TITLE: Why is there no Borel function mapping every countable set of reals outside itself?
QUESTION [19 upvotes]: A choice function maps every set (in its domain) to an element of itself. This question concerns existence of an  anti-choice function defined on the family of countable sets of reals. In an answer to a question about uncountability proofs it was suggested that while Cantor's diagonal method furnishes a Borel function mapping each countable sequence $S$ of real numbers to a number not in the seqence, the same is not true for countable sets of reals. This seems surprising (but believable) to me and an important insight, if true. But why is it true? Put more provocatively (if this is fair, if not explain it as stated)

Let $\mathcal{C}$ be the family of countable subsets of $\mathbb{R}$ .`  Is it the case that for every Borel function $f:\mathcal{C} \rightarrow \mathbb{R}$ there is an $X \in \mathcal{C}$ such that $f(X) \in X$?

REPLY [4 votes]: There is no known forcing-free proof of the following strong version. (Here $E_{cntble}$ is yet another name for the equivalence relation that others call $T_{2}$ or $=^{+}$.)
Theorem Suppose that $E$ is a countable Borel equivalence relation on the standard Borel space $Z$ and that $f: {\mathbb{R}}^{\mathbb{N}} \rightarrow Z$ is a Borel homomorphism from  $E_{cntble}$ to $E$. There there exists $z \in Z$ such that for every $r \in \mathbb{R}$, there exists $x \in  {\mathbb{R}}^{\mathbb{N}}$ with $r \in \{ x_{n} \mid n \in \mathbb{N}\}$ such that $f(x) \mathbin{E} z$.
On the other hand, the analogous statement is false for Borel homomorphisms from $E_{cntble}^{+}$ to $E_{cntble}$. This suggests that there does not exist an easy forcing proof of the general form of Friedman's theorem.