Alphabeta Math
False statementConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-10
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Every set of reals is Borel

Statement

False assertion: every subset of R is Borel.

In ZFC, take an undetermined payoff A⊆NN and the continuous injection e:NN→R supplied below. The set e[A] is a witness that the assertion fails.

Facts & Assumptions

[F1]

Choice produces an undetermined natural-number game supplies A with neither player winning under AC.

[F2]

Continuous injections of sequence spaces into the real line gives the continuous injection e in ZF.

[F3]

Borel hierarchy exhaustion and preservation by continuous pullback makes continuous inverse images of Borel sets Borel in ZFC.

[F4]

Borel games are determined determines every Borel natural-number payoff in ZFC.

Refutation

Given: Work in ZFC throughout this counterexample.

1.1F1F2F3A1

F1 with A1 supplies A and F2 supplies e. Suppose e[A] were Borel in R. By continuity of e, F3 with A1 would make e−1[e[A]] Borel in Baire space. This preimage equals A: if e(x)=e(a) for some a in A, injectivity gives x=a; conversely each a in A maps into e[A].

2.1F1F4A1step 1.1

Then F4 with A1 would give a winning strategy for one player for payoff A, contradicting its defining property from F1. Hence e[A] is not Borel and is the required subset of the real line witnessing the failed assertion. QED.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources