TITLE: Non-principal ultrafilters on ω
QUESTION [13 upvotes]: I thought I had heard or read somewhere that the existence of a non-principal ultrafilter on $\omega$ was equivalent to some common weakening of AC. As I searched around, I read that this is not the case: neither countable choice nor dependent choice are strong enough.
This leads me to two questions:
Where would I find a proof that DC is not strong enough to prove the existence of a non-principal ultrafilter on $\omega$?
Is the assumption that there exists a non-principal ultrafilter on $\omega$ strong enough to show DC or countable choice? I.e. is "there exists a non-principal ultrafilter on $\omega$" stronger than either of countable or dependent choice?

REPLY [5 votes]: There is a nice class of problems that are equivalent to the existence of a non-principal ultrafilter. One such, if I remember correctly, is the existence of a colouring of the infinite subsets of the natural numbers in such a way that no infinite set has all its infinite subsets of the same colour. The obvious proof is to colour the sets in such a way that if you add or take away a single element then you change its colour. To make this proof work, you define two sets to be equivalent if their symmetric difference is finite, and do the colouring in each equivalence class separately. But to get it started you have to pick a set in each equivalence class, and for that the obvious thing to do is use AC.
But you can in fact do it with a non-principal ultrafilter as follows. Given an infinite subset A, define its counting function f(n) to be the cardinality of the intersection of A with {1,2,...,n}. Then take a non-principal ultrafilter α and define F(A) to be the limit along α of $(-1)^{f(n)}$. If you add an element m to A, then f(n) is unchanged up to m and then adds 1 thereafter, so its parity is changed everywhere except on a finite set, which implies that F(A) changes. So F gives you your colouring.
I've never actually thought about the other direction (getting from such a colouring to a non-principal ultrafilter) so I don't know how hard it is. I'm not even 100% sure that it's true, but I'm pretty sure I remember hearing that it was.