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The finite Boolean expansion contradiction in the supported-ideal proof
Statement
Let be a Boolean algebra, let be a proper ideal of , let be a finite set, and let . For define its truth-assignment atom by
If for every , then the finite Boolean expansion of puts in , contradicting propriety.
Here is the exact conditional clopen pattern used in the unary supported-ideal calculation. Let be a finite nonempty index set. For each , let contain pairwise disjoint cells indexed by a finite nonempty set , with all the cells pairwise disjoint as vary. Suppose contains exactly one point of every and no other points. Assume
- whenever meets every , one has ; and
- for every , whenever meets every with , one has .
Then every belongs to , 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.
Boolean algebras and their order supplies distributivity, , and the conventions that the empty join is and the empty meet is .
Boolean ideals, filters, prime ideals and ultrafilters says that an ideal is downward closed and closed under binary joins, and that propriety means .
Proof
If , its only subset is empty, by the empty-meet convention, and therefore .
Suppose the expansion identity holds for a finite set , take , and put . [ih, construct] Every subset of is uniquely either or for a subset . With denoting the atom formed over , the two corresponding atoms over are
Finite distributivity and the induction hypothesis give the following identity.
Steps 1.1 and 2.1 prove for every finite the exact identity below.
With the identity of step 3.1 established, now assume the displayed clopen pattern and fix . If meets every , the first hypothesis puts in , and puts in by downward closure. Otherwise choose an with . The unique point of then lies in for every , so the second hypothesis puts in . Again gives . These two cases are exhaustive, including and .
Thus every term in the finite join of step 3.1 lies in . Repeated binary join closure, with the singleton case covering the empty boundary, puts their join in . This contradicts the propriety condition and proves both assertions.
Depends on
Used by
Nothing in the library uses this result yet.
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Sources
- Miroslav Repický, A proof of the independence of the Axiom of Choice from the Boolean Prime Ideal Theorem, final calculation, pp.545-546 (standard reference, not scraped)