False statementConstruction: Literature-sourcedVerification: AI-generatedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-14

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Weyl equidistribution fails for some rotation angles

Statement

Assume the Axiom of Countable Choice. False claim: $(n α)_{n \geq 0}$ is equidistributed modulo one for every real $α$ .

Facts & Assumptions

Given: Countable choice and the angle $α = 0$ .

[F1]

Equidistribution requires the limiting frequency in $[a, c)$ to be $c - a$ (Equidistribution modulo one).

[F2]

The valid Weyl theorem assumes that the angle is irrational (Weyl equidistribution for irrational rotations).

[F3]

Countable choice is the standing assumption of [F2] (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

Refutation

technique · constructive

1.1

For every $n \geq 0$ , ${n α} = 0$ .

givenconstructalgebra

2.1

Therefore the proportion of the first $N$ terms in $[0, 1 / 2)$ is $1$ for every $N \geq 1$ , whereas the interval length is $1 / 2$ .

step 1.1algebra

3.1

This violates [F1] and refutes the claim. The counterexample is rational, so it does not meet—and does not challenge—the irrationality hypothesis in [F2]. The arithmetic refutation itself makes no choice; countable choice is present only to compare it with [F2].

F1F2F3step 2.1discharge-construct∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

14 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

Charles Walkden, Ergodic Theory lecture notes (standard reference, not scraped)