EN
In ZF (i.e. Zermelo-Fraenkel set theory without the Axiom of Choice AC), we investigate the relationship between UF(ω) (there exists a free ultrafilter on ω) and the statements "there exists a free ultrafilter on every Russell-set" and "there exists a Russell-set A and a free ultrafilter ℱ on A". We establish the following results:
1. UF(ω) implies that there exists a free ultrafilter on every Russell-set. The implication is not reversible in ZF.
2. The statement there exists a free ultrafilter on every Russell-set" is not provable in ZF.
3. If there exists a Russell-set A and a free ultrafilter on A, then UF(ω) holds. The implication is not reversible in ZF.
4. If there exists a Russell-set A and a free ultrafilter on A, then there exists a free ultrafilter on every Russell-set.
We also observe the following:
(a) The statements BPI(ω) (every proper filter on ω can be extended to an ultrafilter on ω) and "there exists a Russell-set A and a free ultrafilter ℱ on A" are independent of each other in ZF.
(b) The statement "there exists a Russell-set and there exists a free ultrafilter on every Russell-set" is, in ZF, equivalent to "there exists a Russell-set A and a free ultrafilter ℱ on A". Thus, "there exists a Russell-set and there exists a free ultrafilter on every Russell-set" is also relatively consistent with ZF.