LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passverified 2026-08-09 (gpt-5.6-terra-codex-subscription)

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.

Every uniformity has a base of symmetric entourages

Statement

If $U$ is a uniformity on $X$ , then its symmetric entourages form a filter base: for every $E \in U$ there is a symmetric $D \in U$ with $D \subseteq E$ . More generally, for every entourage $E$ and every integer $n \geq 1$ , there is a symmetric entourage $D$ whose $n$ -fold composite satisfies $D^{\circ n} \subseteq E$ .

Facts & Assumptions

Given: A uniformity $U$ on $X$ , an entourage $E \in U$ , and an integer $n \geq 1$ .

[A1]

A uniformity is a filter whose members are closed under inverse and admit square roots (Uniform space in the entourage formulation).

[L1]

A nonempty, proper family that refines every pair of its members is a filter base (Filter base and the filter it generates).

Proof

technique · direct

1.1A1choose

Choose $R \in U$ with $R \circ R \subseteq E$ , and put $S : = R \cap R^{- 1}$ .

1.2A1choose

Put $E_{0} : = E$ . By finitely iterating the square-root axiom, choose entourages $E_{1}, \dots, E_{n}$ such that $E_{k + 1} \circ E_{k + 1} \subseteq E_{k}$ for $0 \leq k < n$ , and put $D : = E_{n} \cap E_{n}^{- 1}$ .

2.1step 1.1A1

The set $S$ is an entourage, since $R, R^{- 1} \in U$ and a filter is closed under intersections; also $S = S^{- 1}$ and $S \subseteq R \circ R \subseteq E$ , because every entourage contains the diagonal.

2.2step 1.2A1algebra

The entourage $D$ is symmetric and $D \subseteq E_{n}$ . Induction on $k$ gives $D^{\circ 2^{k}} \subseteq E_{n - k}$ for $0 \leq k \leq n$ , hence $D^{\circ 2^{n}} \subseteq E$ . Since every entourage contains the diagonal and $n \leq 2^{n}$ , one may insert diagonal factors to obtain $D^{\circ n} \subseteq D^{\circ 2^{n}} \subseteq E$ .

3.1step 2.1L1

Thus symmetric entourages refine every entourage; their intersections are symmetric entourages and none is empty because each contains the diagonal, so they form a filter base by [L1].

4.1step 3.1step 2.2∎

Therefore symmetric entourages form a base and admit the asserted finite-composite control.

Depends on

Dependency tree · 2 levels

Sources

J. Wodzicki, Uniform Structure