Proving that a set is an ultrafilter Let $\mathfrak{A}$ be a finite Boolean algebra, and $A_0$ be an atom of $\mathfrak{A}$. I would like to prove that the set $$U(A_0)=\{A\in\mathfrak{A}:A_0<A\}$$ is the principal ultrafilter on $\mathfrak{A}$ generated by $A_0$.
I have already proved that $U(A_0)$ is the principal filter generated by $A_0$. So, $\mathfrak{A}$ being a Boolean algebra, I just have to show that for every $A\in\mathfrak{A}$ either $A$ or $-A$ is in $U(A_0)$.
If $A=1$, this is obvious because $-1=0$ is covered by every atom, in particular $A_0$. How do I prove the property for an arbitrary $A$ of $\mathfrak{A}$ though?

I would show instead that $U(A_0)$ is maximal. If $A\in\mathfrak{A}$, and $A_0\not<A$, then $A_0\land A=0$, since $A_0$ is an atom, and $U(A_0)\cup\{A\}$ is therefore not a filter. It follows that $U(A_0)$ is a maximal filter, i.e., an ultrafilter.
