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

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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.

Every element of $M$ is extremal

Statement

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

Facts & Assumptions

Given: A chain-complete poset $(P, \leq)$ , a progressive $f : P \to P$ , and the smallest admissible set $M$ .

[L1]

If $x \in M$ is extremal then $f (x)$ is extremal (The image of an extremal element is extremal).

[L2]

If $C \subseteq M$ is a chain of extremal elements then $\sup C$ is extremal (A supremum of extremal elements is extremal).

[L3]

$M$ is itself admissible, so it is closed under $f$ and under suprema of its chains, and $M$ is contained in every admissible subset of $P$ (A smallest admissible set exists).

[L4]

A subset is admissible when it is closed under $f$ and under suprema of its chains (Admissible subset (Bourbaki–Witt)).

Proof

technique · direct

1.1

Let $E = {x \in M : x is extremal}$ , so that $E \subseteq M$ by construction.

construct

2.1

$E$ is closed under $f$ : if $x \in E$ then $f (x) \in M$ because $M$ is closed under $f$ , and $f (x)$ is extremal because $x$ is; so $f (x) \in E$ .

step 1.1L3L1

2.2

$E$ is closed under suprema of its chains: if $C \subseteq E$ is a chain then $C \subseteq M$ , so $\sup C \in M$ because $M$ is closed under suprema of its chains, and $\sup C$ is extremal because every element of $C$ is; so $\sup C \in E$ .

step 1.1L3L2

3.1

So $E$ is an admissible subset of $P$ .

step 2.1step 2.2L4

4.1

By minimality of $M$ , $M \subseteq E$ .

step 3.1L3

5.1

With $E \subseteq M$ this gives $E = M$ , so every element of $M$ is extremal.

step 4.1step 1.1∎

Remarks

No base case is needed. One might expect a separate argument that $⊥$ is extremal, but closure under suprema of chains applied to the empty chain already puts $⊥$ into $E$ , and $⊥$ is extremal vacuously because nothing lies strictly below it. This is the last of the four places on this page where the empty-chain convention of Chain-complete poset removes a case rather than creating one; the others are Admissible subset (Bourbaki–Witt), The cut at an extremal element is closed under chain suprema and A supremum of extremal elements is extremal.
This is the second and last use of minimality, after Everything in $M$ is comparable to an extremal element. Everything downstream is bookkeeping.
Combined with comparability, the lemma says that for any two elements of $M$ the comparability conclusion applies, which is precisely total ordering (The smallest admissible set is a chain).

Depends on

Used by

The smallest admissible set is a chain Lemma

Dependency tree · two levels

9 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)

Every element of MM is extremal