Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck 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.

Every cluster point of an ultrafilter is a limit of that ultrafilter

Statement

Every cluster point of an ultrafilter is a limit of that ultrafilter.

Facts & Assumptions

Given: An ultrafilter U on X and a cluster point p of it.

[A1]

Up means every neighbourhood of p belongs to U, while clusterhood means every such neighbourhood meets every member of U (Convergence and cluster points of a filter on a topological space).

[A2]

For every subset S, an ultrafilter contains S or its complement (Characterisation of ultrafilters: every set or its complement).

Proof

technique · contradiction
1.1

Assume for a contradiction that U does not converge to p. Then some neighbourhood N of p is not in U.

A1assume-contra
2.1

By [A2], XNU. But N must meet every member of U by clusterhood, whereas N(XN)=.

step 1.1A1A2
3.1

This contradiction proves Up.

step 2.1discharge-contradiction

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 19 results over 7 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources