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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 $X\setminus x\in F'$.

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  • $\begingroup$ @dfeuer You are correct; I reworded it to make this clearer. $\endgroup$ – Mario Carneiro Dec 12 '13 at 4:19
  • $\begingroup$ Seems to me as unlikely without some choice, but I'm still looking for an example for this failure. $\endgroup$ – Asaf Karagila Dec 12 '13 at 10:23
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Suppose we are in a model where the following is true:

  1. There are no free ultrafilters on $\Bbb N$.
  2. There exists an amorphous set.

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.

So you do need to use some part of the axiom of choice to prove that, and indeed the cleanest part is the ultrafilter lemma.

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    $\begingroup$ 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). $\endgroup$ – Mario Carneiro Dec 13 '13 at 2:17

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