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Fixed Boolean values come from initial collapse layers

Statement

Let B=RO(P), and let Bm be the complete subalgebra of Boolean values fixed by Hm. If uBm, then

u={pm:pP and pBu}.

Here a condition and its restriction are identified with their canonical nonzero regular-open values. Consequently Bm is exactly the complete subalgebra generated by conditions using only layers n<m.

Facts & Assumptions

Given: The Feferman–Levy forcing, its regular-open completion, m<ω, and an Hm-fixed uB.

[F1]

The Feferman–Levy symmetric collapse system defines Hm as the automorphisms acting identically on all layers below m, and allows arbitrary coordinate permutations in every layer at least m. Hereditarily symmetric names have bounded layer support fixes the use of this one bounded stabilizer for a name.

[F2]

Choice-free regular open completion of forcing preorders gives the dense embedding of P into B+=B{0} and the Boolean order and compatibility correspondence.

[F3]

Symmetry lemma for forcing automorphisms gives invariance of the ordinary forcing relation under automorphisms of P. Independently, the explicit regular-open construction in F2 is functorial: for an automorphism π of P, the map UπU is an automorphism of RO(P), because it preserves downward openness, closure, interior, complements, and arbitrary joins.

Proof

technique · contradiction, using a finite fresh-coordinate permutation
1.1

Fix pBu and suppose pm̸Bu. Density of the embedding in F2 gives qP with qpm and qB¬u.

F2assume-contra
1.2

For every layer nm appearing in q, choose a finite permutation of its i-coordinates which moves all upper-layer coordinates of q away from the finitely many upper-layer coordinates of p; extend it by the identity elsewhere. The resulting π lies in Hm. Below layer m, q extends pm and π is the identity; above it, the moved domain of πq is disjoint from the domain of p. Thus p and πq are compatible. This is a finite construction in finitely many represented layers.

F1construct
2.1

Since u is fixed by Hm, the regular-open automorphism described in F3 sends qB¬u to πqB¬u. A common extension of p and πq would lie below both u and ¬u, contradicting Boolean incompatibility. Hence pmBu.

F2F3step 1.1step 1.2discharge-contradiction
3.1

Let v be the displayed join. Step 2.1 gives vu. Conversely every pBu satisfies ppmv, and density below the regular open u gives u={pP:pBu}v. Therefore u=v.

F2step 2.1
4.1

Every initial-layer condition is fixed by Hm, so the complete subalgebra it generates is contained in Bm. The equality in step 3.1 writes every member of Bm as a join of such conditions, yielding the reverse inclusion and the final assertion.

F3step 3.1discharge-contradiction: step 1.1

Depends on

Used by

Dependency tree · two levels

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Sources