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 ?