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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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.

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The smallest admissible set is a chain

Statement

Let $(P, \le)$ be a chain-complete poset, $f : P \to P$ progressive, and $M$ the smallest admissible set. Then $M$ is a chain (Chain in a poset): any two elements of $M$ are comparable.

Facts & Assumptions

Given: A chain-complete poset $(P, \le)$ , a progressive $f : P \to P$ , the smallest admissible set $M$ , and two elements $x, y \in M$ .

[L1]

Every element of $M$ is extremal (Every element of $M$ is extremal).

[L2]

If $x \in M$ is extremal then every $y \in M$ satisfies $y \le x$ or $f(x) \le y$ (Everything in $M$ is comparable to an extremal element).

[L3]

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

[L4]

A subset is a chain when any two of its elements are comparable (Chain in a poset).

[L5]

$\le$ is a partial order, in particular transitive: $u \le v$ and $v \le w$ imply $u \le w$ (Partial order and partially ordered set).

Proof

technique · direct

1.1

The element $x$ is extremal, because every element of $M$ is.

2.1

Applying comparability at $x$ to the element $y$ , either $y \le x$ or $f(x) \le y$ .

step 1.1L2

3.1

In the second case progressivity gives $x \le f(x)$ , so $x \le y$ by transitivity.

step 2.1L3L5

4.1

So in either case $x$ and $y$ are comparable, and since $x$ and $y$ were arbitrary, $M$ is a chain.

step 2.1step 3.1L4∎

Remarks

This is where the two halves of the argument meet. Comparability (Everything in $M$ is comparable to an extremal element) was conditional on extremality, and Every element of $M$ is extremal removes the condition; neither alone gives a chain.
$M$ being a chain is exactly what makes $\sup M$ available in Bourbaki–Witt fixed point theorem. Chain-completeness supplies suprema for chains only, so without this lemma there would be no reason for $\sup M$ to exist at all.
Note that $M$ is a chain but $P$ need not be. The construction carves a totally ordered piece out of an arbitrary chain-complete poset, and the fixed point is found at the top of that piece.

Depends on

Used by

Bourbaki–Witt fixed point theorem Theorem

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 13 results over 10 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click 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)