ExampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-10

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.

Transporting an elementary-chain map through collapse

Example

For collapses $π_{α} : M_{α} \to {\overset{ˉ}{M}}_{α}$ of an elementary membership chain, the maps are $j_{α β} = π_{β} \circ ι_{α β} \circ π_{α}^{- 1}$ . Their action on an ordinal is an order-type embedding, and equality with inclusion is an additional condition. In ZFC the two-stage chain $X ≺ V_{κ + ω}$ constructed below gives an explicit failure of inclusion.

Facts & Assumptions

[F1]

Elementary chains and compatible collapses: A nonempty set-ordinal elementary chain of actual membership structures satisfying Extensionality has a union elementary over every stage, and the union has a transitive collapse. Conjugating the inclusions by stage and union collapses gives coherent elementary embeddings; these are not asserted to be inclusions of the transitive images.

[F2]

What the collapse fixes: Let $π : X \to \overset{ˉ}{X}$ be a collapse of actual membership as above. It fixes every transitive subset $A \subseteq X$ pointwise. If $α \in X$ is an actual ordinal, $π (α)$ is the order type of $X \cap α$ . In particular, if $X \cap α$ is transitive, $π (α) = X \cap α$ .

[F3]

Countable elementary submodels and their collapses: In ZFC, if an infinite set membership structure $M$ satisfies Extensionality, then for every at most countable $A \subseteq M$ there is a countably infinite $X ≺ M$ containing $A$ , and $X$ has a countable transitive collapse. To retain a set $a \in M$ as one parameter, use $A = {a}$ .

[F4]

Hartogs: an ordinal that does not inject into a given set: For every set $A$ there is an ordinal (def-ordinal) that does not inject into $A$ , that is, admits no injective function into $A$ . The least such ordinal is the Hartogs number $ℵ (A)$ , and it is exactly

$ℵ (A) = {o t (S, R) : S \subseteq A and R well-orders S},$

the set of order types (thm-mostowski-collapse) of the well-ordered subsets of $A$ .

The proof is choice free. That is the whole point of the theorem: in ZF alone, with no assumption that $A$ can be well ordered, one still gets an ordinal too long to be laid inside $A$ .

[F5]

The Axiom of Choice: The Axiom of Choice (AC) is the following statement.

Every family of nonempty sets has a choice function (def-choice-function).

Written out: for every set $F$ all of whose members are nonempty, there exists a function $g$ with domain $F$ satisfying $g (S) \in S$ for all $S \in F$ .

An equivalent formulation is that a product of nonempty sets is nonempty: if $X_{i} \neq \emptyset$ for every $i \in I$ , then $\prod_{i \in I} X_{i} \neq \emptyset$ . Here $\prod_{i \in I} X_{i}$ is the set of functions $f$ with domain $I$ such that $f (i) \in X_{i}$ for every $i \in I$ ; when a family of nonempty sets is indexed by itself, such an $f$ is precisely a choice function for it.

Verification

Given: A chain and its actual collapse maps, with a named ordinal at an earlier stage; ambient ZFC for the concrete witness.

1.1

For a named actual ordinal $ξ \in M_{α}$ , put $τ_{α} = ot (M_{α} \cap ξ)$ . F2 gives $π_{α} (ξ) = τ_{α}$ , so direct substitution in F1 yields $j_{α β} (τ_{α}) = π_{β} (ξ) = ot (M_{β} \cap ξ) = τ_{β}$ . On a predecessor $η \in M_{α} \cap ξ$ , its order position $ot (M_{α} \cap η)$ is sent to $ot (M_{β} \cap η)$ . These equalities specify the induced order embedding.

F1F2given

2.1

For three stages the calculation is $j_{β γ} (j_{α β} (u)) = π_{γ} (π_{β}^{- 1} (π_{β} (π_{α}^{- 1} (u)))) = π_{γ} (π_{α}^{- 1} (u)) = j_{α γ} (u)$ , with inclusions understood at the displayed domain changes. For two identical stages this computes the identity. If the named ordinal has $τ_{α} \neq τ_{β}$ , step 1.1 moves it, whereas literal inclusion would fix it. Thus inclusion requires additional agreement of collapse values and does not follow from the conjugation formula.

step 1.1algebra

3.1

Here is a chain for which the values differ. In ZFC let $κ = ℵ (ω)$ from F4, and put $θ = κ + ω$ . The transitive infinite set $V_{θ}$ contains $κ$ and satisfies Extensionality: all members of each of its elements remain in its domain, so internal agreement of members is actual agreement. F3, with the singleton parameter set ${κ}$ and the AC assumption F5, supplies a countable $X ≺ V_{θ}$ containing $κ$ . Take the two-stage chain $M_{0} = X$ , $M_{1} = V_{θ}$ . F2 makes $π_{1}$ the identity, and gives $τ = π_{0} (κ) = ot (X \cap κ)$ . This ordinal injects into $ω$ : compose the inverse order isomorphism with a countable enumeration inverse for X. F4 says $κ$ does not inject into $ω$ , so $τ \neq κ$ . Step 1.1 now calculates $j_{01} (τ) = κ \neq τ$ . This elementary transported map moves an element of its domain and therefore is not literal inclusion.

F2F3F4F5step 1.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

16 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

Geschke, Models of Set Theory — §4 collapse interface pp11–12; published chain theorem (standard reference, not scraped)