Is there in ZF a choice function for the set of all countably infinite subsets of the reals?

Question

Is there in ZF set theory without the axiom of choice, a choice function for the set of all countably infinite subsets of $\mathbb{R}$? Or, is there a model of ZF where there is no choice function for that set?

Answer 1

As Asaf Karagila observes, the answer to your question is negative since the existence of a Vitali set (which is not provable in $\mathsf{ZF}$ alone) is a consequence of the choice principle you describe. However, it's worth noting that we can get a $\mathsf{ZF}$-provable choice principle by controlling the "shape" of the countable sets of reals involved. Specifically, in the construction of a Vitali set, the countable sets in question each "look like $\mathbb{Q}$," and this turns out to be necessary in the following sense:

$\mathsf{ZF}$ does prove that there is a choice function for the set of nonempty sets of reals which have isolated points.

Proof: Let $\mathcal{I}$ be the set of nonempty sets of reals with isolated points, and fix a list $(a_i,b_i)_{i\in\omega}$ of the rational open intervals. For each $X\in\mathcal{I}$, let $i$ be the smallest such that $(a_i,b_i)\cap X$ has exactly one element and let $p_X$ be that element; then the function $X\mapsto p_X$ is the desired choice function. $\quad\Box$

So we only run into trouble when our sets of reals are "$\mathbb{Q}$-like," as in the construction of the Vitali set.