PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-01

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Countable unions and subsets of manifold null sets are null

Statement

Assume the Axiom of Countable Choice. Every subset of a manifold null set is null, and every countable union of manifold null sets is null.

Facts & Assumptions

Given: The Axiom of Countable Choice, null subsets $E, E_{1}, E_{2}, \dots$ of a smooth manifold $M$ , and a subset $A \subseteq E$ .

[L1]

A countable chart cover detects manifold nullity (A countable chart cover detects manifold null sets).

[F1]

In Euclidean space, subsets of null sets are null (Measure zero and content zero in $R^{m}$ by countable and finite cube covers).

[A1]

Countable Choice permits one null cover to be selected for each member of a sequence (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[L2]

The set $N \times N$ is countable, and the geometric budgets $ε 2^{- m - 1}$ sum to $ε$ ( $N \times N \approx N$ , For $∣ r ∣ < 1$ , $\sum_{k \geq 0} r^{k} = 1 / (1 - r)$ , and for $∣ r ∣ \geq 1$ the series diverges).

Proof

technique · direct

1.1

Fix a countable smooth atlas detecting nullity as in [L1]. For each chart $(U_{j}, φ_{j})$ , the set $φ_{j} (A \cap U_{j})$ is contained in the null set $φ_{j} (E \cap U_{j})$ , so [F1] makes it null. Thus $A$ is null.

F1L1given

1.2

Fix a chart index $j$ and $ε > 0$ . For each $m \geq 1$ , nullity of $φ_{j} (E_{m} \cap U_{j})$ supplies cube covers with total volume at most $ε 2^{- m}$ . By [A1], choose these covers simultaneously. By [L2], their doubly indexed union is a single countable cube cover, and its total volume is at most $\sum_{m \geq 1} ε 2^{- m} = ε$ . Hence $⋃_{m \geq 1} φ_{j} (E_{m} \cap U_{j})$ is null.

A1L2givenchoosealgebra

2.1

Since $φ_{j} ((⋃_{m \geq 1} E_{m}) \cap U_{j}) = ⋃_{m \geq 1} φ_{j} (E_{m} \cap U_{j}),$ step 1.2 and [L1] show that $⋃_{m \geq 1} E_{m}$ is null in $M$ .

L1step 1.2algebra

3.1

Therefore subsets and countable unions of manifold null sets are null.

step 1.1step 2.1∎

Depends on

Used by

A null set has dense complement in a positive-dimensional manifold Proposition
Morse-Sard for smooth manifolds Theorem

Dependency tree · two levels

53 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Marco Gualtieri, Topology I: Smooth Manifolds, cumulative notes (standard reference, not scraped)