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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:yB. For zy define its truth-assignment atom by

ez=(azd(a))(ayz¬d(a)).

If ezI for every zy, 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 iK, let Ui contain pairwise disjoint cells Ui,j indexed by a finite nonempty set Ji, with all the cells Ui,j pairwise disjoint as i,j vary. Suppose y contains exactly one point of every Ui,j and no other points. Assume

  • whenever zy meets every Ui, one has azd(a)I; and
  • for every iK, whenever wy meets every Ui,j with jJi, one has aw¬d(a)I.

Then every ez 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¬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 1I.

Proof

technique · induction on the finite set $y$
1.1

If y=, its only subset is empty, e=1 by the empty-meet convention, and therefore zyez=1.

F1base
1.2

Suppose the expansion identity holds for a finite set y0, take ay0, and put y=y0{a}. [ih, construct] Every subset of y is uniquely either w or w{a} for a subset wy0. With ew0 denoting the atom formed over y0, the two corresponding atoms over y are

ew=(¬d(a))ew0,ew{a}=d(a)ew0.

2.1

Finite distributivity and the induction hypothesis give the following identity.

F1step 1.2ih

zyez=wy0((¬d(a)ew0)(d(a)ew0))=(¬d(a)d(a))wy0ew0=11=1.

3.1

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

step 1.1step 2.1

1=zy[(azd(a))(ayz¬d(a))].

4.1

With the identity of step 3.1 established, now assume the displayed clopen pattern and fix zy. If z meets every Ui, the first hypothesis puts pz=azd(a) in I, and ezpz puts ez in I by downward closure. Otherwise choose an iK with zUi=. The unique point of yUi,j then lies in yz for every jJi, so the second hypothesis puts nz=ayz¬d(a) in I. Again eznz gives ezI. These two cases are exhaustive, including z= 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 1I and proves both assertions.

F2step 3.1step 4.1discharge-induction

Depends on

Used by

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Sources