# Krein Milman theorem without the axiom of choice

The Krein-Milman theorem asserts that in a locally convex topological vector space, a nonvoid compact convex subset is the closed convex envelope of its extreme points. But I would like to know when it is possible to avoid using Zorn's lemma, in more simple settings (outside of $\mathbb{R}^n$), because for example, in ergodic theory, one can prove quite easily with this theorem that if $X$ is a compact metric space, and $f:X\rightarrow X$ is continuous, and $M_f(X)$ is the set of $f$-invariant probability measures, then the $f$-invariant, ergodic probability measures are the extreme points of $M_f(X)$. The point is that one can prove the existence of $f$-invariant probability measures whithout the axiom of choice.

• Do you know this paper by Bell and Fremlin? – Asaf Karagila Jan 21 '15 at 10:49
• Should I post it as an answer? Or maybe you should (I have to go now for a while) – Asaf Karagila Jan 21 '15 at 11:03

In fact, this paper asserts that the assertion : "For every normed vector space $X$ over the reals, there exists at least one extreme point in the unit ball of the continuous dual of $X$" is equivalent to the axiom of choice. So it doesn't matter how nice the space $X$, we can't prove the Krein-Miman for $M(X)=C(X)'$ in ZF. As a corollary, the ultrafilter lemma and the Krein Milman theorem imply (so is equivalent to) the axiom of choice, so the Krein Milman theorem cannot be provable only with the ultrafilter lemma, as the ultrafilter lemma is not equivalent to the axiom of choice. In particular, it shows that we can't prove the Krein Milman theorem without at least a weak form of axiom of choice, and as the ultrafilter is not enough, maybe we really need the axiom of choice in full generality, but the paper doesn't answer this question.
• @ François G. Dorais. This means that we don't have an answer finally, if the $(*)$ assertion must hold for every normed vector space, this doen't mean that we can't prove the Krein-Milman theorem in simple settings (like my example) without using the axiom of choice. The observation is clear when we look more closely at the proof of $(*)$ implies $AC$. I will edit the answer later. – Paul-Benjamin Jan 21 '15 at 22:16