if $B$ is a boolean algebra and $a\neq b$ in $B$ there exist an ultrafilter containing $a$ but not $b$.

Question

Suppose $a$ and $b$ are distinct elements in a boolean algebra $B$. I am trying to show there is an ultrafilter on $B$ containing $a$ but not $b$. Does this follow from the fact (is this a fact?) that $a = \wedge \{F:F \text{ is an ultrafilter on B and } a\in F\}$?

Answer 1

As pointed out in the comments, it is not true in general that if $a \neq b$ that there is some ultrafilter containing $a$ and not containing $b$. A simple counterexample would be the case where $a < b$.

In fact, this is the only obstacle. So something we can prove (using the axiom of choice) is the following:

Let $B$ be a Boolean algebra, and let $a, b \in B$. Then there is an ultrafilter $F$ with $a \in F$, $b \not \in F$ if and only if $a \not \leq b$.

One direction is trivial: if $F$ is an ultrafilter that contains $a$, but does not contain $b$, then we cannot have $a \leq b$ because otherwise $F$ would have to contain $b$.

Now for the other direction, suppose that $a \not \leq b$. We claim that $a \wedge \neg b \neq 0$. Suppose for a contradiction that $a \wedge \neg b = 0$, then $b \vee a = (b \vee a) \wedge (b \vee \neg b) = b \vee (a \wedge \neg b) = b$. But that means that $a \leq b$, which we assumed is not the case. So indeed $a \wedge \neg b \neq 0$. Now consider the principal filter $F' = \{c \in B : c \geq a \wedge \neg b\}$. Then clearly $a \in F'$ and $\neg b \in F'$. Extend $F'$ to an ultrafilter $F$ (using the axiom of choice). So now we have $a \in F$, but we cannot have $b \in F$ because we already have $\neg b \in F$.