A filter $F$ of a Boolean algebra B is an ultrafilter if and only if for all $b\in B$, either $b\in F$ or $\neg b \in F$ but not both.

Question

I'm having a hard time proving the direct implication. This is my progress so far:

We prove the converse. Assume that for all $b\in B$, either $b\in F$ or $\neg b\in F$. Let $F'$ be a filter that properly contains $F$. By assumption, there exists $c\in B$ such that $c,\neg c\in F'$. Therefore, $0=c\land \neg c\in F'$. But $0\leq a$ for all $a\in B$. Hence $a\in F'$, implying that $F'=B$. This argument can be used to prove (a part of) the direct implication: if $F$ is an ultrafilter and $b,\neg b \in F$, then $F=B$, which is a contradiction.

But as you can see here, there is no guarantee here that either $b\in F$ or $\neg b\in F$. I tried to assume that both are not in $F$, and then construct a proper filter $U$ that is "generated" by $F$ and $b$, and then show that $\neg b\not\in F$, but I have no idea how to do it.

Answer 1

Suppose $F$ obeys

$$\forall b \in B: b \in F \text{ or } \lnot b \in F\tag{1}$$

Now suppose $F' \supsetneq F$ is a filter too. Let $c \in F'$ so that $c \notin F$ witness the proper inclusion. Then by $(1)$ applied to $c$ we have that $c \notin F$ hence $\lnot c \in F$ and by the inclusion $\lnot c \in F'$ too, which is a contradiction as then $c \land \lnot c = 0 \in F'$ too. So we cannot have a properly larger filter than $F$ and $(1)$ implies maximality of $F$.

OTOH if $F$ is maximal, assume there is some $c \in B$ such that $c \notin F$ and $\lnot c \notin F$ too. Then define $$F' = \{x \in B\mid \exists f_x \in F: f_x \land c \le x\}$$

Then this is a filter:

• $0 \in F'$ would imply $f_0 \land c \le 0$ for some $f_0 \in F$, but then $f_0 \land c=0$ and so $f_0 \le \lnot c$ but this would imply $\lnot c \in F$ which is not the case by assumption. So $F$ is proper (not equal to $B$).
• If $x,y \in F'$ we have $f_x,f_y \in F$ so that $f_x \land c \le x$ and $f_y \land c \le y$. Then $f_x \land f_y \in F$ too and also $$(f_x \land f_y) \land c = (f_x \land c) \land (f_y \land c) \le x \land y$$ showing that $x \land y \in F'$ too.
• If $x \in F'$ and $x \le y$ by the definition it's clear that $y \in F'$ too (use the same $f_x$ for $y$ too).

Use $f_x = x$ to see that $F \subseteq F'$ and the inclusion is proper as $c \in F'$ is witnessed by taking $f_x=1 \in F$ (using $F \neq \emptyset$).

Then $F'$ contradicts the maximality of $F$. So no such $c$ can exist and $F$ obeys $(1)$.