Are sets in ZFC that aren't in certain models of ZF without AC included as classes or some other Platonic concept?

Question

The Banach-Tarski theorem says you can split a sphere into five pieces with certain properties; that is, there's a map from the sphere to {1, 2, 3, 4, 5}, or in other words a subset of the set $\Re \times \Re \times \Re \times \{ 1, 2, 3, 4, 5 \}$. Platonically, all subsets of a set should exist, but this one doesn't if one is not working in certain models of ZF that don't support the Axiom of Choice. So does this subset exist as a class or some other form in these models, or am I being led astray by my biases in what sets exist?

Answer 1

A subclass of a set is a set. This is the essence of bounded comprehension (or separation) which is part of the axioms of $\sf ZF$.

Arguably, if the Banach–Tarski paradox fails, there are two options:

1. You can add by forcing a bijection between $\Bbb R$ and $\omega_1$ without adding new reals, in which case you bring back more than the necessary choice for proving the Banach–Tarski paradox, and thus introduce "solutions to the Banach–Tarski partition". And therefore in some sense the answer to your question is positive, but in a technically more difficult sense.

2. You cannot add such sets using forcing, which tells you something very deep. In particular, it means, that Dependent Choice must fail (as $\sf DC$ would imply the first scenario holds), and that measure theory as a whole is kinda shoddy; and if you want to introduce something like this, you will end up adding more real numbers to your universe as well, and the whole thing sort of falls apart.

Your bias comes from the fact that Platonism is incompatible with independence results. But let's go one step below. If you only accept $\Bbb Q$ to exist, you could argue that $\sqrt2$ doesn't exist. You can still describe it pretty well, though, so you might argue that it does exist, and thus extend your universe a bit. But now what about $e$ or $\pi$? Well, you could argue that they are somehow "geometric numbers" and therefore exist, so you extend your universe a bit more, again.

But now what about Liouville's number, or Chaitin's $\Omega$, and so on? We start to run into more numbers that "don't yet exist", even though we have some "vague descriptions of them". So you can say that all of these things exist, great. What about $\sqrt{-1}$, though?

Existence is always relative to a model. Platonistic naivete makes you think that everything is entirely decided, but it's not. It's just that you've decided some things and not others, and you ignore the other things. But different Platonists might disagree with you (e.g., finitism and ultrafinitism or intuitionistic and constructivists schools of thought).

The solution, of course, is to say that you believe certain things are true, and you simply don't know the answers for others. If you believe that the universe satisfies $\sf ZFC$, then you can either believe that there are inner models of $\sf ZF$ with certain properties, or not. If there are, then the sets which exist, exist in the full universe, but maybe the inner models do not know them (this is similar to $\sqrt2$ not being in $\Bbb Q$).

But maybe you don't believe that. Maybe you only want to accept $V=L$, which implies no inner models exist, so any class model of $\sf ZF$ is the whole universe. In that case, you're going to have to think about set-models of $\sf ZFC$ which may or may not exist, again, depending on your Platonistic beliefs. And if they do exist, then by the virtue of being set-models, they definitely "miss out" on most objects in the universe.