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.

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

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

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

[A1]

$U \to p$ 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).

technique · contradiction

1.1A1assume-contra

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

2.1step 1.1A1A2

By [A2], $X ∖ N \in U$ . But $N$ must meet every member of $U$ by clusterhood, whereas $N \cap (X ∖ N) = \emptyset$ .

3.1step 2.1discharge-contradiction∎

This contradiction proves $U \to p$ .

Depends on