4.4 Elementary Families
Definition 4.4.1 (Elementary Family).label Let $X$ be a set and $\cali{E}\suf 2^{X}$, then $\cali{E}$ is an elementary family if:
- (P1)
$\emp\in\cali{E}$
- (P2)
For any $E,F\in\cali{E},E\cap F\in\cali{E}$
- (E)
For any $E,F\in\cali{E}$ with $E\suf F$, there exists $\curl{E_i}_{1}^{n}\suf\cali{E}$ such that $F\del E=\dcups{n}{i=1}E_{i}$
If $X\in\cali{E}$ then (E) is equivalent to
- (E’)
For any $E\in\cali{E}$, there exists $\curl{E_i}_{1}^{n}\suf\cali{E}$ such that $E^{c}=\dcups{n}{i=1}E_{i}$
Proof. Suppose (E’) holds. Let $E,F\in\cali{E}$ with $E\suf F$ and pick $\curl{E_i}_{1}^{n}\suf\cali{E}$ such that $E^{c}=\dcups{n}{i=1}E_{i}$ then $F\del E=F\cap E^{c}=\dcups{n}{i=1}F\cap E_{i}$ and $F\cap E_{i}\in \cali{E}$ for each $i$ by (P2)$\square$
Proposition 4.4.2.label Let $X$ be a set and $\cali{E}\suf 2^{X}$ be an elementary family and
then $\cali{A}$ is a ring. If $X\in\cali{E}$ then $\cali{A}$ is an algebra.
Proof. Let $A,B\in\cali{A}$ then $A=\dcups{n}{i=1}A_{i},B=\dcups{m}{i=1}B_{i}$ for some $\curl{A_i}_{1}^{n},\curl{B_i}_{1}^{m}\suf\cali{E}$. If $A\cap B=\emp$, then
so $\cali{A}$ is closed under disjoint unions.
(A1): By (P1) and (E1) there exists $\curl{E_i}_{1}^{n}\suf\cali{E}$ such that $X=\emp^{c}=\dcups{n}{i=1}E_{i}\in\cali{A}$
(A2’): Let $A=\dcups{n}{i=1}A_{i},B=\dcups{m}{i=1}B_{i}\in\cali{A}$ with $A\suf B$ then
By (E) and (P2) $\caps{n}{j=1}B_{i}\del A_{j}$ is a finite disjoint union of elements of $\cali{E}$ hence $B\del A\in\cali{A}$.
(A3): Let $A=\dcups{n}{j=1}A_{j},B=\dcups{m}{i=1}B_{i}\in\cali{A}$ then
so $\cali{A}$ is closed under intersections. Thus by (A2’)
and hence
$\square$
Proposition 4.4.3.label Let $X,Y$ be sets, $\cali{E}\suf 2^{X},\cali{F}\suf 2^{Y}$ be elementary families, then the collection of rectangles
is an elementary family.
Proof. (P1): $\emp=\emp\ti\emp$
(P2): For any $A\ti B,C\ti D\in\cali{R}(\cali{E},\cali{F})$,
(E): For any $A\ti B,C\ti D\in\cali{R}(\cali{E},\cali{F})$,
Let $\curl{A_j}_{1}^{n}\suf\cali{E}$ such that $A\del C=\dcups{n}{j=1}A_{j}$ and $\curl{B_i}_{1}^{m}\suf\cali{F}$ such that $B\del D=\dcups{m}{i=1}B_{i}$, then
are all finite disjoint unions of elements of $\cali{R}(\cali{E},\cali{F})$. Therefore $(A\ti B)\del (C\ti D)\in\cali{R}(\cali{E},\cali{F})$.$\square$
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