LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)verified 2026-07-26 (claude-opus-5)

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.

A smallest admissible set exists

Statement

Let $(P, \leq)$ be a chain-complete poset and $f : P \to P$ progressive (Chain-complete poset). Then there is a smallest $f$ -admissible subset $M \subseteq P$ (Admissible subset (Bourbaki–Witt)): $M$ is admissible, and $M \subseteq A$ for every admissible $A \subseteq P$ .

Facts & Assumptions

Given: A chain-complete poset $(P, \leq)$ and a progressive map $f : P \to P$ .

[A1]

A subset $A \subseteq P$ is admissible when (C1) $f (x) \in A$ for every $x \in A$ , and (C2) $\sup C \in A$ for every chain $C \subseteq A$ (Admissible subset (Bourbaki–Witt)).

[L1]

Every chain of $P$ has a least upper bound in $P$ (Chain-complete poset).

Proof

technique · direct

1.1A1

$P$ satisfies (C1), because $f$ is a map from $P$ to $P$ , so $f (x) \in P$ for every $x \in P$ .

1.2A1L1

$P$ satisfies (C2), because every chain $C \subseteq P$ has a least upper bound in $P$ .

2.1step 1.1step 1.2A1

So $P$ is admissible, and the collection $A$ of all admissible subsets of $P$ is nonempty.

3.1step 2.1construct

Define $M = ⋂ A$ , the intersection of all admissible subsets of $P$ .

4.1step 3.1A1

Let $x \in M$ . For every $A \in A$ we have $x \in A$ , hence $f (x) \in A$ by (C1); since this holds for every $A \in A$ , $f (x) \in M$ , so $M$ satisfies (C1).

4.2step 3.1A1L1

Let $C \subseteq M$ be a chain. For every $A \in A$ we have $C \subseteq A$ , hence $\sup C \in A$ by (C2); since this holds for every $A \in A$ , $\sup C \in M$ , so $M$ satisfies (C2).

4.3step 3.1

If $A \subseteq P$ is admissible then $A \in A$ , so $M \subseteq A$ because $M$ is the intersection of a collection containing $A$ .

5.1step 4.1step 4.2A1

$M$ is admissible.

6.1step 5.1step 4.3∎

$M$ is admissible and contained in every admissible subset, so it is the smallest one.

Remarks

Uniqueness is automatic: two smallest admissible sets each contain the other, so they are equal. This licenses writing $M$ for "the" smallest admissible set throughout Extremal element and its cut (Bourbaki–Witt) and the lemmas that follow.
$M$ is never empty. Condition (C2) applied to the empty chain puts $⊥ = \sup \emptyset$ into every admissible set, so $⊥ \in M$ .
Minimality is used exactly twice in what follows, in Everything in $M$ is comparable to an extremal element and in Every element of $M$ is extremal, and both times in the same shape: to prove that all of $M$ has some property, one shows that the elements of $M$ with that property again form an admissible set, which minimality then forces to be all of $M$ . That pattern is what replaces transfinite recursion in this proof of Bourbaki–Witt fixed point theorem. The remaining lemmas use only admissibility of $M$ , progressivity of $f$ , and the order axioms.

A smallest admissible set exists

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources