Alphabeta Math
False statementConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-14
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BPI well-orders every set

False statement

The Boolean Prime Ideal Theorem implies that every set can be well-ordered.

Why this is false

Whenever the basic Cohen symmetric construction in F1 is supplied, its model satisfies BPI and contains an infinite Dedekind-finite set of reals, which cannot be well-ordered. Independently, F2 gives the exact syntactic nonimplication conditional on Con(ZF).

Facts & Assumptions

Given: Assume Con(ZF) for the conditional nonimplication.

[F1]

The basic Cohen model satisfies BPI and fails Choice supplies the basic Cohen model and its infinite Dedekind-finite symmetric set A.

[F2]

Relative consistency of BPI without Choice over ZF supplies the exact syntactic consistency implication from ZF to ZF+BPI+¬AC.

[F3]

The Axiom of Choice defines AC as the assertion that every family of nonempty sets has a choice function.

Proof

technique · conditional countermodel
1.1

In the F1 model, suppose A had a well-order. Recursively choose the least member not chosen earlier. If the recursion stopped, A would be finite; if it did not, it would inject ω into A. Both alternatives contradict that A is infinite and Dedekind-finite. Hence A is not well-orderable although BPI holds.

F1assume-contra
2.1

Universal well-orderability implies F3 directly. Given a family F of nonempty sets, well-order F and assign to every XF its least member; Replacement produces the resulting choice function. Therefore, if ZF+BPI proved universal well-orderability, it would prove AC. This contradicts the consistency of ZF+BPI+¬AC supplied by F2 and gives the syntactic conditional counterexample.

F2F3step 1.1discharge-contradiction

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources