TheoremBase

The Disjointing Sequence of a Sequence of Sets

For a sequence (An)A_n) of subsets of X, the disjointing sequence BnB_n = AnA_n minus the earlier AkA_k consists of pairwise disjoint subsets BnB_n of AnA_n with the same union, every point of AnA_n lying in some BmB_m with m at most n; its terms and the auxiliary sets CnC_n stay in any σ-algebra containing every AnA_n.

Statement

In the setting of Routine Facts Used Without Citation: Arithmetic, Maps, Predicativity and Indices, let XX be a set and let P(X)\mathcal{P}(X) be its power set. Sequences are indexed by the set N\mathbb{N} of natural numbers, with the number 11 of The Set of Natural Numbers and the Number One (ii), the sum n+1n+1 of Addition on Omega (i), the order ≤\le and the strict order <<; differences and intersections of sets are those of The Boolean Operations on Classes, Disjointness, and the Universal Class (i), and ∅\emptyset is the empty set. For a sequence (Dn)(D_{n}) in P(X)\mathcal{P}(X), ⋃n∈NDn\bigcup_{n\in\mathbb{N}}D_{n} is the union of (Dn)(D_{n}) as a family indexed by N\mathbb{N}. Let (An)(A_{n}) be a sequence in P(X)\mathcal{P}(X), and let (Cn)(C_{n}) and (Bn)(B_{n}) be the sequences in P(X)\mathcal{P}(X) of Sequences and Maps of Sets: Complements, Traces, Two-Term Sequences, the Disjointing Sequence, and Sequences in a Subset (iv), so that C1=XC_{1}=X, Cn+1=Cn∩(X∖An)C_{n+1}=C_{n}\cap(X\setminus A_{n}) and Bn=An∩CnB_{n}=A_{n}\cap C_{n} for every n∈Nn\in\mathbb{N}.

For every n∈Nn\in\mathbb{N} and every set xx: x∈Cnx\in C_{n} if and only if x∈Xx\in X and x∉Akx\notin A_{k} for every k∈Nk\in\mathbb{N} with k<nk<n; and x∈Bnx\in B_{n} if and only if x∈Anx\in A_{n} and x∉Akx\notin A_{k} for every k∈Nk\in\mathbb{N} with k<nk<n.

Bn⊆AnB_{n}\subseteq A_{n} and Bn⊆CnB_{n}\subseteq C_{n} for every n∈Nn\in\mathbb{N}.

For all m,k∈Nm,k\in\mathbb{N} with m≠km\neq k, the sets BmB_{m} and BkB_{k} are disjoint, and Bm∩Bk=∅B_{m}\cap B_{k}=\emptyset.

For every n∈Nn\in\mathbb{N} and every x∈Anx\in A_{n} there is m∈Nm\in\mathbb{N} with m≤nm\le n and x∈Bmx\in B_{m}. Consequently ⋃n∈NBn=⋃n∈NAn\bigcup_{n\in\mathbb{N}}B_{n}=\bigcup_{n\in\mathbb{N}}A_{n}.

Let F\mathcal{F} be a σ\sigma-algebra on XX with An∈FA_{n}\in\mathcal{F} for every n∈Nn\in\mathbb{N}. Then Cn∈FC_{n}\in\mathcal{F} and Bn∈FB_{n}\in\mathcal{F} for every n∈Nn\in\mathbb{N}, and (Bn)(B_{n}) is a sequence in F\mathcal{F}.

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