LemmaStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17

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The monotone class generated by an algebra is closed under complements

Statement

If $A$ is an algebra of subsets of $X$ , then the generated monotone class $m_{X} (A)$ is closed under complements relative to $X$ .

Facts & Assumptions

Given: An algebra $A$ on $X$ and $M : = m_{X} (A)$ .

[L1]

An algebra is closed under complements (Algebras of subsets).

[L2]

The family $M$ is the smallest monotone class containing $A$ (The generated monotone class exists and is minimal).

Proof

technique · direct

1.1

Put $C : = {E \in M : X ∖ E \in M}$ . If $(E_{n})$ increases in $C$ , then $X ∖ E_{n}$ decreases, so monotone closure of $M$ places both $⋃_{n} E_{n}$ and its complement $⋂_{n} (X ∖ E_{n})$ in $M$ . The decreasing case is the same with union and intersection interchanged. Hence $C$ is a monotone class.

L2algebra

2.1

If $E \in A$ , then $X ∖ E \in A \subseteq M$ by [L1] and [L2], so $A \subseteq C$ . Minimality in [L2] gives $M \subseteq C$ , which is the asserted complement closure.

step 1.1L1L2∎

Depends on

Used by

The monotone class generated by an algebra equals the sigma-algebra it generates Theorem

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 4 results over 3 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, proof of Theorem 2.10 (standard reference, not scraped)