# If a filter has a unique ultrafilter extending it, then it is that ultrafilter (prove without $\sf{AC}$)

I am not certain if $\sf AC$ (or more conservatively, $\sf UF=$ there is an ultrafilter extending any given filter) is necessary to prove the following statement:

For filters $F,G$ with $\bigcup F=\bigcup G$, say that $F$ extends $G$ if $F\supseteq G$. If $G$ has a unique ultrafilter extension $F$, then $F=G$.

My approach: Let $X=\bigcup F=\bigcup G$, and suppose $F$ is an ultrafilter extension of the filter $G$ with $x\in F$, $x\notin G$. Then $G\cup\{X\setminus x\}$ is a filter subbase, and $$H=\Big\{y\subseteq X\ \Big|\ \exists^{\rm fin}t\subseteq G\cup\{X\setminus x\}:\bigcap t\subseteq y\Big\}$$ is a filter that extends $G$. Now $X\setminus x\in H$ and $X\setminus x\notin F$, so $F$ cannot be an extension of $H$. Thus any ultrafilter extending $H$ would be a counterexample to the uniqueness of ultrafilters extending $G$.

Is there a way to make this final step without having to invoke $\sf UF$, by somehow taking advantage of the given ultrafilter extension $F$? I am envisioning some small modification of $F$ to change it into another ultrafilter for which $x\in F$ and $X\setminus x\in F'$.

• @dfeuer You are correct; I reworded it to make this clearer. – Mario Carneiro Dec 12 '13 at 4:19
• Seems to me as unlikely without some choice, but I'm still looking for an example for this failure. – Asaf Karagila Dec 12 '13 at 10:23

1. There are no free ultrafilters on $\Bbb N$.
First of all note that if $A$ is amorphous, then $A$ carries exactly one free ultrafilter, all the cofinite subsets. Then in this model $A\cup\Bbb N$ has only one unique ultrafilter, all those containing a cofinite subset of $A$.
Consider now the filter $F=\{A\cup M\mid M\text{ is a cofinite subset of }\Bbb N\}$. Then $F$ is not free and can be extended to only one free ultrafilter, but consider now the filter $G$ generated by adding $\Bbb N$ to $F$, that filter cannot be extended anymore to free ultrafilters.
• I'm always impressed by your ability to construct counterexamples in $\sf ZF+\neg AC$. Good job! This is also the closest I've seen to a "constructable" free ultrafilter (the cofinite subsets of an amorphous set). – Mario Carneiro Dec 13 '13 at 2:17