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 ANN and the continuous injection e:NNR 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.

1.1

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 e1[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].

F1F2F3A1
2.1

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.

F1F4A1step 1.1

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources