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We work in G¨odel-Bernays class theory. And we say that a structure ⟨M, A⟩

arXiv:math/9211205v1 [math.LO] 24 Nov 1992MINIMAL UNIVERSESSy D. Friedman*M.I.T.We work in G¨odel-Bernays class theory. And we say that a structure ⟨M, A⟩is a model of ZFCif M is a model of ZFC and obeys replacement for formulaswhich are permitted to mention A ⊆M as a unary predicate.

An inner model Mis minimal if there is a class A such that ⟨M, A⟩is amenable yet has no transitiveproper elementary submodel. M is strongly minimal on a club if there is a club Csuch that ⟨M, C⟩is amenable and α ∈C −→⟨V Mα , C ∩α⟩is not a model of ZFC.Strong minimality on a club implies minimality, but not conversely.

It is consistentfor L to be strongly minimal on C = ORD and if 0# exists, L is not minimal yetL[0#] may or may not be minimal.If M1 ⊆M2 are inner models then M2 is a locally generic extension of M1 if everyx ∈M2 belongs to a set-generic extension of M1. Our main result states that if Vis strongly minimal on a club and 0# exists then some inner model is both minimaland a locally generic extension of L. V can always be made strongly minimal ona club by forcing a strongly minimalizing club without adding sets (Theorem 1).Thus if 0# exists then there does exist an inner model which is both minimal anda locally generic extension of L, definable in a forcing extension of V that addsno sets.

A special case is when V = the minimal model of ZFC + 0# exists, inwhich case there is an inner model which is minimal and does not contain 0#. Thisanswers a question of Mack Stanley.Theorem 1.

(Folklore) There is a class forcing to add a club C such that ⟨V, C⟩is a model of ZFC and α ∈C −→⟨Vα, C ∩α⟩is not a model of ZFC.Proof. Conditions are bounded closed sets p such that α ∈p −→⟨Vα, p ∩α⟩isnot a model of ZFC.

Conditions are ordered by end extension. To preserve ZFCit’s enough to show that if ⟨Di|i < λ⟩is a Σn definable sequence of open denseclasses then the intersection of the Di’s is dense.

Given a condition p, first extend if*Research supported by NSF contract # 8903380.1

2necessary so that p contains an ordinal greater then λ and the parameters defining⟨Di|i < λ⟩and then build a canonical Σn-elementary chain of models ⟨Vαi|i < λ⟩and extensions pi of p in Vαi+1 −Vαi such that pi meets Di. Then at limit stages¯λ ≤λ, p¯λ is a condition since α¯λ is Vα¯λ-definably singularized by ⟨αi|i < ¯λ⟩.⊣Theorem 2.Suppose V is strongly minimal on a club and 0# exists.

Then thereis a minimal locally generic extension of L.Theorem 2 is proved using backwards Easton forcing, where 0#, C are used toselect the appropriate (minimal) almost generic extension, C being a strongly min-imalizing club.We first describe the building blocks of this backwards Eastoniteration, which are designed to produce “generic stability systems”.Definition.A stability system p consists of a successor ordinal |p| = α(p) + 1and functions fk = f pk, k > 0 such that(a)Dom f1 = Lim ∩|p|, f1(α) ≤α for α ∈Dom f1, f1(α) = lim ⟨f1(¯α)|¯α ∈Lim ∩α⟩for α ∈Lim2 ∩|p|. Define α <1 β ⇐⇒α < β and α < γ ≤β, γ ∈Dom f1 −→f1(γ) ≥α.

Then α ∈Dom f1 −→f1(α) ≤1 α. (b)Dom fk+1 = {α < |p|α a

Defineα 0, γ < κ.

The forcing P(κ, ℓ, γ) consistsof all stability systems p such that γ ≤ℓα(p) < κ. Extension of conditions is definedby: q ≤p ←→f qk ⊇f pk for all k and α(p) ≤qℓ−1 α(q), where ≤q0=≤.

We will seethat ≤is transitive.Lemma 1. For any stability system and k; ≤pk is a tree ordering and α ⩽β ≤pk γ,α ≤pk+1 γ −→α ≤pk+1 β.

Also {α|α

Suppose that the result holds for k and

3we wish to show that ≤k+1 is a tree ordering. Reflexivity is clear since we mean≤k+1 to include = .

Antisymmetry is clear since α ≤k+1 β −→α ≤β. Supposeα ≤k+1 β ≤k+1 γ and we want α ≤k+1 γ.

Since we have by definition α ≤k β ≤k γby induction we know α ≤k γ. Suppose α < δ ≤k γ, δ ∈Dom fk+1.

If δ > β thensince β ≤k+1 γ we have fk+1(δ) ≥β ≥α. If δ ≤β then δ ≤k β since ≤k is a treeordering and both δ and β are ≤k γ.

Since α ≤k+1 β we have fk+1(δ) ≥α. Sowe have shown that ≤k+1 is transitive.

Now suppose α ≤β are both ≤k+1 γ. Byinduction α ≤k β. If α < δ ≤k β, δ ∈Dom fk+1 then δ ≤k γ since ≤k is transitiveso fk+1(δ) ≥α since α ≤k+1 γ.

