TheoremStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17‡ rests on unproved material (inherited)

How statement and proof provenance work

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.

‡ Rests on 3 statements not proved in this library, by way of the results it cites. This item cites no such statement directly; it depends on results that do. The unproved premises it inherits are ‡ Cohen's first model: an infinite Dedekind-finite set of reals, ‡ Sierpiński 1947: the generalised continuum hypothesis implies the Axiom of Choice and ‡ The continuum hypothesis and its generalisation are independent of ZFC. Each is recorded with a citation to the literature and is not established here, because the track that would prove it has not yet been developed in this library. Everything else in this proof is proved here.

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 $n \in N$ with $n \geq 1$ ,

$∣ B (R^{n}) ∣ = c : = ∣ P (N) ∣ .$

Facts & Assumptions

Given: The Axiom of Choice and a natural number $n \geq 1$ .

[L1]

The rationals are countably infinite: $Q \approx N$ ( $Q$ is countably infinite).

[L2]

Rational open boxes generate $B (R^{n})$ , and closed subsets of $R^{n}$ are Borel (For n at least one, open sets, closed sets, compact sets, open balls, boxes, rational open boxes, and rational half-open boxes generate the Borel sigma-algebra on R^n).

[L3]

An infinite family of cardinality $κ$ generates at most $κ^{ℵ_{0}}$ sets under the Axiom of Choice (Assuming the Axiom of Choice, an infinite family E generates at most |E|^aleph-zero sets).

[L4]

Equinumerous sets have equinumerous power sets, and function spaces are transported by bijections (Disjoint union, cartesian product, function space and power set respect equinumerosity, and for ordinals $α, β$ the sets $α ⊔ β$ and $α \times β$ carry explicit well-orders, so their cardinalities exist in ZF).

[L5]

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

[L6]

Under the Axiom of Choice, $2^{ℵ_{0}} = ∣ P (N) ∣ = c$ (Assuming the Axiom of Choice, $2^{κ} = ∣ P (κ) ∣$ , and Cantor's theorem in cardinal form: $κ < 2^{κ}$ ).

[L7]

Under the Axiom of Choice every set can be well ordered, so the cardinalities needed here and the exponent $ℵ_{0}^{ℵ_{0}}$ are defined (The well-ordering theorem, Cardinal sum $κ \oplus λ$ , product $κ \otimes λ$ and exponentiation $κ^{λ}$ , and why they are written apart from the ordinal operations).

[L8]

The product of the countably infinite cardinal with itself satisfies $ℵ_{0} \otimes ℵ_{0} = ℵ_{0}$ (Hessenberg: $κ \otimes κ = κ$ for every infinite cardinal $κ$ , proved in ZF from the canonical well-order of $κ \times κ$ ).

Proof

technique · direct

1.1

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 $q \mapsto (q, q + 1)^{n}$ injects $Q$ into that family when $n \geq 1$ . Thus it is countably infinite, and [L2] and [L3] give $∣ B (R^{n}) ∣ \leq ℵ_{0}^{ℵ_{0}}$ .

L1L2L3L8construct

1.2

Characteristic functions inject $P (N)$ into $N^{N}$ . Conversely, the graph map injects $N^{N}$ into $P (N \times N)$ , which is equinumerous with $P (N)$ by [L4] and [L8]. Hence [L5] and [L6] give $ℵ_{0}^{ℵ_{0}} = c$ .

L4L5L6L7L8

1.3

For $S \subseteq N$ , let $E_{k} (S)$ be the singleton ${(k, 0, \dots, 0)}$ when $k \in S$ and the empty set otherwise. The point is defined because $n \geq 1$ , each $E_{k} (S)$ is closed and hence Borel by [L2], and $Ψ (S) : = ⋃_{k \in N} E_{k} (S)$ is Borel. Distinct subsets give distinct unions, so $Ψ$ injects $P (N)$ into $B (R^{n})$ .

L2construct

2.1

Steps 1.1 and 1.2 give an injection $B (R^{n}) \to P (N)$ , while step 1.3 gives the reverse injection. Applying [L5] proves equality with $c$ .

step 1.1step 1.2step 1.3L5∎

Depends on

Used by

FALSE: every subset of the real line is Borel False statement

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 157 results over 30 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

T. Tao, An Introduction to Measure Theory, Exercise 1.4.16 (standard reference, not scraped)
D. H. Fremlin, Measure Theory, Chapter 56, Section 561A and result 567E(b) (standard reference, not scraped)