Let $X$ be an infinite set. A nonempty collection of subsets, say $F$, of $X$ is called a filter if
$\emptyset \notin F$
$A, B \in F$ implies $A \cap B \in F$
$A \in F$, $A \subseteq B$ implies $B \in F$
and is called an ultrafilter if in addition,
for all $A \subseteq X$, $A \in F$ or $A^c \in F$
holds.
A (ultra)filter is called free if for all $a \in X$, $\{a\} \notin F$.
I want to prove that every free filter $F$ is contained in a free ultrafilter.
My try:
I tried to use Zorn's lemma to the collection $\mathcal C := \{A: A$ is a free filter containing $F \}$.
However, I can't prove that the maximal element $U$ is an ultrafilter.
If $U$ were not an ultrafilter, there will exist an $A \neq \emptyset$ such that $A$, $A^c \notin U$. Then we will be able to extend $U \cup \{ A \}$ into a filter. However, I cannot prove that this filter is free, so that it contradicts the maximality of $U$.
Thank you!
Best Answer
First of all, as Asaf Karaglia wrote, a filter $\mathscr F$ is free if $\,\bigcap \mathscr F \not\in\mathscr F$. Clearly, a free filter does not possess finite sets.
It is not hard to show that, if a filter $\mathscr{F}$ is free, then adding to it all the complements of finite sets of $X$, and taking all the the finite intersections of its sets, you produce another free filter $\mathscr{F}'\supset\mathscr{F}$.
Using the Zorn's Lemma, on the filters, bigger than $\mathscr F'$, you obtain an ultrafilter which does not contain finite sets, and hence it is a free ultrafilter.