In the ultrafilter lemma, is the ultrafilter unique?

Question

The ultrafilter lemma says: Every filter $F$ is contained in a ultrafilter.

Question: Is this ultrafilter unique? Or can one find a filter $F$ such that there are several ultrafilters that contain $F$?

Answer 1

No, it isn't even vaguely unique. For any set $X$, the set $\{X\}$ is a filter; this filter is not just contained in several ultrafilters, it's contained in all of them. So, for example, let $X$ be $\mathbb{N}$, and let $F$ be the filter $\{\mathbb{N}\}$. Let $U_1$ be the ultrafilter consisting of exactly those subsets of $\mathbb{N}$ that include $1$, and let $U_2$ be the ultrafilter consisting of exactly those subsets of $\mathbb{N}$ that include $2$. Then $F \subseteq U_1$, $F \subseteq U_2$, and $U_1 \neq U_2$.

More generally, though, let $F$ be any filter on a set $X$ that is not an ultrafilter. Then there is some set $Y \subseteq X$ so that neither $Y$ nor $X \setminus Y$ are in $F$. But we can certainly obtain a filter $F'$ containing $F \cup \{Y\}$ (by closing $F \cup \{Y\}$ under supersets and finite intersections). We can also obtain a filter $F''$ containing $F \cup \{X \setminus Y\}$. By the Lemma, there are ultrafilters $U'$ containing $F'$ and $U''$ containing $F''$; these both extend $F$, but they differ on $Y$, so they are different ultrafilters. So, in fact, the answer is that the ultrafilter is never unique, except when $F$ is already an ultrafilter.

Answer 2

This ultra-filter is not necessarily unique. Consider the cofinite filter on $\mathbb{N}$, where $S\in F\iff |\mathbb{N}\setminus S|<\infty$. Let $F_e$ be the filter generated by $F$ and the set of even numbers, and $F_o$ be the filter generated by $F$ and the set of odd numbers. These can both be extended to an ultrafilter, $U_e$ and $U_o$ respectively, by the lemma. Both $U_e$ and $U_o$ are extensions of the original filter, $F$.