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CounterexampleConstruction: AI-generatedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passverified 2026-08-03 (gpt-5.6-sol-codex-subscription)
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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 A:=2={,{}}={0,1}A := 2 = \{\varnothing,\{\varnothing\}\} = \{0,1\} and B:=[2]1={{0},{1}}B := [2]^{1} = \{\{0\},\{1\}\}, the set of one-element subsets of 22. Both have cardinality 22, and there are exactly two bijections between them.

Facts & Assumptions

Given: A=2={0,1}A = 2 = \{0,1\} with 0=0 = \varnothing and 1={0}1 = \{0\} (The natural numbers N\mathbb{N} (von Neumann)), and B=[2]1B = [2]^{1}, the set of 11-element subsets of 22.

[L1]

n=n\lvert n\rvert = n for a natural nn, and A\lvert A\rvert is the unique natural equinumerous with AA (The cardinality A\lvert A\rvert of a finite set).

[L2]

[X]k=(Xk)\lvert [X]^{k}\rvert = \binom{\lvert X\rvert}{k} and (n1)=n\binom{n}{1} = n (The set [A]k[A]^{k} of kk-element subsets and the binomial coefficient (nk):=[n]k\binom{n}{k} := \lvert [n]^{k}\rvert).

[L5]

Claim 3 of The pigeonhole principle on N\mathbb{N}: a finite set is equinumerous with exactly one natural number.

Counterexample

technique · direct
1.1

The two sets and their cardinalities. A=2A = 2 has A=2\lvert A\rvert = 2 by [L1]. The elements of B=[2]1B = [2]^{1} are the one-element subsets of {0,1}\{0,1\}, namely {0}\{0\} and {1}\{1\}, so B=(21)=2\lvert B\rvert = \binom{2}{1} = 2 by [L2]. Hence A=B\lvert A\rvert = \lvert B\rvert.

givenL1L2
2.1

Two distinct bijections. Let f:ABf : A \to B be f(0)={0}f(0) = \{0\}, f(1)={1}f(1) = \{1\}, and let g:ABg : A \to B be g(0)={1}g(0) = \{1\}, g(1)={0}g(1) = \{0\}. Each is injective, its two values being distinct, and each is surjective, its image being all of BB; so both are bijections by [L4]. They are distinct, since f(0)={0}{1}=g(0)f(0) = \{0\} \ne \{1\} = g(0).

step 1.1L4construct
3.1

There are exactly two. By [L3] the set of bijections ABA \to B has 2!=22! = 2 elements, so ff and gg of step 2.1 are all of them.

step 1.1L3
4.1

The refuted claim fails: A=B\lvert A\rvert = \lvert B\rvert holds, and there are two bijections ABA \to B, not one. The cardinality A\lvert A\rvert of a finite set asserts only that some bijection AAA \to \lvert A\rvert exists; [L5] makes the resulting natural number unique, not the witnessing map.

step 2.1step 3.1L1L5

Remarks

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