Could you please sharpen that? Are you saying that it's false that 𝔐 ⊨ "𝐴 is finite" iff for finitely many 𝑎 ∈ |𝔐|, 𝔐 ⊨ 𝑎 ∈ 𝐴 unless we restrict to standard transitive models?

@Trebor Actually it's enough to restrict attention to $\omega$-models, but this is a very minor point; your overall response is absolutely correct (and should be an answer).

@Trebor just to check I’ve understood: you’re saying we could have the following: (i) M says that $A$ is uncountable; and (ii) M says of only finitely many things $a$ in its domain that $a$ is a member of $A$? I will just point out that $A$ need not even be a set, since M may be a nonstandard model, so I don’t know what “$A$ can be uncountable” means.

@ac2357 Trebor is saying (correctly) that we can have a model $\mathfrak{M}$ with an element $A$ such that $\mathfrak{M}$ thinks that $A$ is countable but in reality (i.e. in $V$) $A$ is uncountable. A bit more precisely, despite $A$ being truly uncountable there is in $\mathfrak{M}$ an injection from $A$ to what $\mathfrak{M}$ thinks is $\omega$.

@NoahSchweber that is not anything to do with what I’m asking. I’m aware that “countable” is not absolute in the sense that “finite” is absolute. My question is about “finite”.

@ac2357 Sorry, I mistyped - "countable" in my previous comment should be "finite" (and "injection from $A$ to $\omega^\mathfrak{M}$" should be "injection from $A$ to an element of $\omega^\mathfrak{M}$"). The situation is really the same, which is why I mixed up the two: basically, any "unbounded" notion is going to behave extremely flexibly between arbitrary models due to compactness. You only get absoluteness for finiteness if you restrict attention to $\omega$-models (or further, e.g. transitive models).

Okay, that’s helpful thanks. I still see no role for “in reality” here, since in reality $A$ may not even be a set. The two clauses either side of the “iff” are about what the model says.

@ac2357 The right hand side of the iff very much does refer to reality: "for finitely many" is not interpreted in $\mathfrak{M}$ but in $V$. If you really want to avoid reference to a real state-of-affairs, here's one way to phrase things: ZFC proves the statement "if there is a (set) model of ZFC in the first place, then there is a (set) model $M$ of ZFC with an element $A$ (which is also a set) such that $M\models$ "$A$ is finite" but $A$ is uncountable." The statement in quotes can be phrased directly in the language of set theory, and the parentheticals are implicit in the quantifiers.