Alphabeta Math
False statementConstruction: AI-adaptedVerification: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-29
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  • 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.
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FALSE: every injection of a set into itself is a bijection

Statement

FALSE. The statement

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

for all sets A.

The claim is plausible because it is true for finite A: that is clause 4 of A subset of a finite set is finite, with ∣B∣≤∣A∣, and equality holds if and only if B=A. What is easy to miss is that the proof of that clause uses finiteness twice, at the transport ∣f[A]∣=∣A∣ 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 with 0=∅ and successor σ (The natural numbers N (von Neumann)), and A:=N, f:=σ.

[L1]

(N,0,σ) satisfies the Peano axioms: σ(n)≠0 for every n, and σ 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 is exactly N∖{0}.

[L3]

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

[L4]

For finite A, every injection A→A is a bijection (A subset of a finite set is finite, with ∣B∣≤∣A∣, and equality holds if and only if B=A, clause 4), the proof going through ∣f[A]∣=∣A∣ and clause 3 of the same theorem (The cardinality ∣A∣ of a finite set).

Refutation

technique · direct
1.1

The witness is the successor map σ:N→N. It is injective by [L1].

givenL1L3
2.1

It is not surjective: 0 is not in its image, since σ(n)≠0 for every n by [L1]. Equivalently, its image is N∖{0} by [L2], a proper subset of N.

step 1.1L1L2L3
3.1

So σ is an injection of 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 A is finite, and N is not finite by [L5]. In the proof of [L4] the hypothesis is spent at the transport of cardinality along the bijection A→f[A], which presupposes A finite, and then at the conclusion f[A]=A from ∣f[A]∣=∣A∣, which is the clause of A subset of a finite set is finite, with ∣B∣≤∣A∣, and equality holds if and only if B=A that fails here: σ[N] is a proper subset of N equinumerous with it.

step 3.1L4L5∎

Remarks

  • A set for which the statement fails is called Dedekind-infinite, and the refutation above exhibits N as one. Claim 5 of The pigeonhole principle on N says that no natural number is Dedekind-infinite, which is the finite half of the same picture.

  • The relation between the two notions of infinity — "not finite" and "Dedekind-infinite" — is a genuine question of set theory without choice, and it is treated on the countability page rather than here.

  • The surjective half fails too. The map N→N sending 0 and 1 to 0 and n≥2 to n−1 is surjective and not injective, so neither half of clause 4 of A subset of a finite set is finite, with ∣B∣≤∣A∣, and equality holds if and only if B=A survives the loss of finiteness.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

26 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