LemmaStatement: 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.

A small transitive set bounds its ranks

Statement

In ZF, if a transitive set $T$ injects into an ordinal $λ < κ$ , where $κ$ is an infinite initial ordinal, then $rank (t) < κ$ for every $t \in T$ . No regularity or Choice is assumed.

Facts & Assumptions

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

[F1]

Let $κ$ be an infinite initial ordinal. In ZF set $H_{κ} = {x : \exists λ < κ \exists j (j : TC ({x}) ↪ λ)} .$ This initially defines a class. Its root-inclusive transitive closure contains $x$ as an element. When $TC ({x})$ is well-orderable its hereditary cardinality means the least ordinal equinumerous with it; under Choice this exists for every set. The injection formulation above is used without Choice. For infinite $κ$ , replacing $TC ({x})$ by $TC (x)$ gives the same class. Indeed $TC ({x}) = {x} \cup TC (x)$ by the finite-stage formula. Adding one point to a set injecting into finite $λ$ gives an injection into $λ + 1 < κ$ ; for infinite $λ$ , keep indices at least $ω$ , shift natural indices by one, and use index zero for the added point, obtaining an injection into $λ$ . Restriction gives the converse. We retain the root-inclusive convention throughout. Conventions and prerequisites: prop-transitive-closure-minimality, def-cardinal. (Hereditary size and H_kappa)

[F2]

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)

Proof

1.1

By Replacement, $A = {rank (t) : t \in T}$ is a set of ordinals. It is downward closed. To see this, suppose $β < α \in A$ but $β \notin A$ , and choose the least attained rank $γ > β$ . Take $t \in T$ of rank $γ$ . Every $u \in t$ lies in $T$ by transitivity, has rank below $γ$ , and cannot have rank equal to $β$ or strictly between $β$ and $γ$ . Thus all have rank below $β$ , making $rank (t) \leq β$ , a contradiction. Consequently $A$ is an ordinal.

2.1

A supplied injection $j : T ↪ λ$ well-orders $T$ by its image order. For each rank in $A$ , take the preimage having least $j$ -value. Replacement produces this uniquely specified section, so $A$ injects into $λ$ . Necessarily $A < κ$ : otherwise restricting this injection to $κ \subseteq A$ would inject $κ$ into $λ < κ$ . The image, ordered as a subset of $λ$ , has order type at most $λ$ , and its bijection with $κ$ would contradict initiality. The order-type bound follows by induction along the enumeration of that subset: its element at position $ξ$ is at least $ξ$ .

F1step 1.1

3.1

For $t \in T$ its rank belongs to the ordinal $A < κ$ , so it is below $κ$ . If $T$ is empty then $A = 0$ and the conclusion about its members is vacuous. No supremum of fewer-than- $κ$ arbitrary ordinals was assumed to be below $κ$ ; the bound came from the injection.

step 1.1step 2.1∎

Depends on

Used by

Hₖappa is a transitive subset of Vₖappa 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

Weiss, An Introduction to Set Theory (2014) — Exercise 34(2), p.101. (standard reference, not scraped)