I'm interested in knowing whether the idempotent ultrafilter proof of Hindman's theorem actually requires the full axiom of choice.
At first I wondered if the theorem itself at all required any choice, but even Neil Hindman himself said that the elementary proof he came up with was hard to read, so I didn't want to get into that, but as it turns out this answer (Examples of theorems that haven't been proven without AC in practice but can be proven without it in principle) already answers this : Hindman's theorem can be proved in second order arithmetic so doesn't require any use of AC.
But if we focus on the easier proof with an idempotent ultrafilter $\mathcal{U}$ on $\mathbb{N}$, then we can try to locate where AC is used. If I'm not mistaken it's used only to show : $\gamma\mathbb{N}$ is a nonempty compact topological space ($\gamma X $ denotes the set of nonprincipal ultrafilters ); any semitopological compact semigroup has an idempotent.
The first statement, although it requires some amount of choice, doesn't require full AC, it only requires the ultrafilter lemma, or equivalently, the BPI, which is known to be strictly weaker than AC (although unprovable in ZF, if the latter is consistent - note that here Tychonov doesn't require AC, since $2$ is Hausdorff)
So essentially my question boils down to : is there a "choice principle" X (BPI, ACD, countable choice,...; formally one could say "a statement that is provable in ZFC) such that BPI+X is strictly weaker than AC, and ZF+ BPI+X proves that any semitopological compact semigroup has an idempotent ?
Another way to put it : does BPI + "any semitopological compact semigroup has an idempotent" imply AC ?
