LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-25verified 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 supremum of extremal elements is extremal

Statement

Let $(P, \leq)$ be a chain-complete poset, $f : P \to P$ progressive, $M$ the smallest admissible set, and $C \subseteq M$ a chain every element of which is extremal (Extremal element and its cut (Bourbaki–Witt)). Then $\sup C$ is extremal.

Facts & Assumptions

Given: A chain-complete poset $(P, \leq)$ , a progressive $f : P \to P$ , the smallest admissible set $M$ , a chain $C \subseteq M$ of extremal elements, and an element $y \in M$ with $y < \sup C$ .

[A1]

Every $z \in C$ is extremal: for every $w \in M$ with $w < z$ , $f (w) \leq z$ (Extremal element and its cut (Bourbaki–Witt)).

[L1]

For an extremal $z$ and every $w \in M$ , either $w \leq z$ or $f (z) \leq w$ (Everything in $M$ is comparable to an extremal element).

[L2]

$M$ is an admissible subset of $P$ , so $M \subseteq P$ , every chain contained in $M$ is a chain of $P$ , and $M$ is closed under suprema of its chains (A smallest admissible set exists, Admissible subset (Bourbaki–Witt)).

[L3]

$\sup C$ is an upper bound of $C$ and is below every upper bound of $C$ (Upper bound, least upper bound, and strict upper bound).

[L4]

$f$ is progressive: $z \leq f (z)$ for every $z \in P$ (Chain-complete poset).

[L5]

$\leq$ is a partial order: it is reflexive ( $u \leq u$ ), transitive ( $u \leq v$ and $v \leq w$ imply $u \leq w$ ) and antisymmetric ( $u \leq v$ and $v \leq u$ imply $u = v$ ), and its strict form $u < v$ means $u \leq v$ together with $u \neq v$ (Partial order and partially ordered set).

Proof

technique · direct

1.1

Write $s = \sup C$ ; it lies in $M$ because $C$ is a chain contained in $M$ and $M$ is closed under suprema of its chains.

L2construct

1.2

It suffices to show $f (y) \leq s$ for the given $y \in M$ with $y < s$ .

suffices: f(y) le s

2.1

If $y$ were an upper bound of $C$ then $s \leq y$ , since $s$ is below every upper bound; but $y < s$ means $y \leq s$ together with $y \neq s$ , and $s \leq y$ with $y \leq s$ would force $y = s$ by antisymmetry, a contradiction. So $y$ is not an upper bound of $C$ .

step 1.2L3L5

3.1

Hence there exists $x \in C$ with $x \leq̸ y$ ; fix one.

step 2.1choose

4.1

The element $x$ is extremal, so comparability gives $y \leq x$ or $f (x) \leq y$ .

step 3.1A1L1

5.1

The alternative $f (x) \leq y$ is impossible: progressivity gives $x \leq f (x)$ , so transitivity would yield $x \leq y$ , contradicting $x \leq̸ y$ . Hence $y \leq x$ .

step 4.1step 3.1L4L5

6.1

Moreover $y \neq x$ , since $y = x$ would give $x \leq y$ by reflexivity, again contradicting $x \leq̸ y$ . So $y < x$ .

step 5.1step 3.1L5

7.1

Extremality of $x$ applied to $y$ gives $f (y) \leq x$ .

step 6.1step 3.1A1

8.1

Since $x \in C$ and $s$ is an upper bound of $C$ , we have $x \leq s$ , so $f (y) \leq s$ by transitivity, and $s$ is extremal.

step 7.1step 3.1L3L5∎

Remarks

Step 2.1 is the subtle one, and it is where chain-completeness does real work. The move from $y < s$ to " $y$ is not an upper bound of $C$ " is exactly leastness of the supremum, closed off by antisymmetry: leastness gives $s \leq y$ , and it is antisymmetry that turns that together with $y \leq s$ into $y = s$ , contradicting $y \neq s$ . If $\sup C$ were merely some upper bound of $C$ , the step would fail and the lemma with it, which is why the Bourbaki-Witt hypothesis asks for least upper bounds rather than upper bounds.
The empty chain is covered without comment: $\sup \emptyset = ⊥$ , and there is no $y \in M$ with $y < ⊥$ , so the condition holds vacuously.
Nothing here needs $C$ to have a largest element, and in the intended application it does not have one.

Depends on

Used by

Every element of M is extremal Lemma

Dependency tree · two levels

10 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

Mathlib, Order.BourbakiWitt (standard reference, not scraped)
Bourbaki-Witt Principle (Menemui Matematik 39(1), 2017) (standard reference, not scraped)
Bourbaki–Witt theorem (Wikipedia) (standard reference, not scraped)