Let $X$ be any set. Let $\mathcal{F}$ be family of subsets of $X$ closed under arbitrary intersections and complements. Let $P(x)=\bigcap \{A\in\mathcal{F}:x\in A\}$. I need to prove that family $\{P(x)\}_{x\in X}$ defines a partition of $X$.
Clearly, $\bigcup_{x\in X} P(x) =X$. But how can I show that $P(x)\cap P(y)\ne \emptyset \implies P(x)=P(y)$?
Let $x$, $y$ $\in \mathcal{X}$ be such that $P(x)\cap P(y) \neq \emptyset$. Then there is some $c \in P(x)$, $P(y)$. Assume for a contradiction that $P(x) \neq P(y)$ . Then there is $A \in F$ such that $x,c \in A$ but $y \notin A$ (or there is $B \in F$ such that $y,c \in B$ but $x \notin B$. This case follows similarly) Since $F$ is closed under taking complements, $A^c \in F$. Note that $y \in A^c$ but $x,c \notin A^c$. Also for any set $S \in F$ containing $y$, $S \cap A^c \in F$, since $F$ is closed under intersections, but $y \in S \cap A^c$ and $c \notin S \cap A^c$. Then $c \notin P(y)$ and this is a contradiction so $P(x) = P(y)$.