ExampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passaudited 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.

An ultrafilter selects exactly one cell of a finite disjoint list whose union it contains

Example

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

$⋃_{i \in n} s (i) \in U,$

then there is a unique $i \in n$ such that $s (i) \in U$ . The selected cell is necessarily nonempty. In particular, for a finite partition of $X$ into nonempty cells, $U$ 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 $U$ on $X$ , a natural number $n$ , and a list $s : n \to P (X)$ such that $s (i) \cap s (j) = \emptyset$ whenever $i \neq j$ , and whose union belongs to $U$ .

[L1]

For every $n \in N$ and every list $s : n \to P (X)$ , if $⋃_{i \in n} s (i) \in U$ , then $s (i) \in U$ for some $i \in n$ (Ultrafilters are prime: a union in $U$ has a member in $U$ ).

[F1]

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

[F2]

In the von Neumann natural numbers, $n$ is the set of its predecessors, so a map $s : n \to P (X)$ is a finite list indexed by $i \in n$ (The natural numbers $N$ (von Neumann)).

Verification

technique · direct

1.1

By [L1], there is an index $i \in n$ with $s (i) \in U$ .

givenL1

1.2

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

givenF1

2.1

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

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 $X$ , their union is $X \in U$ , so step 3.1 says that exactly one partition cell belongs to $U$ .

step 3.1F1F2∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

14 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

Ultrafilter (set theory) (Wikipedia) (standard reference, not scraped)
Ultrafilter (Wikipedia) (standard reference, not scraped)