So α ≤k+1 β and we have shown that ≤k+1 is atree ordering.If α ⩽β ≤k γ, α ≤k+1 γ then α ≤k β since ≤k is a tree ordering. If α < δ ≤k β,δ ∈Dom fk+1 then δ ≤k γ since ≤k is transitive so fk+1(δ) ≥α since α ≤k+1 γ. Soα ≤k+1 β.Finally we show that {α|α

Suppose it holds for k and ¯α is a limit of {α|α

Then fk+1(γ) ≥α forall α

So ¯α

If α ∈Dom fk, fk(α) < αthen fk(α) = largest ¯α

Suppose α ∈Dom fk, fk(α) < α. We know that fk(α)

If fk(α) < β < α then β ≮k α since fk(α) ≱β. So fk(α) = largest ¯α

Suppose α ∈

So α = lim{fk+1(¯α)|¯α

(For, we need only first choose ¯α′0 to guaranteefk+1(¯α) ≥α0 for all ¯α beyond ¯α′0 and then minimize fk+1(¯α′0) to get ¯α0.) Thenfk+1(¯α0)

If β ≤¯α0 then fk+1(β) ≥fk+1(¯α0)since fk+1(¯α0)

So fk+1(¯α0)

4Lemma 3. Suppose r ≤q ≤p in P(κ, ℓ, γ).

Then r ≤p.Proof. We need to check that α(p) ≤rℓ−1 α(r).

But α(p) ≤qℓ−1 α(q) and so α(p) ≤rℓ−1α(q), and α(q) ≤rℓ−1 α(r). So the result follows from Lemma 1.⊣Lemma 4.

Suppose p ∈P(κ, ℓ, γ) and α(p) ≤α < κ. Then there exists q ≤p,α(q) = α.Proof.

For limit λ ∈(α(p), α] define f qk(λ) = λ. It is routine to verify that theresulting q is a condition and extends p.⊣Lemma 5.

Suppose p0 ≥p1 ≥· · · is a sequence of conditions in P(κ, ℓ, γ) of length< κ. Then there is p ≤each pi, α(p) = Siα(pi).Proof.

Assume that the pi’s are distinct. Let α = Siα(pi).

We must define f pk(α).We do so by induction on k > 0. If α /∈Lim, f p1 (α) is undefined.

If α ∈Lim2, letf p1 (α) = lim ⟨f pi1 (¯α)|¯α ≤α(pi), ¯α limit⟩. If α ∈Lim −Lim2 then let f p1 (α) = α.Assuming f pk(α) is defined (and f pk ↾α = Sif pik ) it makes sense to ask if α ∈

If not then f pk+1(α) is undefined. If α ∈

If α ∈

Suppose α(pi) <β ≤pk α, β ∈Dom f pk+1. If β < α then we can choose j so that β ≤pjkα(pj) andthen f pjk+1(β) ≥α(pi) since pj ≤pi.

If β = α then f pk+1(β) < α(pi) can only resultif f pk+1(¯α) < α(pi) for some ¯α α(p) but then ¯α

P is the iteration with Easton supports over L where P0 = the trivialforcing, Pλ = inverse limit at singular λ, direct limit at regular λ, Pκ+1 = Pκ ∗˙Qκwhere ˙Qκ is a term for the trivial forcing unless κ is regular. For regular κ, ˙Qκ is aterm for the following forcing in L[Gκ], Gκ denoting the Pκ-generic: choose a pair

5(ℓκ, γk) with ℓκ > 0, γκ < κ and apply the forcing P(κ, ℓκ, γκ). Now Pκ ⊩˙Qκ is< κ-closed and has cardinality κ, so P preserves cofinalities.Our goal is to build G = ⟨Gα|α ∈ORD ⟩so that Gα is Pα-generic over L andto select ordinals αi ∈[i, i∗), i < i∗adjacent Silver indiscernibles such that (writingGα+1 = Gα ∗gα) :1. i < j in I = Silver indiscernibles, p ∈gi, q ∈gj, α(q) ≥i −→f pk ⊆f qk for allk.

let fk = S{f pk|p ∈gi for some i ∈I}.2. For i ∈I, ℓi = least ℓsuch that the ⟨0#, C⟩, i −Σℓstables in C are boundedin i, where C = the given strongly minimalizing club for V. (α is B, β −Σℓstable if⟨Lα[B], B ∩α⟩is a Σℓ-elementary submodel of ⟨Lβ[B], B ∩β⟩).

Also γi = αj wherej = S{⟨0#, C⟩, i −Σℓi stables in C} ≥0. (By convention, α0 = 0.)3.

For i ∈I, fk(i) = i if k < ℓi and fℓi(i) = γi.4. For i ∈I, i ≤ℓi αi (where ≤k is defined from the fk’s) and αi ∈Dom fℓi+1, fℓi+1(αi) =αj where j = S{⟨0#, C⟩, i −Σℓi+1 stables in C}.Suppose that the fk’s have been constructed to obey 1–4 above and we nowprove Theorem 2.

