CorollaryStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passverified 2026-08-05 (gpt-5.6-sol-codex-subscription)‡ rests on unproved material (inherited)

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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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

‡ Rests on 3 statements not proved in this library, by way of the results it cites. This item cites no such statement directly; it depends on results that do. The unproved premises it inherits are ‡ Cohen's first model: an infinite Dedekind-finite set of reals, ‡ Sierpiński 1947: the generalised continuum hypothesis implies the Axiom of Choice and ‡ The continuum hypothesis and its generalisation are independent of ZFC. Each is recorded with a citation to the literature and is not established here, because the track that would prove it has not yet been developed in this library. Everything else in this proof is proved here.

Absorption: for cardinals $\kappa, \lambda$ with $\kappa$ infinite and $\lambda \le \kappa$ , $\kappa \oplus \lambda = \kappa$ , and $\kappa \otimes \lambda = \kappa$ when $\lambda \ne 0$

Statement

Let $\kappa$ be an infinite cardinal and $\lambda$ a cardinal with $\lambda \le \kappa$ (Cardinal (initial ordinal) and cardinality). Then

$\kappa \oplus \lambda = \kappa, \qquad \text{and} \qquad \kappa \otimes \lambda = \kappa \ \text{ whenever } \lambda \ne 0$

(Cardinal sum $\kappa \oplus \lambda$ , product $\kappa \otimes \lambda$ and exponentiation $\kappa^{\lambda}$ , and why they are written apart from the ordinal operations). The exception at $\lambda = 0$ is not an artefact: $\kappa \otimes 0 = 0$ .

This is a theorem of ZF, inherited from Hessenberg: $\kappa \otimes \kappa = \kappa$ for every infinite cardinal $\kappa$ , proved in ZF from the canonical well-order of $\kappa \times \kappa$ , which is choice free. In particular the ordinary arithmetic of infinite cardinals collapses completely for $\oplus$ and $\otimes$ : below the level of exponentiation, the larger argument simply swallows the smaller one.

Facts & Assumptions

Given: An infinite cardinal $\kappa$ and a cardinal $\lambda \le \kappa$ , in ZF.

[L1]

$\kappa \otimes \kappa = \kappa$ for every infinite cardinal $\kappa$ (Hessenberg: $\kappa \otimes \kappa = \kappa$ for every infinite cardinal $\kappa$ , proved in ZF from the canonical well-order of $\kappa \times \kappa$ ).

[L2]

$\kappa \oplus \lambda = \lvert \kappa \sqcup \lambda\rvert$ and $\kappa \otimes \lambda = \lvert \kappa \times \lambda\rvert$ , with $\kappa \sqcup \lambda = (\{0\} \times \kappa) \cup (\{1\} \times \lambda)$ (Cardinal sum $\kappa \oplus \lambda$ , product $\kappa \otimes \lambda$ and exponentiation $\kappa^{\lambda}$ , and why they are written apart from the ordinal operations).

[L3]

For cardinals, $\kappa \le \lambda$ iff $\kappa \preceq \lambda$ ; $A \preceq B$ with both well-orderable gives $\lvert A\rvert \le \lvert B\rvert$ ; the unit laws $\kappa \otimes 1 = \kappa$ and $\kappa \otimes 0 = 0$ hold; and $\oplus$ , $\otimes$ are monotone in each argument (claims (a), (d), (e) of Commutativity, associativity, distributivity and monotonicity of $\oplus$ and $\otimes$ , the unit laws, the two exponent laws, and $\kappa \le \lambda$ if and only if $\kappa$ injects into $\lambda$ ).

[L4]

For a well-orderable set $X$ , $\lvert X\rvert$ is the least ordinal equinumerous with $X$ ; and $\lvert \alpha\rvert = \alpha$ exactly when $\alpha$ is a cardinal (A set equinumerous with some ordinal has a least such ordinal, that ordinal is a cardinal, and equinumerous sets get the same one; no choice principle is used).

[L5]

Every natural number is a cardinal and $\omega$ is a cardinal, so $0, 1, 2 \in \omega \subseteq \kappa$ (Every natural number and $\omega$ are cardinals, every infinite cardinal is a limit ordinal, and on the natural numbers the cardinal operations are the published finite counting operations, with $\lvert A \rvert$ in the finite sense equal to $\lvert A \rvert$ in the cardinal sense, $\omega$ is the least limit ordinal).

