Proof ultrafilter theorem without axiom of choice?

Question

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 ?

Answer 1

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.