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

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A countable entourage base can be replaced in ZF by a decreasing symmetric base whose next triple composite lies in the preceding member

Statement

In ZF, every countably based uniformity has a decreasing symmetric base (En) with En+1∘3⊆En.

Facts & Assumptions

Given: A countable entourage base B.

[L1]

Symmetric entourages form a base and have square roots (Every uniformity has a base of symmetric entourages).

[L2]

A nonempty subset of N has a least element (The well-ordering principle).

[L3]

Recursion constructs a sequence from a specified starting value and successor map (The recursion theorem).

Proof

technique · constructive
1.1

Use the finite listing or bijection supplied by countability to write the given base as (Cn), repeating its last member in the finite case. Put Bn=⋂i≤n(Ci∩Ci−1). Then (Bn) is a canonically defined decreasing symmetric cofinal base.

L1construct
1.2

Define indices recursively. Put r0=0, and let rn+1 be the least k>rn such that Bk∘3⊆Brn; then put En=Brn.

L1L2L3construct
2.1

Each required set of indices is nonempty: choose a symmetric entourage D with D∘3⊆Brn, then use cofinality and decreasingness to find k>rn with Bk⊆D. Thus the recursion is defined. The inequalities rn+1>rn give decreasingness and cofinality, while the defining clause gives triple control.

step 1.1step 1.2L1L2
3.1

Therefore (En) is the asserted normal base in ZF.

step 2.1discharge-construct∎

Depends on

Used by

Dependency tree · two levels

16 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