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:

  1. (P1)

    $\emp\in\cali{E}$

  2. (P2)

    For any $E,F\in\cali{E},E\cap F\in\cali{E}$

  3. (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

  1. (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

\begin{align*}\cali{A}=\curl{\dcups{n}{i=1}E_i:\curl{E_i}_1^n\suf\cali{E}\text{ pairwise disjoint},n\in\N^+}\end{align*}

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

\begin{align*}A\sqcup B=\dcups{n}{i=1}A_{i}\sqcup\dcups{m}{i=1}B_{i}\in\cali{A}\end{align*}

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

\begin{align*}B\del A=\parens{\dcups{m}{i=1}B_i}\cap \parens{\caps{n}{j=1}A_j^c}=\dcups{m}{i=1}\parens{B_i\cap\caps{n}{j=1}A_j^c}=\dcups{m}{i=1}\caps{n}{j=1}B_{i}\del A_{j}\end{align*}

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

\begin{align*}A\cap B=\braks{\dcups{n}{j=1}A_j}\cap\braks{\dcups{m}{i=1}B_i}=\dcups{n}{j=1}\dcups{m}{i=1}A_{j}\cap B_{i}\in\cali{A}\end{align*}

so $\cali{A}$ is closed under intersections. Thus by (A2’)

\begin{align*}A\del B=A\del (B\cap A),B\del A=B\del (A\cap B)\in\cali{A}\end{align*}

and hence

\begin{align*}A\cup B=(A\del B)\sqcup (B\del A)\sqcup (A\cap B)\in\cali{A}\end{align*}

$\square$

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

\begin{align*}\cali{R}(\cali{E},\cali{F})=\curl{E\ti F:E\in\cali{E},F\in\cali{F}}\end{align*}

is an elementary family.

Proof. (P1): $\emp=\emp\ti\emp$
(P2): For any $A\ti B,C\ti D\in\cali{R}(\cali{E},\cali{F})$,

\begin{align*}(A\ti B)\cap (C\ti D)=(A\cap C)\ti (B\ti D)\in\cali{R}(\cali{E},\cali{F})\end{align*}

(E): For any $A\ti B,C\ti D\in\cali{R}(\cali{E},\cali{F})$,

\begin{align*}(A\ti B)\del (C\ti D)=(A\del C)\ti (B\del D)\sqcup(A\del C)\ti (B\cap D)\sqcup (A\cap C)\ti (B\del D)\end{align*}

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

\begin{align*}(A\del C)\ti (B\del D)&=\dcups{n}{j=1}\dcups{m}{i=1}A_{j}\ti B_{i}\\ (A\del C)\ti (B\cap D)&=\dcups{n}{j=1}A_{j}\ti (B\cap D)\\ (A\cap C)\ti (B\del D)&=\dcups{m}{i=1}(A\cap C)\ti B_{i}\end{align*}

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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