Explicit Construction of a Function from any $A\in2^{\mathbb{R}}\setminus{\{\emptyset}\}$ to $A$

Question

The axiom of choice states that for any set of nonempty sets $X$ there is exists a function $f \colon X \rightarrow \bigcup X$ so that $\forall A \in X \, ( f(A) \in A )$. Constructing such function for $X=2^{\mathbb{N}}\setminus{\{\emptyset}\}$, the set of all nonempty subsets of $\mathbb{N}$, is easy (by taking the minimum of the group $A$). Simillar ideas apply for constructing an explicit function from $2^{\mathbb{Z}}\setminus{\{\emptyset}\}$ and $2^{\mathbb{Q}}\setminus{\{\emptyset}\}$. Is there a known construction of such function for $2^{\mathbb{R}}\setminus{\{\emptyset}\}$, i.e., a function that given any nonempty subset $A\subseteq\mathbb{R}$ returns some element in $A$?

Answer 1

This cannot be done in any good way: it as Wojowu says, it is consistent with ZF that no such function exists at all (let alone a nicely definable one).

One way to see this is to note that there is a model of ZF with an infinite Dedekind-finite set of reals (Cohen's original model of the failure of AC has this property, for example). But the existence of a choice function on $2^\mathbb{R}\setminus \emptyset$ is incompatible with this: if such an $f$ existed and $D\subseteq\mathbb{R}$ is infinite, we get an $\omega$-sequence of distinct elements of $D$ given recursively by $$a_0=f(D), a_{i+1}=f(D\setminus\{a_0,...,a_i\}),$$ demonstrating that $D$ is not Dedekind-finite.

Of course this blackboxes Cohen's result, but that's fairly necessary at this point: his proof (and forcing/symmetric submodels arguments in general) is quite hard.