False statementConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-13

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False: every orbit of an ergodic system is dense

Statement

Assuming countable choice, an ergodic probability-preserving continuous map need not have every orbit dense. Lebesgue doubling on the circle is ergodic, but the forward orbit of zero is the singleton ${0}$ .

Facts & Assumptions

[F1]

Doubling is ergodic for Lebesgue probability. Doubling is ergodic for Lebesgue measure.

[F2]

The circle is represented by $[0, 1)$ , with doubling and the circle metric. The circle, rotations and the doubling map.

Refutation

Given: Assuming countable choice, an ergodic probability-preserving continuous map need not have every orbit dense. Lebesgue doubling on the circle is ergodic, but the forward orbit of zero is the singleton ${0}$ .

1.1

The map of [F2] satisfies $D (0) = 0$ , so induction gives $D^{n} (0) = 0$ for every $n \geq 0$ . Its orbit is exactly ${0}$ . The circle ball of radius $1 / 8$ centered at $1 / 2$ is a nonempty open set disjoint from this orbit, since $d (0, 1 / 2) = 1 / 2$ . Thus the orbit is not dense.

2.1

Nevertheless [F1] proves that doubling is ergodic for Lebesgue probability, so the displayed orbit refutes the every-point assertion. The countable-choice assumption is needed for that measure-theoretic supplier; the fixed-orbit and open-ball calculations in step 1.1 are choice-free.

1.1F1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

19 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

Sarig Proposition 1.5, fixed-point specialization (standard reference, not scraped)