>>10825541There is a countable transitive model of ZFC (assuming it is first-order logic).
What does this mean?
You can think of any axiomatic theory as an algebra, i.e. you have a bunch of axioms, and you prove these axioms are true for some set.
Peano arithmetic, for instance, is an axiomatic theory of arithmetic, and the usual is a model for it, therefore everything you can prove in Peano arithmetic is true in . But there are of course plenty of models of arithmetic.
Same thing goes for set theory. There are models of ZFC, i.e. sets which satisfy all the axioms of ZFC. The models of ZFC are of course defined inside ZFC, so you can see it as a russian doll containing itself. Moreover, there are transitive models of ZFC, i.e. models where the "" relation is modelled by itself. In a transitive model, you have the set of integers, intersection of sets is the real intersections, and everything behaves like the real stuff up to a point.
Take a transitive model of ZFC. It is infinite because you have all the integers inside, and it can be uncountable. But the set of theorems of ZFC is countable (because a proof is a bunch of text, which is countable), so the theorem of Löwenheim-Skolem states there are elements of the model you can remove without breaking any of the theorems. In fact you can end up with a countable model of ZFC.
This may seem to contradict the Cantor diagonal argument, because, if everything is countable, then is countable. Actually, is said to be uncountable because there is no one-to-one correspondence between and , which is the case in the countable model: some of the bijections between and itself have been removed from the model.
Philosophically speaking, it could mean there is only one infinity, and the different infinite cardinals only show the absence of a bijection.
