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LemmaStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)verified 2026-07-26 (claude-opus-5)
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A nonempty set is at most countable iff it is a surjective image of N\mathbb{N}

Statement

Let AA be a nonempty set. Then AA is at most countable (Finite, countably infinite, countable, uncountable) if and only if there is a surjection s:NAs : \mathbb{N} \to A (Injection, surjection, bijection).

Moreover, from any such surjection an injection ANA \to \mathbb{N} is obtained explicitly, without any choice, by

j(a):=min{kN:s(k)=a}.j(a) := \min\{\, k \in \mathbb{N} : s(k) = a \,\}.

This is the working form of countability used everywhere below: to prove a nonempty set countable it suffices to list its elements, repetitions and all.

No choice principle is used. The backward direction is where an appeal to choice would be natural ("for each aa pick some kk with s(k)=as(k) = a") and it is avoided outright, because min\min is canonical: every nonempty set of naturals has a least element (The well-ordering principle), so j(a)j(a) is determined by aa and ss alone.

Facts & Assumptions

Given: A nonempty set AA. For aAa \in A and a function s:NAs : \mathbb{N} \to A write s1({a})={kN:s(k)=a}s^{-1}(\{a\}) = \{\, k \in \mathbb{N} : s(k) = a \,\}.

[L1]

AA is at most countable when AnA \approx n for some nNn \in \mathbb{N} or ANA \approx \mathbb{N}; A0=A \approx 0 = \varnothing holds only for A=A = \varnothing (Finite, countably infinite, countable, uncountable, The natural numbers N\mathbb{N} (von Neumann)).

[L2]

Bijections, injections, surjections, images and the symmetry and transitivity of \approx; an injection is a bijection onto its image (Injection, surjection, bijection, Equinumerous sets, ABA \approx B and ABA \preceq B).

[L3]

Well-ordering: every nonempty subset of N\mathbb{N} has a least element (The well-ordering principle).

[L4]

Every subset of an at most countable set is at most countable (Every subset of an at most countable set is at most countable).

[L5]

For naturals, kn    k<nk \in n \iff k < n, so a natural number nn is the set {kN:k<n}\{\, k \in \mathbb{N} : k < n \,\} of naturals below it; in particular 0n0 \in n whenever n0n \ne 0 (On N\mathbb{N} the order is membership: m<n    mnm < n \iff m \in n, proved earlier on this page from the additive order of Order on the natural numbers on the von Neumann naturals of The natural numbers N\mathbb{N} (von Neumann)).

Proof

technique · direct
1.1

For the forward implication assume AA is at most countable; since AA \ne \varnothing we have ANA \approx \mathbb{N}, or AnA \approx n for some nNn \in \mathbb{N} with n0n \ne 0, and in either case fix a bijection ψ\psi from N\mathbb{N}, respectively from nn, onto AA.

assume-hypgivenL1L2
1.2

For the converse implication assume a surjection s:NAs : \mathbb{N} \to A is given.

assume-hyp
2.1

If ψ\psi is defined on N\mathbb{N} it is itself a surjection NA\mathbb{N} \to A; if ψ\psi is defined on n0n \ne 0, then 0n0 \in n by [L5] and the function sψ:NAs_\psi : \mathbb{N} \to A with sψ(k)=ψ(k)s_\psi(k) = \psi(k) for k<nk < n and sψ(k)=ψ(0)s_\psi(k) = \psi(0) for knk \ge n is a surjection, since every element of AA is ψ(k)\psi(k) for some knk \in n. In both cases a surjection NA\mathbb{N} \to A exists.

step 1.1L2L5construct
2.2

For each aAa \in A the set s1({a})s^{-1}(\{a\}) is a nonempty subset of N\mathbb{N}, because ss is surjective, so [L3] provides its least element and j(a):=mins1({a})j(a) := \min s^{-1}(\{a\}) defines a function j:ANj : A \to \mathbb{N}; no selection is made, since the least element is uniquely determined.

step 1.2L3construct
3.1

jj is injective: if j(a)=j(a)=kj(a) = j(a') = k then s(k)=as(k) = a and s(k)=as(k) = a', because j(a)s1({a})j(a) \in s^{-1}(\{a\}) and j(a)s1({a})j(a') \in s^{-1}(\{a'\}), so a=aa = a'.

step 2.2L2
4.1

Hence jj is a bijection of AA onto j[A]Nj[A] \subseteq \mathbb{N}, so Aj[A]A \approx j[A]; the subset j[A]j[A] of the at most countable set N\mathbb{N} is at most countable by [L4], and transitivity of \approx transfers this to AA.

step 3.1L1L2L4
5.1

The forward implication is step 2.1 and the converse is step 4.1, so for nonempty AA countability and the existence of a surjection NA\mathbb{N} \to A are equivalent, with jj of step 2.2 the promised injection ANA \to \mathbb{N}.

step 2.1step 3.1step 4.1

Remarks

  • The hypothesis AA \ne \varnothing cannot be dropped in the forward direction: \varnothing is finite, hence at most countable, but no function N\mathbb{N} \to \varnothing exists at all. The converse direction needs no such hypothesis, since a surjection onto AA already forces AA \ne \varnothing.

  • Combining the two directions: a nonempty AA is at most countable if and only if ANA \preceq \mathbb{N} (Equinumerous sets, ABA \approx B and ABA \preceq B). The forward direction of that reformulation is immediate, and the backward direction is step 4.1.

  • The lemma is what licenses the informal phrase "enumerate AA as a0,a1,a2,a_0, a_1, a_2, \dots, possibly with repetitions". Repetitions are exactly what distinguishes a surjection from a bijection, and allowing them is what makes the criterion easy to apply: the enumerations built in A product of two at most countable sets is at most countable and Countable unions of at most countable sets, assuming ACω\mathrm{AC}_\omega repeat.

Depends on

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