13.3 Elementary Families
Definition 13.3.1 (Elementary Family). Let $X$ be a set and $\ce \subset 2^{X}$, then $\ce$ is an elementary family if:
$\emptyset \in \ce$.
For any $A, B \in \ce$, $A \cap B \in \ce$.
For any $E, F \in \ce$ with $E \subset F$, there exists $\seqf{E_j}\subset \ce$ such that $E \setminus F = \bigsqcup_{j = 1}^{n} E_{j}$.
If $X \in \ce$, then (E) may be replaced with
For any $E \in \ce$, there exists $\seqf{E_j}\subset \ce$ such that $E^{c} = \bigsqcup_{j = 1}^{n} E_{j}$.
Proposition 13.3.2. Let $X$ be a set and $\ce \subset 2^{X}$ be an elementary family and
then is a ring. If $X \in \ce$, then $\ce$ is an algebra.
Proof. Firstly, let $A, B \in \alg$ with $\seqf{A_j}, \seqf[m]{B_j}\subset \ce$ and $A = \bigsqcup_{j = 1}^{n} A_{j}$ and $B = \bigsqcup_{j = 1}^{m}B_{j}$. If $A \cap B = \emptyset$, then
so $\alg$ is closed under disjoint unions.
(A1): By (P1) $\emptyset \in \ce$. By (E1), there exists $\seqf{E_j}\subset \ce$ such that $X = \emptyset^{c} = \bigsqcup_{j = 1}^{n}E_{j} \in \alg$.
(A2’): Let $A = \bigsqcup_{j = 1}^{n} A_{j} \in \alg$ and $B \in \ce$. By including additional empty sets, assume without loss of generality that there exists $\bracs{E_{i, j}| 1 \le i \le n, 1 \le j \le m}$ such that $B \setminus A_{i} = \bigsqcup_{j = 1}^{m} E_{i, j}$ for each $1 \le j \le n$. In which case,
Thus if $B \in \alg$ with $B = \bigsqcup_{j = 1}^{n} B_{j}$, then
(A3): Let $A = \bigsqcup_{j = 1}^{n} A_{j}, B = \bigsqcup_{j = 1}^{m} B_{j} \in \alg$, then
so $\alg$ is closed under intersections. Thus using (A2), $A \cup B = A \setminus B \sqcup B \setminus A \sqcup A \cap B$.$\square$