Proof of The Disjointing Sequence of a Sequence of Sets

Proof of (i). Let $S$ be the class of all $n\in\mathbb{N}$ such that, for every set $x$, $x\in C_{n}$ if and only if $x\in X$ and $x\notin A_{k}$ for every $k\in\mathbb{N}$ with $k<n$. This class exists because its defining formula quantifies over no class variable and is therefore predicative (Routine Facts Used Without Citation: Arithmetic, Maps, Predicativity and Indices (iv)), and the condition $n\in S$ on $n$, with $S$ as a class parameter, is atomic, hence predicative, so that Induction for Predicative Formulas, and the Order of ℕ below 1 and below n+1 (i) applies to it. Then $1\in S$: $C_{1}=X$, and as there is no $k\in\mathbb{N}$ with $k<1$ (Induction for Predicative Formulas, and the Order of ℕ below 1 and below n+1 (ii)) the condition on the $A_{k}$ with $k<1$ is empty. Let $n\in S$. By The Boolean Operations on Classes, Disjointness, and the Universal Class (i), $x\in C_{n+1}=C_{n}\cap(X\setminus A_{n})$ if and only if $x\in C_{n}$ and $x\notin A_{n}$ (as $C_{n}\subseteq X$), that is, as $n\in S$, if and only if $x\in X$, $x\notin A_{k}$ for every $k\in\mathbb{N}$ with $k<n$, and $x\notin A_{n}$. As $k<n+1$ if and only if $k<n$ or $k=n$ (Induction for Predicative Formulas, and the Order of ℕ below 1 and below n+1 (iii)), this says that $x\in X$ and $x\notin A_{k}$ for every $k\in\mathbb{N}$ with $k<n+1$; so $n+1\in S$. By induction (Induction for Predicative Formulas, and the Order of ℕ below 1 and below n+1 (i)), $n\in S$ for every $n\in\mathbb{N}$, which is the first assertion. For the second, $x\in B_{n}=A_{n}\cap C_{n}$ if and only if $x\in A_{n}$ and $x\in C_{n}$, by The Boolean Operations on Classes, Disjointness, and the Universal Class (i); as $A_{n}\in\mathcal{P}(X)$, $x\in A_{n}$ implies $x\in X$, so by the first assertion this holds if and only if $x\in A_{n}$ and $x\notin A_{k}$ for every $k\in\mathbb{N}$ with $k<n$.

Proof of (ii). $B_{n}=A_{n}\cap C_{n}$ is contained in $A_{n}$ and in $C_{n}$ by The Boolean Operations on Classes, Disjointness, and the Universal Class (i).

Proof of (iii). Let $m,k\in\mathbb{N}$ with $m\neq k$. By Arithmetic and Order of the Natural Numbers (vii), $m<k$ or $k<m$; as $B_{m}\cap B_{k}$ and $B_{k}\cap B_{m}$ have the same elements, we may take $m<k$. If $x\in B_{m}\cap B_{k}$, then $x\in B_{k}$ gives $x\notin A_{m}$ by (i), while $x\in B_{m}\subseteq A_{m}$ by (ii), which is impossible. So $B_{m}\cap B_{k}$ has no element, that is, $B_{m}$ and $B_{k}$ are disjoint by The Boolean Operations on Classes, Disjointness, and the Universal Class (ii), and $B_{m}\cap B_{k}$ equals the set with no element, $\emptyset$, by Elements Are Sets, and the Empty Set and the Pair of Two Sets Exist and Are Unique (ii).

Proof of (iv). Let $n\in\mathbb{N}$ and $x\in A_{n}$. The subset $T=\{k\in\mathbb{N}:x\in A_{k}\}$ of $\mathbb{N}$ has the element $n$, so by Arithmetic and Order of the Natural Numbers (xv) it has an element $m$ with $m\le v$ for every $v\in T$; in particular $x\in A_{m}$ and $m\le n$. Let $k\in\mathbb{N}$ with $k<m$. If $x\in A_{k}$, then $k\in T$, so $m\le k$, that is, $m<k$ or $m=k$ by Arithmetic and Order of the Natural Numbers (vi), and either contradicts $k<m$ by Arithmetic and Order of the Natural Numbers (vii). So $x\notin A_{k}$ for every $k\in\mathbb{N}$ with $k<m$, and $x\in B_{m}$ by (i).

By Indexed Families of Sets and Their Union, Intersection and Product (ii), the elements of $\bigcup_{n\in\mathbb{N}}B_{n}$ are the $x$ with $x\in B_{n}$ for some $n\in\mathbb{N}$, and likewise for $\bigcup_{n\in\mathbb{N}}A_{n}$. An element of some $B_{n}$ lies in $A_{n}$ by (ii), and an element of some $A_{n}$ lies in some $B_{m}$ by what was just shown. So the two unions have the same elements and are equal.

Proof of (v). Let $\mathcal{F}$ be as there. By σ-Algebra and Measurable Space (i), $\mathcal{F}\subseteq\mathcal{P}(X)$, $X\in\mathcal{F}$, and $X\setminus A\in\mathcal{F}$ for every $A\in\mathcal{F}$. Let $R$ be the class of all $n\in\mathbb{N}$ with $C_{n}\in\mathcal{F}$. As for $S$, this class exists because its defining formula quantifies over no class variable (Routine Facts Used Without Citation: Arithmetic, Maps, Predicativity and Indices (iv)), and the condition $n\in R$, with $R$ as a class parameter, is atomic, hence predicative. Then $1\in R$, as $C_{1}=X\in\mathcal{F}$. If $n\in R$, then $X\setminus A_{n}\in\mathcal{F}$, as $A_{n}\in\mathcal{F}$, so $C_{n+1}=C_{n}\cap(X\setminus A_{n})\in\mathcal{F}$ by Closure Properties of σ-Algebras, Intersections and Traces of σ-Algebras, and the Generated σ-Algebra (ix), and $n+1\in R$. By induction (Induction for Predicative Formulas, and the Order of ℕ below 1 and below n+1 (i)), $n\in R$, that is $C_{n}\in\mathcal{F}$, for every $n\in\mathbb{N}$. Hence $B_{n}=A_{n}\cap C_{n}\in\mathcal{F}$ for every $n\in\mathbb{N}$, by Closure Properties of σ-Algebras, Intersections and Traces of σ-Algebras, and the Generated σ-Algebra (ix). Finally, $(B_{n})$ is a sequence in $\mathcal{P}(X)$ whose terms lie in the subset $\mathcal{F}$ of $\mathcal{P}(X)$, so it is a sequence in $\mathcal{F}$ by Sequences and Maps of Sets: Complements, Traces, Two-Term Sequences, the Disjointing Sequence, and Sequences in a Subset (v).