Is it circular to define the Von Neumann universe using "sets"?

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I was just reading the Wikipedia page on the Von Neumann universe, where it is stated that this universe "is often used to provide an interpretation or motivation of the axioms of ZFC." However, later on in the article, the Von Neumann universe is formally constructed using "sets," starting with "the empty set." This bothers me greatly, because as I understand it, ZFC is merely a theory describing how the "sets" in a model must behave; it is not itself a model.

For example ZFC, does not specify the particular object that the "empty set" is; it merely stipulates that its models must satisfy the axiom of the empty set (through either including this axiom directly or deducing it from the other axioms), i.e., the axiom,

$$ \exists x\forall y(\lnot y\in x) \text. $$

So, how is it possible to construct the Von Neumann universe from the very theory it's supposed to be a model for? Where does the definition of the empty set actually come in? I would appreciate any clarification.


Update: Asaf Karagila writes here that

If you want to consider the "simple" foundational approach to theories like $\sf ZFC$, then sets are primitive objects and $V$ is a given universe to begin with.

The axioms of $\sf ZFC$ tell you what sort of properties $V$ and its $\in$ relation satisfy. [...]

What the von Neumann hierarchy gives you is the understanding that if $V$ is already given, then we can write this wonderful filtration of $V$ into a very nice hierarchy. Additional theorems like the reflection theorem also tell you more about this hierarchy and its deep connection with the structure of $V$ as a universe of sets.

So am I correct in understanding that the Von Neumann universe is a construction that we put together once we assume that we have some model $V$ of ZFC?

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You cannot construct what is not there.

When we construct the real numbers from the rational numbers, we don't invent them out of thin air. We use material around us: sets of rational numbers (or sets of functions which are sets of sets of rational numbers). And similarly when we want to construct a dual vector space, we don't wave our hands and whisper some ancient texts from the Necronomicon ex Mortis. We use the sets at our disposal (and the assumptions they satisfy certain properties) to show we can define a structure with the wanted properties.

So the real numbers, and dual vector spaces, and all the other mathematical constructions, have existed in your fixed universe of sets before you began your work. What we do, if so, is not as much as constructing as we are defining them and using our axioms to argue that as the definition "makes sense" (whatever that means in the relevant context), such objects exist.

"Okay, Asaf, but what does all that have to do with my question?", you might be asking yourself, or me, at this point. Well, if you don't interrupt me, I might as well tell you.

The von Neumann universe is a way to represent a universe of $\sf ZF$ as constructed from below. But it is using the pre-existing sets of the universe. What is clever in this construction is that it exhaust all the sets of the universe. And if the universe only satisfied $\sf ZF-Reg$, then the result is the largest transitive class which will satisfy $\sf ZF$.

But what happens in different models of set theory? Well, we can prove from $\sf ZF$ that the von Neumann hierarchy, which has a relatively simple definition in the language of set theory, exhausts the universe. So each different model will have a different von Neumann hierarchy. And models which are not well-founded, will have a non-well-founded von Neumann hierarchy.

So yes, we first need a model of $\sf ZF$ in order to construct this hierarchy, but we don't need it inside the theory. We need it in the meta-theory. Namely, if you are working with $\sf ZF$, then you most likely assume it is consistent in your meta-theory, where you formalize your arguments and do things like induction on formulas. And that is enough to prove the existence of the von Neumann hierarchy; because once you work inside $\sf ZF$, the whole universe is given to you!