quantification domain of set theory formulas

Question

Let ZFC set theory, what is the domain of quantification of a formula like $\forall x\phi(x)$? If the domain is the whole Von Neumann Hierarchy $V$ why it is not a problem that it doesn't form a set?

Answer 1

You can see Kenneth Kunen, The Foundations of Mathematics (2009), page 16 :

[The context is an informal discussion of ] Axiom 1. Extensionality :

$\forall x y [\forall z(z \in x \leftrightarrow z \in y) \rightarrow x = y]$.

This says that a set is determined by its members, so that if $x,y$ are two sets with exactly the same members, then $x, y$ are the same set. Extensionality also says something about our intended domain of discourse, or universe, which is usually called $V$. Everything in our universe must be a set, since if we allowed objects $x, y$ which aren't sets, such as a duck ($D$) and a pig ($P$), then they would have no members, so that we would have

$\forall z[z \in P \leftrightarrow z \in D \leftrightarrow z \in \emptyset \leftrightarrow FALSE]$,

whereas $P, D, \emptyset$ are all different objects. So, physical objects, such as $P, D$, are not part of our universe.

Now, informally, one often thinks of sets or collections of physical objects, such as $\{ P, D \}$, or a set of ducks, or the set of animals in a zoo. However, these sets are also not in our mathematical universe. Recall that in writing logical expressions, it is understood that the variables range only over our universe, so that a statement such as "$\forall z \ldots $" is an abbreviation for "for all $z$ in our universe $\ldots$". So, if we allowed $\{ P \}$ and $\{ D \}$ into our universe, then $\forall z(z \in {P} \leftrightarrow z \in {D})$ would be true (since $P, D$ are not in our universe), whereas $\{ P \} \ne \{ D \}$.

More generally, if $x, y$ are (sets) in our universe, then all their elements are also in our universe, so that the hypothesis "$\forall z (z \in x \leftrightarrow z \in y)$" really means that $x, y$ are sets with exactly the same members, so that Extensionality is justified in concluding that $x = y$. So, if $x$ is in our universe, then $x$ must not only be a set, but all elements of $x$, all elements of elements of $x$, etc. must be sets.

See also, after the discussion of Russell's Paradox [page 18] :

Theorem 1.6.6. There is no universal set: $\forall z \exists R[R \notin z]$.

So, although we talk informally about the universe, $V$, we see that there really is no such object.

If we want to study models of $\mathsf {ZFC}$, we have to do it into a theory $\mathsf {ZFC}^+$ "stronger" than $\mathsf {ZFC}$, i.e. into a theory capable of proving the existence of a set $Z$ [which is an object of the universe of $\mathsf {ZFC}^+$] "large enough" to act as $V$ for $\mathsf {ZFC}$, i.e. a set of $\mathsf {ZFC}^+$ containing all the objects necessary to satisy the axioms of $\mathsf {ZFC}$.

Answer 2

Set theory doesn't happen in vacuum. There's still first-order logic in the meta-level (which is often either some set theory, or a weak number theoretic theory; depending on the philosophical bent of the mathematician).

The quantifiers are objects of the meta-theory, not of $\sf ZFC$. We define their meaning from outside of set theory.

What might be confusing is the fact that $\sf ZFC$ can "internalize" first-order logic, and reinterpret it as sets and define what is a structure and so one and so forth. In which case, a universal quantifier is defined as a set and is interpreted only on a given structure.

But the quantifiers in the axioms of $\sf ZFC$, or generally in the language of set theory, are not internal to the universe of set theory, but rather external and live in a larger universe (in case the meta-theory is a set theory), or they are syntactic objects (in case the meta-theory is a number theory). Those are two different "planes of existence".

Answer 3

I've found what I was looking for in these two answers regarding the same topic:

1. https://math.stackexchange.com/a/402387/113758
2. https://math.stackexchange.com/a/150237/113758