TheoremStatement: AI-adaptedProof: AI-generatedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07

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.

Rank characterizes hierarchy membership

Statement

In ZF, for every set $x$ and ordinal $α$ ,

$x \in V_{α} ⟺ rank (x) < α, x \subseteq V_{α} ⟺ rank (x) \leq α .$

Thus $rank (x)$ is the least $α$ with $x \subseteq V_{α}$ , and $rank (x) = α$ iff $x \in V_{α + 1} ∖ V_{α}$ .

Facts & Assumptions

Given: Work in ZF unless the statement explicitly weakens or supplements it; fix the objects and hypotheses of the statement.

[F1]

Now assume ZF, including Foundation. Membership on the universe is well-founded and setlike, so its ordinal rank is defined for every set. Write $rank (x) = \sup {rank (y) + 1 : y \in x} .$ The empty supremum is $0$ , so $rank (\emptyset) = 0$ . If $y \in x$ , then $rank (y) < rank (x)$ . This is the Foundation-dependent special case of relation rank. The earlier construction of $V_{α}$ did not require Foundation. Conventions and prerequisites: def-rank-of-a-well-founded-relation, thm-foundation-equivalent-to-hierarchy-exhaustion. (Membership rank under Foundation)

[F2]

In ZF without Foundation, every $V_{α}$ is transitive and $α \leq β$ implies $V_{α} \subseteq V_{β}$ . Also $V_{α} \cap Ord = α$ , and both $α$ and $V_{α}$ belong to $V_{α + 1} ∖ V_{α}$ . (Transitivity and growth of hierarchy stages)

Proof

1.1F1F2

Induct on $α$ for the membership equivalence, for all sets $x$ at once. At zero neither membership in the empty stage nor rank below zero holds. At a successor $β + 1$ , $x \in V_{β + 1}$ iff every $y \in x$ lies in $V_{β}$ , iff every $rank (y) < β$ , iff $\sup_{y \in x} (rank (y) + 1) \leq β$ , iff $rank (x) < β + 1$ . All equivalences include the empty supremum case.

2.1F1F2step 1.1

At a nonzero limit $λ$ , membership means membership in some $V_{β}$ with $β < λ$ . By induction this implies rank below $λ$ . Conversely if $ρ = rank (x) < λ$ , then $ρ + 1 < λ$ and induction puts $x$ in $V_{ρ + 1}$ , hence in $V_{λ}$ .

3.1F1step 1.1step 2.1∎

Now $x \subseteq V_{α}$ iff each $y \in x$ has rank below $α$ , iff the supremum of their successor ranks is at most $α$ . This proves the subset equivalence. Taking $α = rank (x)$ gives the least-stage assertion; applying membership at $α + 1$ and at $α$ gives the successor-shell assertion.

Depends on

Used by

The universe is the class union of its stages Corollary
Balogh finite restriction data Definition
Elementary hull transfer for bounded cofinality strata Lemma
Least witness ranks give choice-free bounds Lemma
Minimum-rank selection and Collection Proposition
Ranks of ordinals and hierarchy stages Proposition
Hₖappa is a transitive subset of Vₖappa Theorem
Ranks agree and hierarchy membership is absolute Theorem

Dependency tree · two levels

6 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

Marks, Set Theory, Berkeley edition — 7.3 p.34; Weiss Exercise 33 p.100. (standard reference, not scraped)