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LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)verified 2026-07-26 (claude-opus-5)
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  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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A smallest admissible set exists

Statement

Let (P,≤) be a chain-complete poset and f:P→P progressive (Chain-complete poset). Then there is a smallest f-admissible subset M⊆P (Admissible subset (Bourbaki–Witt)): M is admissible, and M⊆A for every admissible A⊆P.

Facts & Assumptions

Given: A chain-complete poset (P,≤) and a progressive map f:P→P.

[A1]

A subset A⊆P is admissible when (C1) f(x)∈A for every x∈A, and (C2) sup⁡C∈A for every chain C⊆A (Admissible subset (Bourbaki–Witt)).

[L1]

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

Proof

technique · direct
1.1

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

A1
1.2

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

A1L1
2.1

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

step 1.1step 1.2A1
3.1

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

step 2.1construct
4.1

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

step 3.1A1
4.2

Let C⊆M be a chain. For every A∈A we have C⊆A, hence sup⁡C∈A by (C2); since this holds for every A∈A, sup⁡C∈M, so M satisfies (C2).

step 3.1A1L1
4.3

If A⊆P is admissible then A∈A, so M⊆A because M is the intersection of a collection containing A.

step 3.1
5.1

M is admissible.

step 4.1step 4.2A1
6.1

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

step 5.1step 4.3∎

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⁡∅ into every admissible set, so ⊥∈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.

Depends on

Used by

Dependency tree · two levels

4 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