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LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-14
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  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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Schema of continuity in the basic Cohen model

Statement

Let VZF, let P=Add(ω,ω) be the forcing in The basic Cohen symmetric system, and let G be V-generic. Define ai(n)=b exactly when some pG has p(i,n)=b, and put A={ai:iω}. Let xˉV be a finite tuple, let h:mω be injective, put sj=ah(j), and fix a formula φ(xˉ,s,A). If

V[G]φ(xˉ,s,A),

then there are pairwise disjoint basic clopen sets Uj2ω, with sjUj, such that

V[G]φ(xˉ,t,A)

whenever tAm and tjUj for every j<m.

Consequently, if finitely many parameters are ordinal-definable in V[G] from A and a fixed finite tuple u of distinct members of A, they may be held fixed while a finite tuple of distinct members of A, disjoint from rng(u), is varied through pairwise disjoint basic clopen neighbourhoods which also avoid rng(u). Here “ordinal-definable from A,u” means unique definability using the predicate A, the tuple u, and finitely many ordinal parameters, in the coded sense of Ordinal definability and HOD.

Facts & Assumptions

Given: The ZF forcing extension, tuples, formula, and displayed truth in the Statement. Existence of the particular generic G is a hypothesis. Once it and the finite tuples are fixed, the argument below makes only finitely many explicit extensions and permutations; no form of Choice is used.

[F1]

The basic Cohen symmetric system gives the finite-coordinate forcing and the action πa˙i=a˙π(i), πA˙=A˙.

[F2]
[F3]

Forcing theorem and Monotonicity, density, and decision for forcing give the truth lemma, persistence, and density closure used below.

[F4]

Symmetry lemma for forcing automorphisms transports forced formulas under finite permutations of the first coordinate.

[F5]

Ordinal definability and HOD supplies coded unique definitions from ordinal parameters.

Proof

technique · direct, with a local contradiction
1.1

If m=0, take the empty family of clopens: there is one empty tuple, and the conclusion is the given truth. Assume m>0. By the truth lemma choose pG which forces φ(xˉˇ,s˙,A˙), where s˙j=a˙h(j). It is enough to show that below every such p there is a condition forcing the asserted clopen-box conclusion, because density closure and the truth lemma then put that conclusion in V[G].

F3given
1.2

Extend p to a condition p and choose kω so that dom(p)=k×k, rng(h)k, and the rows pi=p({i}×k) are pairwise distinct for i<k. This is a finite construction: first enlarge the rectangle, then give each pair of rows a fresh column on which their bits differ. Put Uj=[ph(j)]. The Uj are pairwise disjoint because their defining binary strings are incompatible, and p forces a˙h(j)Uj.

F1F2construct
2.1

Suppose, towards a contradiction, that some rp forces that a tuple t˙A˙m lies in j<mUj but fails φ(xˉˇ,t˙,A˙). By finitely many applications of the membership forcing clause and density, strengthen r so that t˙j=a˙z(j) for a ground-model map z:mω. The disjointness of the Uj and F2 make z injective. Moreover, if z(j)<k, then rp and ra˙z(j)[ph(j)] give pz(j)=ph(j); the pairwise distinct rows imply z(j)=h(j). Thus every z(j)h(j) lies outside k.

F2F3step 1.2assume-contra
3.1

Let π interchange h(j) and z(j) whenever they differ and fix every other coordinate. The transpositions are disjoint by the last conclusion of step 2.1. The symmetry lemma gives

πr¬φ(xˉˇ,s˙,A˙),

because πa˙z(j)=a˙h(j), while check names and A˙ are fixed. On every row below k not in rng(h), πr agrees with r and hence with p. On row h(j), the part of πr below column k is the old z(j)-row of r, which equals ph(j) because r forced a˙z(j)Uj. Since p has domain k×k, p and πr are compatible. [F1, F4, step 2.1]

4.1

A common extension of p and πr would force both φ(xˉˇ,s˙,A˙), by persistence from pp, and its negation, by step 3.1. This is impossible. Hence p forces that every tuple from A in the displayed clopen box satisfies φ. Since such a p is available below every p forcing the original instance, step 1.1 and density closure prove the first assertion in V[G].

F3step 1.1step 3.1discharge-contradiction
5.1

For the consequence, choose fixed formulas and ordinal parameters which uniquely define the finitely many supported parameters from A,u. Replace their occurrences in the desired assertion by those definitions and conjoin uniqueness. Apply the first assertion to the concatenated tuple us. Keep the coordinates belonging to u at their original values and retain only the clopens belonging to s. All clopens in the larger box are pairwise disjoint, so the retained ones avoid rng(u); unique definability restores the fixed parameters after every permitted substitution. The construction is finite and makes no choice from an arbitrary family.

F5step 4.1

Depends on

Used by

Dependency tree · two levels

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Sources