ExampleConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck 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 extending the Fréchet filter on $N$ is free, and its existence uses the ultrafilter lemma

Example

There exists an ultrafilter $U$ on $N$ containing the Fréchet filter $F_{F r}$ , and every such extension is free. Its existence here is obtained from the ultrafilter lemma. That lemma is proved from Zorn's lemma and therefore from the Axiom of Choice; it does not construct or distinguish the resulting ultrafilter.

Facts & Assumptions

Given: The Fréchet filter $F_{F r}$ on $N$ .

[L1]

The Fréchet filter is a proper filter, and $N ∖ {n} \in F_{F r}$ for every $n \in N$ because this complement contains a tail (The subsets of $N$ containing a tail form the Fréchet filter, and it is proper and not an ultrafilter).

[L2]

Every filter on a set is contained in an ultrafilter. The proof uses Zorn's lemma and hence the Axiom of Choice (The ultrafilter lemma, from the Axiom of Choice: every filter extends to an ultrafilter).

[F1]

An ultrafilter is principal if it has the form ${A \subseteq X : x \in A}$ for some $x \in X$ , and it is free otherwise (Ultrafilter).

[L3]

An ultrafilter contains exactly one of $A$ and $X ∖ A$ for every $A \subseteq X$ (Characterisation of ultrafilters: every set or its complement).

Verification

technique · constructive

1.1

Let $U$ be an arbitrary ultrafilter on $N$ with $F_{F r} \subseteq U$ ; at least one such $U$ exists by applying [L2] to $F_{F r}$ .

L1L2construct

2.1

For every $n \in N$ , the set $N ∖ {n}$ lies in $U$ .

step 1.1L1

3.1

By [L3], step 2.1 forces ${n} \notin U$ for every $n \in N$ .

step 2.1L3

4.1

A principal ultrafilter at $n$ contains ${n}$ , so step 3.1 shows that $U$ is not principal and hence is free.

step 3.1F1

5.1

Since $U$ was arbitrary, every ultrafilter extending the Fréchet filter is free, and step 1.1 supplies one by the ultrafilter lemma, with the choice cost stated in [L2].

step 1.1step 4.1L2discharge-construct∎

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: 44 results over 14 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

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

An ultrafilter extending the Fréchet filter on NN is free, and its existence uses the ultrafilter lemma