TITLE: Existence of mathematical objets constructed using the axiom of choice
QUESTION [9 upvotes]: Let consider the Vitali set $V \subset \mathbb R$, which is constructed using the axiom of choice. (I could take any other mathematical "object" that can be constructed using the axiom of choice, but I chose the Vitali set just to make the purpose clearer).
On the one hand, the Vitali set exists thanks to the axiom of choice. On the other hand, the power set $\mathcal P (\mathbb R)$ exists in ZF without assuming AC. In some sense, the existence of any $A \in \mathcal P (\mathbb R)$ is independant on (i.e. does not require) AC.
But then, I had the following reflection : the existence of the Vitali set $V \in \mathcal P (\mathbb R)$, as an element of this power set, should not require the axiom of choice... There must be some fallacy here ; that's why I would like some clarifications.
To be concise : can we say that the axiom of choice "affects" the existence of elements in $\mathcal P (\mathbb R)$ (as it seems to be the case for the existence of the Vitali set), although the construction of $\mathcal P (\mathbb R)$ (in ZF) has nothing to do with AC?
Thank you!

REPLY [2 votes]: It is important to distinguish between actual objects (in our case 'sets') and definitions thereof. While we can prove in $\operatorname{ZF}$ (with or without choice) that every model of $\operatorname{ZF}$ has to contain an object that satisfies (inside this model) the definition of '$\mathcal P(\mathbb R)$', this doesn't tell us the whole story of what this particular object actually is.
(In the following, let's say that we defined $\mathbb R$ to be the set of all functions $f \colon \omega \to \omega$. It doesn't really matter, but we have to be a bit careful as how to define the reals as a set for some absoluteness reasons... Let's also take $\operatorname{ZFC}$ as our background theory.)
For example: Let $(M; \in)$ be a countable transitive model of (a sufficiently large fragement of) $\operatorname{ZF}$. Then there will be some $x \in M$ such that
$$
(M; \in) \models 'x = \mathcal P (\mathbb R)'
$$
(by which I mean that $M$ thinks that $x$ is the powerset of the reals). As $M$ is countable and transitive, we have that $x \subseteq M$ and thus $x$ has to be countable as well. But (this is the point where it matters how we defined the reals as a set) from the point of view of our background universe, $x$ really consists of subsets of reals, so $x$ is a subset of the 'true' powerset of the reals. It's just the case that it misses most subsets. In particular, $x$ may or may not contain some $y \in x$ such that
$$
(M; \in ) \models 'y \text{ is a Vitali set}'
$$
If $(M; \in)$ satisfies choice, then there will be indeed be such a $y \in x$.