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TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17
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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • 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.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Assuming the Axiom of Choice, the Borel sigma-algebra on R^n has cardinality continuum for n at least one

Statement

Assume the Axiom of Choice. For every n∈N with n≥1,

∣B(Rn)∣=c:=∣P(N)∣.

Facts & Assumptions

Given: The Axiom of Choice and a natural number n≥1.

[L1]

The rationals are countably infinite: Q≈N (Q is countably infinite).

[L3]

An infinite family of cardinality κ generates at most κℵ0 sets under the Axiom of Choice (Assuming the Axiom of Choice, an infinite family E generates at most |E|^aleph-zero sets).

[L5]

If each of two sets injects into the other, then they are equinumerous (The Schröder-Bernstein theorem).

[L7]

Under the Axiom of Choice every set can be well ordered, so the cardinalities needed here and the exponent ℵ0ℵ0 are defined (The well-ordering theorem, Cardinal sum κ⊕λ, product κ⊗λ and exponentiation κλ, and why they are written apart from the ordinal operations).

[L8]

The product of the countably infinite cardinal with itself satisfies ℵ0⊗ℵ0=ℵ0 (Hessenberg: κ⊗κ=κ for every infinite cardinal κ, proved in ZF from the canonical well-order of κ×κ).

Proof

technique · direct
1.1L1L2L3L8construct

By [L1] and repeated use of the product identity in [L8], endpoint tuples show that the rational open boxes form an at most countable family. It is infinite because q↦(q,q+1)n injects Q into that family when n≥1. Thus it is countably infinite, and [L2] and [L3] give ∣B(Rn)∣≤ℵ0ℵ0.

1.2L4L5L6L7L8

Characteristic functions inject P(N) into NN. Conversely, the graph map injects NN into P(N×N), which is equinumerous with P(N) by [L4] and [L8]. Hence [L5] and [L6] give ℵ0ℵ0=c.

1.3L2construct

For S⊆N, let Ek(S) be the singleton {(k,0,…,0)} when k∈S and the empty set otherwise. The point is defined because n≥1, each Ek(S) is closed and hence Borel by [L2], and Ψ(S):=⋃k∈NEk(S) is Borel. Distinct subsets give distinct unions, so Ψ injects P(N) into B(Rn).

2.1step 1.1step 1.2step 1.3L5∎

Steps 1.1 and 1.2 give an injection B(Rn)→P(N), while step 1.3 gives the reverse injection. Applying [L5] proves equality with c.

Depends on

Used by

Dependency tree · two levels

65 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