Alphabeta Math
TheoremStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (openai/gpt-5.4)verified 2026-07-26 (claude-opus-5)
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  • AI-adapted — a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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A product of two at most countable sets is at most countable

Statement

If AA and BB are at most countable (Finite, countably infinite, countable, uncountable) then so is A×BA \times B.

No choice principle is used: the two enumerations are given, and the enumeration of the product is written down from them.

Facts & Assumptions

Given: At most countable sets AA and BB, and the product A×B={(a,b):aA, bB}A \times B = \{\, (a,b) : a \in A,\ b \in B \,\}.

[L1]

Finite, countably infinite and at most countable; 0\varnothing \approx 0, so \varnothing is finite (Finite, countably infinite, countable, uncountable, The natural numbers N\mathbb{N} (von Neumann)).

[L2]

A nonempty set is at most countable if and only if some surjection N\mathbb{N} \to it exists (A nonempty set is at most countable iff it is a surjective image of N\mathbb{N}).

[L3]

There is a bijection β:NN×N\beta : \mathbb{N} \to \mathbb{N} \times \mathbb{N} (N×NN\mathbb{N} \times \mathbb{N} \approx \mathbb{N}, Equinumerous sets, ABA \approx B and ABA \preceq B).

[L4]

A composition of surjections is a surjection (Injection, surjection, bijection).

Proof

technique · direct
1.1

If A=A = \varnothing or B=B = \varnothing then A×B=A \times B = \varnothing, which is finite and hence at most countable.

givenL1
1.2

Assume instead AA \ne \varnothing and BB \ne \varnothing; then [L2] provides surjections f:NAf : \mathbb{N} \to A and g:NBg : \mathbb{N} \to B.

givenL2
1.3

Fix the bijection β:NN×N\beta : \mathbb{N} \to \mathbb{N} \times \mathbb{N} of [L3], in particular a surjection.

L3
2.1

Define t:N×NA×Bt : \mathbb{N} \times \mathbb{N} \to A \times B by t(m,n)=(f(m),g(n))t(m,n) = (f(m), g(n)). It is surjective: any (a,b)A×B(a,b) \in A \times B has a=f(m)a = f(m) and b=g(n)b = g(n) for some m,nm, n, so (a,b)=t(m,n)(a,b) = t(m,n).

step 1.2
3.1

Hence tβ:NA×Bt \circ \beta : \mathbb{N} \to A \times B is a surjection by [L4], and A×BA \times B is nonempty, so it is at most countable by [L2].

step 1.3step 2.1L2L4
4.1

Both cases give the conclusion: A×BA \times B is at most countable whenever AA and BB are.

step 1.1step 3.1L1

Remarks

  • Iterating gives the same conclusion for A1×A2××AkA_1 \times A_2 \times \dots \times A_k for each fixed kk: (A1×A2)×A3(A_1 \times A_2) \times A_3 is a product of two at most countable sets, and so on, so k1k - 1 applications of the theorem settle the case kk. Stating this uniformly in kk, as a single theorem quantified over kk, needs finite sequences of sets and a recursive definition of the kk-fold product, which this library does not yet have; the iterated form above is the honest statement of what is proved.

  • The infinite product is a different matter and is not covered: {0,1}N\{0,1\}^{\mathbb{N}} is a product of countably many two-element sets and is uncountable, by the same diagonal argument as Cantor's theorem: AP(A)A \prec \mathcal{P}(A). Countability is not preserved by infinite products of any kind.

  • Together with Every subset of an at most countable set is at most countable this gives the countability of every set that can be coded by finitely many naturals, which is how Q\mathbb{Q} is countably infinite is proved.

Depends on

Used by

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Sources