Let $A$ be a Boolean algebra and let $Ult(A)$ be its Stone space. Let us say that a set $U\subseteq Ult(A)$ determines an element $a\in A$ if there exists $V\subseteq U$ such that $$\big\{b\in A\!: (\forall u\in V)(b\in u)\ \wedge\ (\forall u\in U\setminus V)(b\notin u)\big\}=\{a\}.$$ Let us further say that $U\subseteq Ult(A)$ is determining if it determines all elements of $A$.
Question: Is there any characterization of determining sets?
For $a\in A$ denote $S(a)=\{u\in Ult(A)\!:a\in u\}$. Then $\{S(a)\!:a\in A\}$ is a basis of the Stone topology on $Ult(A)$.
Claim: A set $U\subseteq Ult(A)$ is determining if and only if it is dense in the Stone topology.
Proof. Let $U\subseteq Ult(A)$ be determining, $a\in A$, $a\neq 0$. Then there is $V\subseteq U$ such that $\det(V,U)=\{a\}$, where $$\det(V,U)=\big\{b\in A\!:(\forall u\in V)(b\in u)\ \wedge\ (\forall u\in U\setminus V)(b\notin u)\big\}.$$ Since $0\in\det(\emptyset,U)$, $V$ is nonempty. For $u\in V$ we have $u\in S(a)\cap U$. This proves one direction.
To prove the opposite, let $U$ be dense and $a\in A$. We have to find $V\subseteq U$ such that $\det(V,U)=\{a\}$. Let $V=\{u\in U\!:a\in U\}$. Clearly, $a\in\det(V,U)$. If $b\neq a$ then either $a\wedge b'\neq 0$ or $a'\wedge b\neq 0$. Hence there exists $u\in U$ containing exactly one of elements $a,b$, thus $b\notin\det(V,U)$. It follows that $\det(V,U)=\{a\}$. q.e.d.