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LemmaStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)verified 2026-07-26 (claude-opus-5)
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Every element of MM is extremal

Statement

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

Facts & Assumptions

Given: A chain-complete poset (P,)(P, \le), a progressive f:PPf : P \to P, and the smallest admissible set MM.

[L1]

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

[L2]

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

[L3]

MM is itself admissible, so it is closed under ff and under suprema of its chains, and MM is contained in every admissible subset of PP (A smallest admissible set exists).

[L4]

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

Proof

technique · direct
1.1

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

construct
2.1

EE is closed under ff: if xEx \in E then f(x)Mf(x) \in M because MM is closed under ff, and f(x)f(x) is extremal because xx is; so f(x)Ef(x) \in E.

step 1.1L3L1
2.2

EE is closed under suprema of its chains: if CEC \subseteq E is a chain then CMC \subseteq M, so supCM\sup C \in M because MM is closed under suprema of its chains, and supC\sup C is extremal because every element of CC is; so supCE\sup C \in E.

step 1.1L3L2
3.1

So EE is an admissible subset of PP.

step 2.1step 2.2L4
4.1

By minimality of MM, MEM \subseteq E.

step 3.1L3
5.1

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

step 4.1step 1.1

Remarks

Depends on

Used by

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Sources