TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-17

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A countable partition generates exactly the unions of its blocks, and the resulting sigma-algebra is countable exactly for a finite partition

Statement

Let $(P_{i})_{i \in I}$ be an at most countable partition of $X$ : its blocks are nonempty and pairwise disjoint, and their union is $X$ . Then

$σ_{X} ({P_{i} : i \in I}) = {⋃_{i \in S} P_{i} : S \subseteq I} .$

The map $S \mapsto ⋃_{i \in S} P_{i}$ is a bijection from $P (I)$ onto this sigma-algebra. If $I$ has $k$ members, the sigma-algebra has $2^{k}$ members, including $k = 0$ when $X = \emptyset$ . The generated sigma-algebra is at most countable if and only if the partition is finite.

Facts & Assumptions

Given: An at most countable partition $(P_{i})_{i \in I}$ of $X$ .

[L1]

A generated sigma-algebra is the smallest sigma-algebra containing its generators (Nonempty intersections of sigma-algebras are sigma-algebras, so the generated sigma-algebra exists and is minimal).

[L2]

At most countable means finite or countably infinite (Finite, countably infinite, countable, uncountable), and a sigma-algebra is closed under countable unions (Sigma-algebras).

[L3]

A set is not equinumerous with its power set (Cantor's theorem: $A ≺ P (A)$ ), and injections both ways imply equinumerosity (The Schröder-Bernstein theorem).

[L4]

Every subset of an at most countable set is at most countable (Every subset of an at most countable set is at most countable).

Proof

technique · direct

1.1

Let $U : = {⋃_{i \in S} P_{i} : S \subseteq I}$ . The empty union is empty; the complement of the union indexed by $S$ is the union indexed by $I ∖ S$ ; and countable unions correspond to unions of the indexing subsets. Thus $U$ is a sigma-algebra containing every block.

givenalgebra

2.1

By [L1], $σ_{X} ({P_{i} : i \in I}) \subseteq U$ . Conversely, every $S \subseteq I$ is at most countable by [L4], so $⋃_{i \in S} P_{i}$ is a countable union of generators and belongs to the generated sigma-algebra by [L2]. Hence equality holds.

step 1.1L1L2L4

3.1

Pairwise disjointness and nonemptiness make $S \mapsto ⋃_{i \in S} P_{i}$ injective, and step 2.1 makes it surjective. For $∣ I ∣ = k < \infty$ , its domain has $2^{k}$ members; when $k = 0$ , the partition is possible exactly for $X = \emptyset$ and the sigma-algebra is ${\emptyset}$ .

step 2.1construct

4.1

If $I$ is countably infinite and the generated sigma-algebra were at most countable, step 3.1 would inject $P (I)$ into $N$ . Since $I \approx N$ , [L3] would then force $I \approx P (I)$ , contradicting Cantor's theorem. Together with the finite case of step 3.1, this proves both directions of the final equivalence.

step 3.1L2L3∎

Depends on

Used by

A partition into k nonempty blocks generates a sigma-algebra with 2ᵏ members Example

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 38 results over 14 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

R. F. Bass, Real Analysis for Graduate Students, version 5.0, Examples 2.4-2.6 (standard reference, not scraped)
T. Tao, An Introduction to Measure Theory, Exercise 1.4.10 (standard reference, not scraped)