to the von Neumann hierarchy, since you can prove it. Suppose there were some set that didn't belong to this hierarchy. Then by Foundation, there is an $\in$-minimal such set, let's call it $x$. Every element of $x$ belongs to the von Neumann hierarchy, but $x$ itself (supposedly) does not. But if we let $\alpha = \sup \{ \mathrm{rank}(y)+1 : y \in x \}$, it's clear that $x \subset V_{\alpha}$ and so $x \in V_{\alpha + 1}$, meaning it does belong to the hierarchy after all.

In fact, if $M$ is any transitive model of $ZFC$, it has what it thinks are the operations of power set and union, and what it thinks is the class of all ordinals, and so $M$ can construct what it thinks is the von Neumann hierarchy, and this resulting hierarchy will equal $M$. So in particular, it'll think that its version of the von Neumann hierarchy forms a proper class.

But if $M$ were, say, the $\kappa ^{\mathrm{th}}$ level in the von Neumann hierarchy of some bigger model $N$ of set theory, where $\kappa$ is inaccessible in $N$, then $N$ will think of $M$ as a set whereas $M$ thinks of itself as a proper class. Along these same lines, $N$ will think of $\kappa$ as some ordinal whereas $M$ will think of $\kappa$ as the class of all ordinals. More interestingly, there will be subcollections of $M$ belonging to $N$ which aren't definable over $M$ (by a counting argument), thus they will be subcollections of $M$ which are neither sets in $M$ nor are they what $M$ would consider proper classes.

The moral is that, generally speaking, what gets considered a proper class and what doesn't depends on the context. Given a transitive model $M$ of ZFC, a proper class in $M$ is technically a formula $\phi (x, p)$ with parameters $p$ from $M$ such that there is no member of $M$ consisting of precisely all those members of $M$ which satisfy the given formula (according to $M$), i.e. $$M \not \vDash \exists y (x \in y \leftrightarrow \phi (x, p))$$ and informally it's the collection of those things in $M$ satisfying that formula, i.e. $$\{x \in M | M \vDash \phi (x,p)\}$$ but this collection may exist as a member of some larger model, or it may not.

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The thing is that a proper class is a collection of elements in a universe of set theory which is not a set, i.e. not an element of the universe.

It follows from the Russell's paradox that the collection of all sets is not a set, therefore a proper class.

However, suppose $V$ is a model of set theory, and $\langle M,E\rangle\in V$ is such that $M$ can be a model of set theory when thinking of $E$ as the epsilon relation for this model, then $M$ sees itself as a proper class, but $V$ - the larger universe - sees $M$ as a set.

On the other hand, consider $V$ is a model of set theory, it think of itself as a class, and consider its inner model $L$, the Godel universe of $V$, then $V$ thinks $L$ is a class as well, simply by the fact that the class of ordinals is embedded into $L$.

To conclude - a model of set theory always sees itself as a proper class. The way it sees other models, and they see it can be either as a set or as a class.

As for the collection of the ordinals this is simpler, as this is a well-ordered collection, so it must be isomorphic to some ordinal if it was a set, but then you have that it is a set which is a member of itself, therefore the collection of all ordinals is a proper class.

One more addendum: $V_\omega$ is indeed a model of $\mbox{ZFC}-\mbox{Infinity}$, however its existence as a set cannot be proved without the axiom of infinity, if you assume the existence of the empty set then you can build $V_\omega$ (by reiterating the power set operation) and have a model for the set theory of finitely-hereditary sets, which thinks of itself as a class, for the same reasons as above.

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