LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-31

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.

Assuming dependent choice, every entourage admits a normal symmetric sequence subordinate to it

Statement

Assuming dependent choice, for every entourage $U$ there are symmetric entourages $(E_{n})_{n \in N}$ such that $E_{0} = X \times X$ , $E_{1} \subseteq U$ , the sequence is decreasing, and $E_{n + 1}^{\circ 3} \subseteq E_{n}$ for every $n \in N$ .

Facts & Assumptions

Given: A uniformity $U$ , an entourage $U$ , and dependent choice.

[L1]

Every entourage has a symmetric square root (Every uniformity has a base of symmetric entourages).

[L2]

Dependent choice produces a sequence following any serial relation (The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain).

Proof

technique · constructive

1.1

Given a symmetric entourage $E$ , choose a symmetric $R$ with $R^{\circ 2} \subseteq E$ , and then a symmetric $D$ with $D^{\circ 2} \subseteq R$ . Since every entourage contains the diagonal, $D \subseteq R$ , and hence $D^{\circ 3} \subseteq R^{\circ 2} \subseteq E .$ Thus there exists a symmetric $D \subseteq E$ with $D^{\circ 3} \subseteq E$ .

L1construct

2.1

The relation $E R D$ meaning that $D$ is symmetric, $D \subseteq E$ , and $D \circ D \circ D \subseteq E$ is serial by step 1.1.

step 1.1

3.1

Choose a symmetric entourage $E_{1} \subseteq U$ using [L1]. Dependent choice applied to the serial relation of step 2.1 starting at $E_{1}$ gives $E_{1}, E_{2}, \dots$ . Adjoin $E_{0} = X \times X$ ; then $E_{1}^{\circ 3} \subseteq X \times X = E_{0}$ , and all the required properties hold.

step 2.1L1L2discharge-construct∎

Depends on

Used by

Dependency tree · two levels

11 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

J. Wodzicki, Uniform Structure (standard reference, not scraped)
M. Kunzinger, General Topology (standard reference, not scraped)