False statementConstruction: Literature-sourcedVerification: 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.

FALSE: every subset of the real line is Borel

Statement

Assume the Axiom of Choice. Every subset of $R$ is Borel.

Facts & Assumptions

Given: The Axiom of Choice.

[L1]

Under the Axiom of Choice, $∣ B (R) ∣ = c$ (Assuming the Axiom of Choice, the Borel sigma-algebra on R^n has cardinality continuum for n at least one).

[L2]

For every set $A$ , there is no surjection $A \to P (A)$ , so $A$ is strictly smaller than its power set (Cantor's theorem: $A ≺ P (A)$ ).

[L3]

Every sigma-algebra contains the empty set (Sigma-algebras).

[L4]

The map $b \mapsto Φ ((2 b_{k})_{k})$ is a bijection from ${0, 1}^{N}$ , the set of sequences with values in ${0, 1}$ , onto the Cantor set $C$ (The Cantor set is exactly the set of $\sum_{k \geq 1} a_{k} 3^{- k}$ with every $a_{k} \in {0, 2}$ , and this gives a bijection with ${0, 1}^{N}$ , The Cantor middle-thirds set as the intersection of the sets $C_{n}$ obtained by removing open middle thirds).

[L5]

Under the Axiom of Choice, $2^{∣ A ∣} = ∣ P (A) ∣$ for every set $A$ (Assuming the Axiom of Choice, $2^{κ} = ∣ P (κ) ∣$ , and Cantor's theorem in cardinal form: $κ < 2^{κ}$ ), and $κ^{λ}$ is the cardinality of the set of functions from $λ$ to $κ$ (Cardinal sum $κ \oplus λ$ , product $κ \otimes λ$ and exponentiation $κ^{λ}$ , and why they are written apart from the ordinal operations).

Refutation

technique · contradiction

1.1

Suppose, for contradiction, that every subset of $R$ is Borel, so $P (R) = B (R)$ .

assume-contra

1.2

The Cantor set satisfies $C \subseteq R$ , and [L4] with [L5] gives $∣ C ∣ = ∣ {0, 1}^{N} ∣ = 2^{ℵ_{0}} = ∣ P (N) ∣ = c$ .

L4L5

2.1

Every subset of $C$ is a subset of $R$ , so $P (C) \subseteq P (R)$ . Step 1.1 makes that inclusion land in $B (R)$ , whose cardinality is $c$ by [L1]; hence $∣ P (C) ∣ \leq c = ∣ C ∣$ by step 1.2.

step 1.1step 1.2L1

3.1

By [L2], $∣ C ∣ < ∣ P (C) ∣$ , contradicting step 2.1. Therefore some subset of $R$ is not Borel. The cardinality argument selects no particular subset, but [L3] shows that the omitted subset cannot be empty.

step 1.2step 2.1L2L3discharge-contradiction∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 140 results over 27 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 and Remark 1.4.18 (standard reference, not scraped)