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Boolean Prime Ideal Theorem in the Basic Cohen Model — Examples

1 · Prerequisites

2 · Summary

The finite Boolean expansion example isolates the last algebraic calculation that appears in continuity-style arguments: once every truth-assignment atom lies in a proper ideal, their finite join forces the unit into that ideal.

The false statement records the exact logical consequence of the model. AC proves BPI, while, conditional on Con(ZF), the basic Cohen model proves that BPI does not imply AC. The consistency antecedent is part of the claim.

3 · Logical flowchart

4 · Definitions, theorems and proofs

ExampleConstruction: Literature-sourcedVerification: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-14Open item page →

The finite Boolean expansion contradiction in the supported-ideal proof

Statement

Let $B$ be a Boolean algebra, let $I$ be a proper ideal of $B$ , let $y$ be a finite set, and let $d : y \to B$ . For $z \subseteq y$ define its truth-assignment atom by

$e_{z} = (⋀_{a \in z} d (a)) \land (⋀_{a \in y ∖ z} \neg d (a)) .$

If $e_{z} \in I$ for every $z \subseteq y$ , then the finite Boolean expansion of $1$ puts $1$ in $I$ , contradicting propriety.

Here is the exact conditional clopen pattern used in the unary supported-ideal calculation. Let $K$ be a finite nonempty index set. For each $i \in K$ , let $U_{i}$ contain pairwise disjoint cells $U_{i, j}$ indexed by a finite nonempty set $J_{i}$ , with all the cells $U_{i, j}$ pairwise disjoint as $i, j$ vary. Suppose $y$ contains exactly one point of every $U_{i, j}$ and no other points. Assume

whenever $z \subseteq y$ meets every $U_{i}$ , one has $⋀_{a \in z} d (a) \in I$ ; and
for every $i \in K$ , whenever $w \subseteq y$ meets every $U_{i, j}$ with $j \in J_{i}$ , one has $⋀_{a \in w} \neg d (a) \in I$ .

Then every $e_{z}$ belongs to $I$ , so the contradiction follows. These hypotheses isolate the final finite calculation; they do not assert that the unresolved A-page primality lemma has constructed such a clopen pattern.

Facts & Assumptions

Given: The Boolean algebra, proper ideal, finite set, map, and, for the second assertion, the displayed finite clopen-incidence hypotheses.

[F1]

Boolean algebras and their order supplies distributivity, $b \lor \neg b = 1$ , and the conventions that the empty join is $0$ and the empty meet is $1$ .

[F2]

Boolean ideals, filters, prime ideals and ultrafilters says that an ideal is downward closed and closed under binary joins, and that propriety means $1 \notin I$ .

Proof

technique · induction on the finite set $y$

1.1

If $y = \emptyset$ , its only subset is empty, $e_{\emptyset} = 1$ by the empty-meet convention, and therefore $⋁_{z \subseteq y} e_{z} = 1$ .

F1base

1.2

Suppose the expansion identity holds for a finite set $y_{0}$ , take $a \notin y_{0}$ , and put $y = y_{0} \cup {a}$ . [ih, construct] Every subset of $y$ is uniquely either $w$ or $w \cup {a}$ for a subset $w \subseteq y_{0}$ . With $e_{w}^{0}$ denoting the atom formed over $y_{0}$ , the two corresponding atoms over $y$ are

$e_{w} = (\neg d (a)) \land e_{w}^{0}, e_{w \cup {a}} = d (a) \land e_{w}^{0} .$

2.1

Finite distributivity and the induction hypothesis give the following identity.

F1step 1.2ih

$\begin{aligned} \underset{z \subseteq y}{⋁} e_{z} & = \underset{w \subseteq y_{0}}{⋁} ((\neg d (a) \land e_{w}^{0}) \lor (d (a) \land e_{w}^{0})) \\ = (\neg d (a) \lor d (a)) \land \underset{w \subseteq y_{0}}{⋁} e_{w}^{0} \\ = 1 \land 1 = 1. \end{aligned}$

3.1

Steps 1.1 and 2.1 prove for every finite $y$ the exact identity below.

step 1.1step 2.1

$1 = ⋁_{z \subseteq y} [(⋀_{a \in z} d (a)) \land (⋀_{a \in y ∖ z} \neg d (a))] .$

4.1

With the identity of step 3.1 established, now assume the displayed clopen pattern and fix $z \subseteq y$ . If $z$ meets every $U_{i}$ , the first hypothesis puts $p_{z} = ⋀_{a \in z} d (a)$ in $I$ , and $e_{z} \leq p_{z}$ puts $e_{z}$ in $I$ by downward closure. Otherwise choose an $i \in K$ with $z \cap U_{i} = \emptyset$ . The unique point of $y \cap U_{i, j}$ then lies in $y ∖ z$ for every $j \in J_{i}$ , so the second hypothesis puts $n_{z} = ⋀_{a \in y ∖ z} \neg d (a)$ in $I$ . Again $e_{z} \leq n_{z}$ gives $e_{z} \in I$ . These two cases are exhaustive, including $z = \emptyset$ and $z = y$ .

F2givenstep 3.1cases

5.1

Thus every term in the finite join of step 3.1 lies in $I$ . Repeated binary join closure, with the singleton case covering the empty $y$ boundary, puts their join $1$ in $I$ . This contradicts the propriety condition $1 \notin I$ and proves both assertions.

F2step 3.1step 4.1discharge-induction∎

False statementConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-14Open item page →

BPI is equivalent to the Axiom of Choice

False statement

The Boolean Prime Ideal Theorem is equivalent over ZF to the Axiom of Choice.

Why this is false

Conditional on $Con (Z F)$ , AC strictly implies BPI over ZF: AC proves BPI, while BPI does not prove AC.

Facts & Assumptions

Given: Work over ZF. For the strictness assertion assume $Con (Z F)$ .

[F1]

The Boolean prime ideal principle and The Axiom of Choice state BPI and AC.

[F2]

Relative consistency of BPI without Choice over ZF gives the exact syntactic implication $Con (Z F) \Rightarrow Con (Z F + B P I + \neg A C)$ .

[F3]

AC implies BPI proves in ZF that AC implies BPI, with the exact Zorn argument and degenerate Boolean-algebra case.

Proof

technique · implication plus conditional countermodel

1.1

F3 gives $A C \Rightarrow B P I$ over ZF.

F1F3

1.2

If ZF+BPI proved AC, then adding $\neg$ AC would make ZF+BPI+ $\neg$ AC inconsistent. Under the given consistency hypothesis this contradicts F2.

F2assume-contra

2.1

Thus, conditional on $Con (Z F)$ , the reverse implication fails while the forward implication of step 1.1 holds. The false equivalence is refuted with exactly the stated consistency qualification.

step 1.1step 1.2discharge-contradiction∎

5 · Examples, counterexamples and false statements

None yet.

Boolean Prime Ideal Theorem in the Basic Cohen Model — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

The finite Boolean expansion contradiction in the supported-ideal proof

Statement

Facts & Assumptions

Proof

BPI is equivalent to the Axiom of Choice

False statement

Why this is false

Facts & Assumptions

Proof

5 · Examples, counterexamples and false statements

Sources