Alphabeta Math
ExampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + claude-opus-5[1m])audited 2026-08-24
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.

The ultrafilter algebra on a finite discrete space

Example

Let X be a finite set with the discrete topology. Every ultrafilter on X is principal at a unique point, so the principal-unit map ηX:XβX is a bijection. Its inverse ξX:βXX is the ultrafilter algebra structure and the unique-limit map of the finite discrete space.

For X={0,1}, the only ultrafilters are ηX(0) and ηX(1), and ξX returns the corresponding point.

Facts & Assumptions

Given: A finite set X with the discrete topology.

[L1]

An ultrafilter contains exactly one of A and XA for every AX (Characterisation of ultrafilters: every set or its complement).

[L2]

The principal unit is ηX(x)={AX:xA} (The ultrafilter endofunctor with principal unit and flattening multiplication).

[L4]

The ultrafilter endofunctor β with principal unit η and multiplication μ is a monad, so μXβ(ηX)=1βX (The ultrafilter endofunctor with principal unit and flattening multiplication is a monad).

Verification

technique · direct
1.1

If an ultrafilter on a nonempty finite set contained no singleton, [L1] would put the complement of every singleton into it; their finite intersection is empty, impossible for a filter. Thus it contains some singleton and is principal.

L1construct
2.1

It cannot contain two distinct singletons because their intersection is empty, so the principal point is unique.

step 1.1L1
3.1

By [L2], ηX is therefore a bijection and ξX=ηX1. In the discrete topology [L3], an ultrafilter converges precisely to the point whose singleton it contains, so ξX is the unique-limit map.

step 2.1L2L3
4.1

The equation ξXηX=1X is immediate. Since ηX is bijective by step 3.1, β(ηX) is bijective with inverse β(ξX). The monad unit law in [L4] says μXβ(ηX)=1βX, so uniqueness of the inverse gives μX=β(ξX). Composing with ξX yields ξXμX=ξXβ(ξX), and ξX is an algebra.

step 3.1L4algebra
5.1

If X=, no ultrafilter exists, so βX= and the unique empty map is an algebra. For X={0,1}, steps 1.1 and 2.1 give exactly ηX(0),ηX(1) and step 3.1 gives their displayed values.

step 1.1step 4.1construct

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

28 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