False statementConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17

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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

1.1assume-contra

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

1.2L4L5

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

2.1step 1.1step 1.2L1

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.

3.1step 1.2step 2.1L2L3discharge-contradiction∎

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.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

48 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

T. Tao, An Introduction to Measure Theory, Exercise 1.4.16 and Remark 1.4.18 (standard reference, not scraped)