Proving that a set is an ultrafilter

Question

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?

Answer 1

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.

Answer 2

I'm not sure what notation you are using for your two binary boolean operations. I will use $\wedge$ for conjunction and $\vee$ for disjunction. I will also use $\neg$ for negation.

One useful fact is that for all $A,B\in\mathfrak A,$ we have $B<A$ if and only if $A\wedge B=B.$ From this, we can derive yet another useful fact: given $A_0\in\mathfrak A,$ we have that $A_0$ is an atom of $\mathfrak A$ if and only if for all $A\in\mathfrak A$ either $A_0\wedge A=0$ or $A_0\wedge A=A_0.$ (Note: the "either...or" means that we can't have both, which rules out $A_0=0,$ in particular.) It is good practice to prove these facts.

Now, assuming that $A_0\not<A,$ as you did, we have $A_0\wedge A\neq A_0$ by our first fact, so $A_0\wedge A=0$ by our second fact. We want to show that $A_0<\neg A,$ or equivalently that $A_0\wedge\neg A=A_0.$

Since $$A_0\wedge\neg A=0\vee(A_0\wedge\neg A)=(A_0\wedge A)\vee(A_0\wedge\neg A)=A_0\wedge(A\vee\neg A),$$ what can we conclude?