The starting place is this question on trying to show that the map $\sigma$ from the set of partitions $\mathcal{P}$ of a finite set $X$ to the set of algebras $\mathcal{A}$ on $X$ defined by $\sigma: P\in \mathcal{P}\mapsto \sigma(P)\in \mathcal{A}$ ,where $\sigma(P)$ is the smallest algebra generated by the partition $P$, is injective.
The answer gives one proof of injectivity. I am interested in how to show surjectivity of the above map $\sigma$. So to be precise, I would like to show that given some algebra $A\in\mathcal{A}$, there exists a partition of $P$ such that $\sigma(P)=A$.
One idea is to define, for each $x\in X$, $A_x$ as the smallest set in $A$ containing $x$ and consider the collection $Y=\{A_x\}_{x\in X}$. I believe I can show that $X=\bigcup_{x\in A}A_x$ since $X=\bigcup_{x\in X}\{x\}\subseteq \bigcup_{x\in X} A_x$ and $A_x\subseteq X$, thus $\bigcup_{x\in X} A_x\subseteq X$. But I still need to show that $A_{x_1}$ and $A_{x_2}$ are disjoint for $x_1\neq x_2$ and that any set in $A$ can be generated from sets in the collection $Y$. Where do I go from here?
It's cleaner to define the partition as consisting of the non-empty elements of $\mathcal{A}$ which are minimal with respect to inclusion, or equivalently which are irreducible with respect to intersection. In the language of Boolean algebra these are called atoms. Now it's clearer how to move forward: if $A, B$ are two atoms then $A \cap B$ is a subset of both $A$ and $B$. By the minimality of $A$ it must be empty or $A$, and in the latter case $A \subseteq B$, so by the minimality of $B$ we must have $A = B$ in this case. So distinct atoms are disjoint. Moreover
So the atoms form a partition. Now, given an arbitrary subset $S \in \mathcal{A}$, we can consider its intersections $S \cap A$ with every atom. Again by minimality these intersections must either be empty or equal to $A$, and in the latter case $A \subseteq S$. Moreover every $s \in S$ is contained in some atom, as above. This gives that
$$S = \bigcup_{A \text{ atom}} (S \cap A).$$