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Abstract

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