In the ultrafilter lemma, is the ultrafilter unique? 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 $F$?
 A: It is generally the case that it is not unique. Consider the following theorem.

Theorem. Let $\cal F$ be a filter on a set $X$. The following are equivalent:
  
  
*
  
*$\cal F$ has a unique extension to an ultrafilter on $X$.
  
*$\cal F$ is an ultrafilter on $X$.
  

Proof. The implication from (2) to (1) is trivial; it remains to show the opposite.
Assume that $\cal F$ is not an ultrafilter on $X$. Then there is some $A\subseteq X$ such that neither $A$ nor $X\setminus A$ are in $\cal F$.
I claim now, that both $\mathcal F\cup\{A\}$ and $\mathcal F\cup\{X\setminus A\}$ extend to filters, and these extend to ultrafilters which are different since one has $A$ and the other doesn't.
To see this, note that for all $B\in\cal F$, $A\cap B\neq\varnothing$. Otherwise, if $B\in\cal F$ and $A\cap B=\varnothing$, then $B\subseteq X\setminus A$, and then $X\setminus A\in\cal F$. This is contrary to the assumption, of course. Therefore $\mathcal F\cup\{A\}$ extends to a filter.
By symmetry the same is true for $X\setminus A$. So we proved that if $\cal F$ has only one extensions to an ultrafilter, it was already an ultrafilter. $\quad\square$

So really, in any case you don't already have an ultrafilter, you can extend into at least two different ultrafilters. However, you cannot guarantee more.
Consider this example: Let $\mathcal U$ be a free ultrafilter on $\Bbb N$. Then there is a natural extension of $\mathcal U$ to a filter on $\Bbb N\cup\{-1\}$: $\mathcal F=\{X\cup\{-1\}\mid X\in\mathcal U\}$.
But now we have exactly two ultrafilters which extend $\cal F$: The free ultrafilter given by $\cal F\cup\{\Bbb N\}$ and the one given by $\cal F\cup\{\Bbb N\}$. It is not hard to check that either choice generates an ultrafilter. One is free, the other principal.
A: 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, $F$.
A: 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.
