ExampleConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-26‡ rests on unproved material (inherited)

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.

‡ Rests on 3 statements not proved in this library, by way of the results it cites. This item cites no such statement directly; it depends on results that do. The unproved premises it inherits are ‡ Cohen's first model: an infinite Dedekind-finite set of reals, ‡ Sierpiński 1947: the generalised continuum hypothesis implies the Axiom of Choice and ‡ The continuum hypothesis and its generalisation are independent of ZFC. Each is recorded with a citation to the literature and is not established here, because the track that would prove it has not yet been developed in this library. Everything else in this proof is proved here.

A left Kan extension along the inclusion of the rationals in the reals

Example

View $Q$ and $R$ as thin categories under their usual order, and let $i : Q ↪ R$ be the inclusion (A preorder is a category with at most one morphism between any two objects, and its functors are exactly monotone maps).

Define a functor $F : Q \to S e t$ by

$F (q) = (- \infty, q) \subseteq R,$

with the structure maps the evident inclusions when $q \leq q^{'}$ .

Then the pointwise left Kan extension of $F$ along $i$ is the functor $L : R \to S e t$ with

$L (x) = (- \infty, x) \subseteq R .$

Facts & Assumptions

Given: The inclusion $i : Q ↪ R$ and the functor $F (q) = (- \infty, q)$ .

[F1]

A preorder gives a thin category, and a monotone map gives a functor (A preorder is a category with at most one morphism between any two objects, and its functors are exactly monotone maps).

[L1]

The comma-category colimit formula computes the left Kan extension value at a real number $x$ (Comma-category limit and colimit formulae compute Kan extensions).

[L2]

Between any two distinct reals there is a rational number (The rationals embed densely in the reals).

Verification

technique · direct

1.1

For a real $x$ , the comma category $(i ↓ x)$ is the preorder of rationals $q \leq x$ . The induced diagram sends such a $q$ to $(- \infty, q)$ and its colimit in $S e t$ is the union $⋃_{q \leq x, q \in Q} (- \infty, q)$ .

F1L1

2.1

This union is exactly $(- \infty, x)$ . If $r < x$ , [L2] gives a rational $q$ with $r < q < x$ , so $r \in (- \infty, q)$ and hence lies in the union. Conversely every $(- \infty, q)$ with $q \leq x$ is contained in $(- \infty, x)$ .

L2step 1.1

3.1

Therefore [L1] gives $L (x) = (- \infty, x)$ for the left Kan extension value at every real $x$ .

L1step 2.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

24 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

E. Riehl, Category Theory in Context, 2nd ed., Example 6.2.10 (standard reference, not scraped)