On the Existence of Free Ultrafilters on $\omega $ and on Russell-sets in ZF
Tom 63 / 2015
Streszczenie
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 $\mathbf{ZF}$.
