Alphabeta Math
False statementConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-29
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FALSE: every injection of a set into itself is a bijection

Statement

FALSE. The statement

every injective function f:AAf : A \to A from a set to itself is a bijection

for all sets AA.

The claim is plausible because it is true for finite AA: that is clause 4 of A subset of a finite set is finite, with BA\lvert B\rvert \le \lvert A\rvert, and equality holds if and only if B=AB = A. What is easy to miss is that the proof of that clause uses finiteness twice, at the transport f[A]=A\lvert f[A]\rvert = \lvert A\rvert and at the step "a subset of the same cardinality as the whole is the whole", and neither survives without it.

Facts & Assumptions

Given: The von Neumann naturals N\mathbb{N} with 0=0 = \varnothing and successor σ\sigma (The natural numbers N\mathbb{N} (von Neumann)), and A:=NA := \mathbb{N}, f:=σf := \sigma.

[L1]

(N,0,σ)(\mathbb{N},0,\sigma) satisfies the Peano axioms: σ(n)0\sigma(n) \ne 0 for every nn, and σ\sigma is injective (The von Neumann naturals form a Peano system).

[L2]

Every nonzero natural is a successor (Every nonzero natural number is a successor), so the image of σ\sigma is exactly N{0}\mathbb{N}\setminus\{0\}.

[L3]

Injection, surjection, bijection (Injection, surjection, bijection): ff is surjective when its image is the whole codomain, and bijective when injective and surjective.

[L4]

For finite AA, every injection AAA \to A is a bijection (A subset of a finite set is finite, with BA\lvert B\rvert \le \lvert A\rvert, and equality holds if and only if B=AB = A, clause 4), the proof going through f[A]=A\lvert f[A]\rvert = \lvert A\rvert and clause 3 of the same theorem (The cardinality A\lvert A\rvert of a finite set).

[L5]

N\mathbb{N} is not finite: N≉n\mathbb{N} \not\approx n for every natural nn (The pigeonhole principle on N\mathbb{N}, claim 4, Finite, countably infinite, countable, uncountable, Equinumerous sets, ABA \approx B and ABA \preceq B).

Refutation

technique · direct
1.1

The witness is the successor map σ:NN\sigma : \mathbb{N} \to \mathbb{N}. It is injective by [L1].

givenL1L3
2.1

It is not surjective: 00 is not in its image, since σ(n)0\sigma(n) \ne 0 for every nn by [L1]. Equivalently, its image is N{0}\mathbb{N}\setminus\{0\} by [L2], a proper subset of N\mathbb{N}.

step 1.1L1L2L3
3.1

So σ\sigma is an injection of N\mathbb{N} into itself that is not a bijection, and the displayed statement is false.

step 1.1step 2.1L3
4.1

Finiteness is exactly the missing hypothesis. By [L4] the statement is true whenever AA is finite, and N\mathbb{N} is not finite by [L5]. In the proof of [L4] the hypothesis is spent at the transport of cardinality along the bijection Af[A]A \to f[A], which presupposes AA finite, and then at the conclusion f[A]=Af[A] = A from f[A]=A\lvert f[A]\rvert = \lvert A\rvert, which is the clause of A subset of a finite set is finite, with BA\lvert B\rvert \le \lvert A\rvert, and equality holds if and only if B=AB = A that fails here: σ[N]\sigma[\mathbb{N}] is a proper subset of N\mathbb{N} equinumerous with it.

step 3.1L4L5

Remarks

Depends on

Used by

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Sources