Exercise 1.4.4 (Finite Boolean algebra is atomic)

Since $B$ is finite with $N:=\#B$, we can enumerate its elements as ${(A_n)}_{n=1}^N$. Define

Sorting out the empty sets, consider

We argue that the elements of this set, enumerated as ${(A_k^{\prime })}_{k=1}^{N^{\prime }}$, are atoms that generate $B$. We verify the atom criteria from Example 1.4.7. First of all, it is obvious that the elements $A_1^{\prime },\dots ,A_{N^{\prime }}^{\prime }$ are disjoint.

Also, it is easy to verify that their union gives us the ambient space

Finally, any set of $B$ can be represented as a union of some of these atom sets. To see why, pick an arbitrary $A_n\in B,A_n\ne \varnothing$. Then

We shortly verify this. If $x\in A_n$ then either $x\in A_n^{\prime }$ or $x\in {\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i\ne n}{A}_i$. In the former case, $x$ is contained in the right-hand side due to $A_n\cap A_n^{\prime }\ne \varnothing$. In the former case, there exists $i\ne n$ such taht $x\in A_i$ and thus $A_i\cap A_n\ne \varnothing$. Now pick an arbitrary $x$ from the right-hand side. Then $x\in A_k$ such that $A_k\cap A_n\ne \varnothing$. But $A_k^{\prime }:=A_k\setminus {\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i\in \{1,\dots ,N\}∕\{k\}}{A}_i$, which is only possible when $n=k$, and we have $x\in A_n$.

Let $B$ be a finite Boolean algebra, so we can model $B$ as a sequence of its elements ${(E_i)}_{i=1}^n$.

Define the set $F^{\prime }=\{F:F\in B\mid F\cap E=\varnothing \text{ or }F\cap E=F\text{ for all }E\in B\}$. Since $F^{\prime }\subseteq B$, it is also finite. We claim $F^{\prime }$ is a partition of $X$.

The elements being disjoint is enforced by the conditions set.

Let $x$ be an arbitrary element of $X$. Since $B$ is a Boolean algebra, it covers $X$, and there must be an element $E_0\in B$ such that $x\in E_0$. Define the function $f:B\times B\to B$ as follows:

Then the sequence $G_0=E_0$ and $G_i=f(G_{i-1},E_i)$ is well defined and finite, ending with $n$.

Since we started with an element containing $x$, $G_n$ must also contain $x$. Now assume to the contrary there is an $E$ different from $G_n$ such that $E\cap G_n\ne \varnothing$. Then $E\cap G_n\in B$, and so is $G_n\setminus E$. $x$ is in either one of them, and then at some point the algorithm would have chosen one of these sets (or one contained in them) over $G_n$, a contradiction.

Thus, $G_n\in F^{\prime }$, and $F^{\prime }$ covers $X$ and thus is a partition. Since it is finite, we can enumerate it and call it ${(F_{\alpha })}_{\alpha \in I}$ with $I=\{1,\dots ,m\}$ for some $m\in \mathbb{N}$.

Now we have to prove that every element of $B$ is in $A({(F_{\alpha })}_{\alpha \in I})$.

Consider an arbitrary $E\in B$. Since $F^{\prime }$ covers $X$, it also covers $E$ in a disjoint manner. Let $F_E={\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{\alpha \in I}{F}_{\alpha }\text{ where }{F}_{\alpha }\cap E\ne \varnothing$, and assume to the contrary $F_E\ne E$. As $F^{\prime }$ covers $X$, we must have $E\subset F_E$. But then there must exist at least one $F_i$ such that $F_i\cap E\subset F_i$ and $F_i\cap E\ne \varnothing$, a contradiction of $F_i$ being a member of $F^{\prime }$. So $E=F_E$, and thus $E\in A({(F_{\alpha })}_{\alpha \in I})$.

Since every $F_i\in B$, which is closed under union, every ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{\alpha \in J}{F}_{\alpha }$, with $J\subseteq I$, is in $B$, and both algebras are equal.