Can non-existent set still be infinite? Can we claim that there are infinite number of objects when the set of the objects does not exist?
For example, there is no set of all sets, but can we still say that there are infinitely many sets (of any kind)? And what would that mean? How would we say that formally?
One way is just to say: look, there are infinitely many subsets of integers, so, of course, there are infinitely many sets of any kind overall. But saying that, we must rely on something like "a superset of an infinite set is infinite" or something similar. And what would that mean to say that there are infinitely many sets?
Another example is from here: provide-different-proofs-for-the-following-equality. The set of all proofs of a given theorem is not defined, but we can clearly describe an infinite set of such proofs (which is NOT a subset of the set of all proofs because such a set does not exist). Can we still claim that there are infinitely many proofs of the theorem, and what would that claim mean exactly?
P.S. If you are aware of any literature discussing this or similar questions, you may just guide me through the literature by providing some references. I would definitely appreciate it.
 A: Let change a bit the question. What does mean finite ?
There are several definitions. One of them is that a set is finite if it has an injection in a finite ordinal (a finite ordinal is an element of the first limit ordinal). Of course, one can extend this definition to collections (a collection is a predicate with one free variable) of set. And it is clear that if a collection has a injection into a finite ordinal, then the collection is itself a set. Indeed, form the injection, ou can construct a bijection onto a smaller ordinal, and then the collection is the image of the ordinal under the inverse map, hence it is a set (replacement axiom).
So if a collection is not a set, it is not finite in this sense.
Another definition for a set (or collection) to be finite is that every injection of the collection into itself is surjective. For sets, and with dependent choice axiom, this is equivalent to the former definition. Things are a bit more complicated, but I think the conclusion should be the same. I can see a proof with the foundation axiom : let C be a collection, finite in that sense. Then every set $C\cap V_\alpha$ is finite. Consider the non-decreasing map $\alpha \mapsto \operatorname{card} C\cap V_\alpha$, form ordinals to finite ordinals. This map is eventually constant, so there exists an $\alpha$ such that $C\cap V_\alpha = C$. Thus, $C$ is a set.
A: Euclid proved there are an infinite number of primes without ever knowing set theory. He formalized it as follows: no (finite) list of primes contains all primes. You can apply similar formalization to all of your examples, showing that no finite list of set contains all sets of no finite set of proofs contains all proofs.
A: Yes, of course. Finitists will argue that it doesn't make sense to talk about infinite sets, or that they don't actually exist. But this argument mostly an exercise in pedantry and doesn't impact the mathematics at all. Of course in the physical universe, we have limitations that prevent us from looking at infinite collections of objects, but mathematics doesn't study the physical universe.
When we talk about some infinite set, like for example "the set of all even numbers", there are two possible interpretations: The classical interpretation (what you're thinking of) is you are discussing the infinitude of even numbers taken as a whole. Whether or not this "actually exists" is an ontological question outside the scope of mathematics. The finitist interpretation is that you are talking about the set that includes every even number; one way to think about it would be to think about the limiting properties of "the set of even numbers less than N" as N gets large. In this sense, the set exists as an idea (a limit of finite sets), even though there do not exist infinitely many even numbers, every even number that exists or could hypothetically exist is being considered when you talk about the set of all even numbers. In either interpretation, the theorems you'll come up with are the same, so practically speaking there's no difference.