[L6]

Ordinals are comparable, $\alpha \subseteq \beta$ iff $\alpha \in \beta$ or $\alpha = \beta$ , trichotomy holds, and $\alpha \subseteq \beta \subseteq \alpha$ forces $\alpha = \beta$ (Trichotomy and well-ordering of the ordinals, Basic closure properties of ordinals).

[L7]

A subset inclusion is an injection and a bijection witnesses $\approx$ (Injection, surjection, bijection, Equinumerous sets, $A \approx B$ and $A \preceq B$ ).

Proof

technique · direct

1.1

The map $\xi \mapsto (0,\xi)$ is an injection $\kappa \to \kappa \sqcup \lambda$ , so $\kappa \le \kappa \oplus \lambda$ by [L2], [L3] and [L4].

L2L3L4L7

1.2

The map $(i,\xi) \mapsto (\xi,i)$ is an injection $\kappa \sqcup \kappa \to \kappa \times \kappa$ , since its image lies in $\kappa \times 2$ and $2 \in \omega \subseteq \kappa$ by [L5], and it is injective because both coordinates are recovered from the image.

L5L7

1.3

From $\lambda \le \kappa$ and monotonicity, $\kappa \oplus \lambda \le \kappa \oplus \kappa$ and $\kappa \otimes \lambda \le \kappa \otimes \kappa$ .

1.4

If $\lambda \ne 0$ then $1 \le \lambda$ by [L6] and [L5], so $\kappa = \kappa \otimes 1 \le \kappa \otimes \lambda$ by the unit law and monotonicity in [L3].

L3L5L6

2.1

$\kappa \oplus \kappa \le \kappa$ : step 1.2 with [L3] gives $\lvert \kappa \sqcup \kappa\rvert \le \lvert \kappa \times \kappa\rvert$ , which by [L2] and [L1] is $\kappa \otimes \kappa = \kappa$ .

step 1.2L1L2L3

3.1

Combining: $\kappa \le \kappa \oplus \lambda \le \kappa \oplus \kappa \le \kappa$ gives $\kappa \oplus \lambda = \kappa$ by [L6]; and for $\lambda \ne 0$ , $\kappa \le \kappa \otimes \lambda \le \kappa \otimes \kappa = \kappa$ gives $\kappa \otimes \lambda = \kappa$ , while $\kappa \otimes 0 = 0$ by [L3].

step 1.1step 1.3step 1.4step 2.1L1L3L6∎

Remarks

What absorption costs. Nothing beyond Hessenberg: $\kappa \otimes \kappa = \kappa$ for every infinite cardinal $\kappa$ , proved in ZF from the canonical well-order of $\kappa \times \kappa$ : the only extra input is the injection of step 1.2, which folds two copies of $\kappa$ into a rectangle of width $2$ . So absorption is choice free wherever Hessenberg's theorem is, that is, for cardinals.

Why the hypothesis is $\lambda \le \kappa$ and not $\lambda < \kappa$ . The case $\lambda = \kappa$ is the interesting one and is used constantly: $\kappa \oplus \kappa = \kappa$ and $\kappa \otimes \kappa = \kappa$ . Stating the corollary with $\le$ avoids a separate appeal to Hessenberg's theorem at every later use.

Absorption destroys cancellation. From $\kappa \oplus \lambda = \kappa$ for every $\lambda \le \kappa$ it follows at once that $\oplus$ cannot be cancellative on infinite cardinals, and the companion false statement FALSE: $\kappa \oplus \mu = \lambda \oplus \mu$ implies $\kappa = \lambda$ records exactly that. The same collapse does not reach exponentiation: assuming the Axiom of Choice, Assuming the Axiom of Choice, $2^{\kappa} = \lvert \mathcal{P}(\kappa) \rvert$ , and Cantor's theorem in cardinal form: $\kappa < 2^{\kappa}$ gives a strict increase at every cardinal.

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 103 results over 30 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

K. Kearnes, Cardinal Arithmetic (Fall 2025 course handout) (standard reference, not scraped)
Cardinal number — cardinal arithmetic (Wikipedia) (standard reference, not scraped)
T. Jech, Set Theory, 3rd millennium ed., Ch. 3 (Cardinal numbers) (standard reference, not scraped)

Absorption: for cardinals κ,λ\kappa, \lambdaκ,λ with κ\kappaκ infinite and λ≤κ\lambda \le \kappaλ≤κ, κ⊕λ=κ\kappa \oplus \lambda = \kappaκ⊕λ=κ, and κ⊗λ=κ\kappa \otimes \lambda = \kappaκ⊗λ=κ when λ≠0\lambda \ne 0λ=0