Alphabeta Math
ExampleConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 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.
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An ultrafilter selects exactly one cell of a finite disjoint list whose union it contains

Example

Let U\mathcal U be an ultrafilter on XX, let n∈Nn\in\mathbb N, and let s:n→P(X)s:n\to\mathcal P(X) be a finite list of pairwise disjoint sets. If

⋃i∈ns(i)∈U,\bigcup_{i\in n}s(i)\in\mathcal U,

then there is a unique i∈ni\in n such that s(i)∈Us(i)\in\mathcal U. The selected cell is necessarily nonempty. In particular, for a finite partition of XX into nonempty cells, U\mathcal U selects exactly one cell.

The empty list causes no exceptional conclusion: its union is ∅\emptyset, so the displayed hypothesis is false for a proper filter.

Facts & Assumptions

Given: An ultrafilter U\mathcal U on XX, a natural number nn, and a list s:n→P(X)s:n\to\mathcal P(X) such that s(i)∩s(j)=∅s(i)\cap s(j)=\emptyset whenever i≠ji\neq j, and whose union belongs to U\mathcal U.

[L1]

For every n∈Nn\in\mathbb N and every list s:n→P(X)s:n\to\mathcal P(X), if ⋃i∈ns(i)∈U\bigcup_{i\in n}s(i)\in\mathcal U, then s(i)∈Us(i)\in\mathcal U for some i∈ni\in n (Ultrafilters are prime: a union in U\mathcal{U} has a member in U\mathcal{U}).

[F1]

A filter omits ∅\emptyset and is closed under pairwise intersection (Filter on a set).

[F2]

In the von Neumann natural numbers, nn is the set of its predecessors, so a map s:n→P(X)s:n\to\mathcal P(X) is a finite list indexed by i∈ni\in n (The natural numbers N\mathbb{N} (von Neumann)).

Verification

technique · direct
1.1

By [L1], there is an index i∈ni\in n with s(i)∈Us(i)\in\mathcal U.

givenL1
1.2

If distinct indices i,j∈ni,j\in n both satisfied s(i),s(j)∈Us(i),s(j)\in\mathcal U, then pairwise disjointness and intersection closure would give ∅=s(i)∩s(j)∈U\emptyset=s(i)\cap s(j)\in\mathcal U, contradicting properness.

givenF1
2.1

This selected cell is nonempty, because s(i)=∅s(i)=\emptyset would put ∅\emptyset in the proper filter U\mathcal U.

step 1.1F1
3.1

Step 1.1 gives existence and step 1.2 gives uniqueness, while step 2.1 shows the selected cell is nonempty.

step 1.1step 2.1step 1.2
4.1

When the listed sets are nonempty and partition XX, their union is X∈UX\in\mathcal U, so step 3.1 says that exactly one partition cell belongs to U\mathcal U.

step 3.1F1F2∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 26 results over 10 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