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.
Two sets of the same finite cardinality between which the bijection is not unique
Statement refuted
Refuted claim: two equinumerous finite sets admit exactly one bijection between them.
The witness is and , the set of one-element subsets of . Both have cardinality , and there are exactly two bijections between them.
Facts & Assumptions
Given: with and (The natural numbers (von Neumann)), and , the set of -element subsets of .
for a natural , and is the unique natural equinumerous with (The cardinality of a finite set).
The set of bijections between two finite sets of common cardinality has exactly elements (A finite set with has exactly bijections onto itself, and bijections onto any set of the same cardinality), and (The factorial and the falling factorial , defined by recursion in ).
A bijection is an injective surjection (Injection, surjection, bijection, Equinumerous sets, and ).
Claim 3 of The pigeonhole principle on : a finite set is equinumerous with exactly one natural number.
Counterexample
The two sets and their cardinalities. has by [L1]. The elements of are the one-element subsets of , namely and , so by [L2]. Hence .
Two distinct bijections. Let be , , and let be , . Each is injective, its two values being distinct, and each is surjective, its image being all of ; so both are bijections by [L4]. They are distinct, since .
There are exactly two. By [L3] the set of bijections has elements, so and of step 2.1 are all of them.
The refuted claim fails: holds, and there are two bijections , not one. The cardinality of a finite set asserts only that some bijection exists; [L5] makes the resulting natural number unique, not the witnessing map.
Remarks
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What is unique and what is not. Cardinality is a well-defined function of the set, but a bijection witnessing an equality of cardinalities need not be unique. Whenever a construction is made "along a bijection", one has to check, as The sum over a finite index set, and its product form does, that the result does not depend on which bijection was used.
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The count grows fast. For there are bijections onto any set of the same cardinality (A finite set with has exactly bijections onto itself, and bijections onto any set of the same cardinality), so the bijection is unique only for .
Depends on
- The cardinality $\lvert A\rvert$ of a finite set
- The pigeonhole principle on $\mathbb{N}$
- A finite set $A$ with $\lvert A\rvert = n$ has exactly $n!$ bijections onto itself, and $n!$ bijections onto any set of the same cardinality
- The set $[A]^{k}$ of $k$-element subsets and the binomial coefficient $\binom{n}{k} := \lvert [n]^{k}\rvert$
- Equinumerous sets, $A \approx B$ and $A \preceq B$
- Injection, surjection, bijection
- The factorial $n!$ and the falling factorial $n^{\underline{k}}$, defined by recursion in $\mathbb{N}$
- The natural numbers $\mathbb{N}$ (von Neumann)
Used by
Nothing in the library uses this result yet.
Dependency tree · next 3 levels
Direct dependencies and their dependencies through the next three levels: 70 results over 25 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
- Cardinality (Wikipedia) (standard reference, not scraped)
- Bijective proof (Wikipedia) (standard reference, not scraped)
- P. Halmos, Naive Set Theory, §13 (standard reference, not scraped)