# Proof ultrafilter theorem without axiom of choice?

This is stated that: the ultrafilter theorem can't be proved without the axiom of choice in Zermelo-Fraenkel.

Is this true? Axiom of choice implies ultrafilter theorem but why is it not possible to prove ultrafilter theorem without axiom of choice?

The ultrafilter theorem written in link: Every Boolean algebra has an ultrafilter on it.

Is it same as: Every filter can be extended to an ultrafilter?

Here filter is subset of power set. But in partial orders:

And is Every filter can be extended to an ultrafilter for sets same as Every filter on poset can be extended to ultrafilter ?

## 1 Answer

The statement at the link is at best rather misleading. The ultrafilter theorem ($\mathsf{UT}$) is strictly weaker than the axiom of choice: if $\mathsf{ZF}$ is consistent, so is $\mathsf{ZF}+\mathsf{UT}+\neg\mathsf{AC}$. (And the result that the union of countably many countable sets is countable is a lot weaker than $\mathsf{AC}$, though it does not follow from $\mathsf{ZF}$ alone.)

If ‘cannot be proved in $\mathsf{ZF}$ without $\mathsf{AC}$’ means simply ‘cannot be proved in $\mathsf{ZF}$’, the statement is true. If it means that $\mathsf{AC}$ is required in order to prove $\mathsf{UT}$, then it’s false, since $\mathsf{UT}$ can hold in models in which $\mathsf{AC}$ is false.

• Thank you!!! And is the UT for filter on sets the same as the UT for filter on poset? – dolan Feb 25 '13 at 15:27
• @dolan: It definitely has the same strength as the UT for Boolean algebras; I’d have to think a bit about the situation for posets in general. – Brian M. Scott Feb 25 '13 at 15:32
• @dolan: It should be okay, since you can embed a poset in a Boolean algebra. – Brian M. Scott Feb 25 '13 at 15:38
• Thank you!!! And is it true that it is impossible to prove UT without axiom of choice? (It is stated in link) – dolan Feb 25 '13 at 15:45
• I bumped the question for retagging; so while at it, I fixed a typo here. – Asaf Karagila Jan 22 '16 at 9:06