The desired minimal, locally generic extension of L is L[⟨Gα|α ∈ORD⟩], witnessed by the amenable class ⟨fk|k ∈ω⟩. The reason for minimality isroughly as follows: there are unboundedly many α

More precisely:Lemma 6. let i < j be indiscernibles, i⟨0#, C⟩, j −Σk stable, i ∈C and (j a limitof ⟨0#, C⟩, j −Σk−2 stables or k ≤2). Then i ≤k αi

By induction on k and for fixed k by induction on j. By property 4, i ≤ℓi αiand clearly ℓi ≥k since i is a limit of ⟨0#, C⟩, i−Σk−1 stables (k > 1) and so ℓi ≥kfollows by property 2.

So we only need to check that αi

Thenβ < j as properties 2, 3 imply that f1(j) ≥αi, since i is ⟨0#, C⟩, j −Σ1 stableand belongs to C. There are no indiscernibles between β and j as otherwise we canapply induction on j. So ¯j ≤β < j where ¯j = I-predecessor to j.

In fact ¯j < β < jsince otherwise αi < β = ¯j and again 2, 3 imply that f1(¯j) ≥αi. If f1(j) < β then

6since f1(j) <1 j we have f1(j) ≤f1(β) and hence f1(β) ≥f1(j) ≥αi, contrary toassumption. So f1(j) ≥β > ¯j and by 3, f1(j) = α¯j, ¯j is ⟨0#, C⟩, j −Σ1 stable andbelongs to C. And ¯j < β ≤α¯j.

But ¯j ≤1 α¯j so f1(β) ≥¯j and ¯j > αi since β ≤α¯jand β > αi. So f1(β) > αi, contradicting our assumption.Suppose the lemma holds for k and we prove it for k+1.

If αi ≮k+1 j then chooseαi < β ≤k j so that β ∈Dom fk+1, fk+1(β) < αi. By 2, 3 we have β

By 2, 3 fk(j) is defined and equal to α¯j where ¯j = S{⟨0#, C⟩, j −Σkstables in C} and since β

Now α¯j = fk(j)

Clearly ℓ≥k since ¯j is⟨0#, C⟩, j−Σk stable, ¯j ∈C and hence ¯j = S{⟨0#, C⟩, ¯j−Σk−1 stables in C}. But ifℓ> k then by 4, ¯j

So ℓ= k, ¯j

Thus αi ≤fk+1(α¯j) < β.This contradicts β ≤k α¯j, fk+1(β) < αi since fk+1(α¯j)

Then for each k there are cofinally many α

The class of all such i is cofinal in the ordinals.⊣Corollary 8. No α is

Choose k large enough so that i = the least ⟨0#, C⟩−Σk stable in C is largerthan α. Then i

So α ≮k ∞.⊣Thus we have minimality, since if ⟨L[⟨Gα|α ∈ORD⟩], ⟨fk|k ∈ω⟩⟩had a transitiveelementary submodel, by Corollary 7 its height α would be

When defining

7Gi∗+1 we also specify αi ∈[i, i∗), where i∗denotes the I-successor to i.Gi0+1, i0 = min I Choose Gi0 to be the L[0#]-least generic for Pi0, using the count-ability of i0. Set ℓi0 = 1, γi0 = 0 = α0 and choose gi0 to be the L[0#]-least genericfor P(i0, 1, 0) as defined in L[Gi0].Gi∗+1, i ∈IFirst choose Gi∗to be the L[0#, C]-least generic for Pi∗extendingGi+1, using the ≤i-closure of Pi+1,i∗in L[Gi+1], where Pi∗= Pi+1 ∗Pi+1,i∗.

(Notethat the dense sets in Pi+1,i∗can be grouped into countably many collections ofsize i, enabling an easy construction of a generic.) The key step involves the choiceof gi∗.Choose n ≥0 so that the ordertype of the ⟨0#, C⟩, i −Σℓstables in C is λ + n,λ limit or 0, where ℓ= ℓi = least ℓsuch that the ⟨0#, C⟩, i −Σℓstables in C arebounded in i.

Let Qi∗be the forcing P(i∗, ℓ+ 1, αj) as defined in L[Gi∗], wherej = S{⟨0#, C⟩, i −Σℓ+1 stables in C} ≥0. Let p0 ∈Qi∗be the “condition” definedby α(p0) = i, f p0k↾i = S{f pk|p ∈gi}, f p0k (i) = i if k < ℓ, f p0ℓ(i) = γi = αj′ wherej′ = S{⟨0#, C⟩, i −Σℓstables in C} ≥0.

We will verify later that p0 is indeed acondition in Qi∗. Choose p1 ≤p0 in Qi∗to meet all dense ∆in L[Gi∗] definable inL[Gi∗] from Gi∗and parameters in (i+1) ∪{j1 · · · jn} where j1 · · · jn are the first nindiscernibles ≥i∗.

Also arrange that α(p1) is a

Choose gi∗to be generic for Qi∗= P(i∗, 1, αi)over L[Gi∗], extending the condition p1 ∈Qi∗.Gi+1, i ∈Lim I Gi = S{Gj|j ∈I ∪i}. