I hope the following questions are not fundamentally stupid to raise.
Let A be a class term in the form of $ A = \{ x | \phi \}$.
Claim $1$: $A$ is a proper class in ZFC if and only if it is also a proper class in ZF.
As I try to prove this, this claim came up:
Claim $2$: Let $x$ and $y$ be sets. Then $x \neq y$ in ZF if and only if $x \neq y$ in ZFC.
Are these claims true? Any hints?
Thanks!
The statement you're trying to prove is false: let $\varphi(x)$ be the formula $$(AC\wedge x\not=x)\vee \neg AC.$$ From AC, it follows that $\varphi$ defines a set (namely $\emptyset$), while from $\neg$AC it follows that $\varphi$ defines a proper class (namely everything). In particular, ZFC proves that $\varphi$ defines a set, while ZF does not.
Meanwhile, in your claim 2, you conflate theories and models. A specific set isn't in a theory. Rather, what you need to talk about are formulas defining sets. Then, as above, the claim is false: let $\psi$ be the formula "is the empty set" and $\theta$ be the formula "is the emptyset if AC holds, and is $\omega$ otherwise".
There is a difference between "in ZF" and "in a model of ZF". And that difference is huge.
You can talk about definitions for sets, e.g. $\{x\mid x=\varnothing\land\sf AC\}$ is a definition which ZF proves to be a set, but it is empty if and only if the axiom of choice fails. On the other hand, $\varnothing$ is defined as $\{x\mid x\in x\}$, and ZF proves that it is always empty.
On the other hand, if $x$ and $y$ are two elements of $M$, a model of ZF, then they are equal if and only if they are equal in any larger or smaller model which knows both of them (and it is transitive with respect to $M$'s membership relation). So that much would be true.
Of course, since classes are in fact "formulas" interpreted in some model, it is easy to write a formula which defines a set in some models and a class in others: e.g. $\{x\mid x=x\land\sf AC\}$. Which is a proper class if and only if the axiom of choice holds.
