CorollaryStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-13

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Topological recurrence on second-countable spaces

Statement

Let $X$ have a countable open basis contained in $A$ , and let $T$ preserve a finite measure $μ$ on $(X, A)$ . Outside one measurable null set, every neighborhood of $x$ is revisited infinitely often by its positive orbit. If the topology is induced by a metric $d$ , there are strictly increasing positive integers $n_{j}$ with $T^{n_{j}} x \to x$ .

Facts & Assumptions

[F1]

Finite-measure recurrence applies to each measurable basis member. Poincare recurrence for finite measure-preserving systems.

[F2]

A basis refines each open neighborhood at its point. Second countability: an at most countable basis for the topology.

[F3]

The exceptional union over the countable basis is null. Finite and countable subadditivity of measures.

[F4]

Each nonempty set of eligible positive return times has a least member. The well-ordering principle.

Proof

Given: Let $X$ have a countable open basis contained in $A$ , and let $T$ preserve a finite measure $μ$ on $(X, A)$ . Outside one measurable null set, every neighborhood of $x$ is revisited infinitely often by its positive orbit. If the topology is induced by a metric $d$ , there are strictly increasing positive integers $n_{j}$ with $T^{n_{j}} x \to x$ .

1.1

For each basis member $B$ define its exceptional set explicitly as $N_{B} = B \cap ⋃_{q \geq 0} T^{- q} (B ∖ ⋃_{n \geq 1} T^{- n} B)$ . The recurrence proof shows that $N_{B}$ is measurable and null. Thus $N = ⋃_{B \in B} N_{B}$ is measurable and null. This is a prescribed family, not a choice of null covers. A finite basis is handled by a finite union, and the empty space has no points to check.

F1F3

2.1

If $x \notin N$ and $U$ is a neighborhood of $x$ , choose an open set $V$ with $x \in V \subseteq U$ and a basis member $B$ with $x \in B \subseteq V$ . Since $x \notin N_{B}$ , infinitely many positive iterates enter $B$ , and hence $U$ .

step 1.1F2

3.1

In the metric case set $n_{0} = 0$ and let $n_{j}$ be the least integer exceeding $n_{j - 1}$ for which $d (T^{n_{j}} x, x) < 1 / j$ , for $j \geq 1$ . Infinitely many visits to the ball make this set nonempty. Least-element recursion supplies the sequence without countable choice; $n_{j} \geq j$ and the displayed bound proves convergence.

step 2.1F4algebra∎

Depends on

Used by

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Dependency tree · two levels

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Sources

E–W Exercise 2.2.3 pp.22–23 (standard reference, not scraped)