TITLE: Amenability and ultrafilters
QUESTION [14 upvotes]: Among hundreds of equivalent definitions of amenability (for discrete, countable, groups), I would like to discuss two which are most common:
A1. A group $G$ is amenable if it admits a Folner sequence. 
A2. A group $G$ is amenable if it admits an invariant mean. 
(See e.g. http://en.wikipedia.org/wiki/Amenable_group or http://terrytao.wordpress.com/2009/04/14/some-notes-on-amenability/) 
However, proofs of equivalence that I know (even for $G={\mathbb Z}$) require either axiom of choice or, at least, existence of a nonprincipal ultrafilter on ${\mathbb N}$. 
Question: Is there a proof that A1 $\iff$ A2 which uses only ZF axioms? Or, maybe $A1\iff A2$ 
implies existence of nonprincipal ultrafilters, maybe in a weakened form? 
This question was discussed a bit in Why are abelian groups amenable? and Why groups that admit Folner Sequences are amenable but not in the above form. 
Note: I am not a logician, but a geometric group-theorist and I frequently use ultrafilters. As the result I am often asked if the results could be proven without ultrafilters. For most proofs my answer usually is: "Yes, if you work much harder and write ugly and long proofs." However, I do not know the answer in the context of amenable groups.

REPLY [14 votes]: Of course, $ZF$ is enough to prove that $\mathbb{Z}$ has a Folner sequence. But, as you point out, $ZF$ is not enough to prove that $\mathbb{Z}$ has an invariant mean. Thus $ZF$ does not prove the equivalence of A1 and A2.
On the other hand, the Hahn-Banach Theorem is enough to prove the equivalence of A1 and A2 for countable discrete groups and Pincus-Solovay have constructed a model of $ZF$ in which the Hahn-Banach Theorem is true but there are no nonprincipal ultrafilters on $\mathbb{N}$. Hence the equivalence of A1 and A2 for countable groups does not imply the existence of nonprincipal ultrafilters on $\mathbb{N}$.