Is Axiom of Choice for convex sets of distributions on naturals necessary? Take any family $(S_i)_{i∈I}$ such that each $S_i$ is a convex set of functions $f : ℕ→[0,1]$ where $\sum_{k∈ℕ} f(k) = 1$. By "convex" we mean that for any $f,g∈S_i$ and any $a,b∈[0,1]$ such that $a+b=1$ we have $a·f+b·g∈S_i$. By AC (axiom of choice) there is a choice function for $S$. But I get a feeling that AC can be eliminated. Can it?
If I am not mistaken, we can construct a $g : ℕ→[0,1]$ uniquely definable from $S_i$ such that each finite prefix of $g$ agrees with some member of $S_i$, say by inductively defining $g(k)$ to be $\max(\frac12(p+q),q-2^{-k})$ where $p,q$ are the $\inf,\sup$ of the interval of possible values of $f(k)$ for $f∈S_i$ that extends $g↾ℕ_{<k}$ (it is an interval due to convexity). But we may not have $g∈S_i$ (there is a simple counter-example), so this does not work.
So is AC actually necessary here? If so, can it be eliminated if we have some stronger condition than convexity? For instance, what if we require that $S_i$ is closed under weighted combination, meaning that for every $w : S_i→[0,1]$ satisfying $\sum_{f∈S_i} w(f) = 1$ we have $h∈S_i$ where $h : ℕ→[0,1]$ given by $h(k) = \sum_{f∈S_i} ( w(f)·f(k) )$? My counter-example for the $g$ above does not work for this stronger condition, but I am still unable to  prove that $g∈S_i$ without using AC...
 A: The existence of a choice function for the set of non-empty convex sets of distributions on $\mathbb N$ is equivalent to the existence of a well-ordering of $\mathbb R.$ This is true even if we strengthen convexity by allowing weighted combinations. The well-orderability of $\mathbb R$ is Form 79 in Howard-Rubin’s “Consequences of the Axiom of Choice” and is known not to follow from ZF, unless ZF is inconsistent.
In one direction, a well-ordering of $\mathbb R$ easily gives you a choice function for the set of non-empty sets of functions $\mathbb N\to[0,1],$ which is sufficient.
For the other direction, assume we have a function $c$ so that $c(S)\in S$ whenever $S$ is a non-empty set of distributions on $\mathbb N$ closed under weighted combinations. We will construct a choice function for the set of non-empty subsets of $\mathbb R$ (Form 79A).
For any non-empty subset $A$ of $\mathbb R$ let $D_A$ denote the set of distributions on $A,$ i.e. functions $f:A\to [0,1]$ with $\sum_{x\in A} f(x)=1.$ The main steps are:

*

*$D_A$ is closed under weighted combinations

*There is a weighted-sum-preserving injection $\Phi_A:D_A\to D_{\mathbb N}$

*From an element $f\in D_A$ we can pick out an element $r(f)\in A$
So $r(\Phi_A^{-1}(c(\Phi_A[D_A])))$ is a choice of element of $A.$ These constructions can be done uniformly in $A$ so that the axiom of choice is not invoked.
Point 1 is obvious from the definition.
For point 2, first inject $D_A$ to $[0,1]^{\mathbb Q}$ by taking $f\in D_A$ to the cumulative distribution function $F$ evaluated at rationals, $F(q)=\sum_{a\leq q}f(a).$ Then use a bijection $\mathbb Q\cong\mathbb N$ to inject to $[0,1]^{\mathbb N}.$ Finally, take $x\in [0,1]^{\mathbb N}$ to the function $y\in D_{\mathbb N}$ defined by $y(n+1)=x(n)2^{-n-1}$ and $y(0)=1-\sum_{n\in \mathbb N} x(n)2^{-n-1}$ (assuming $0\in\mathbb N$).
For point 3, take the maximum modal value, in symbols $r(f)=\max \{x:f(x)=\max f\}$
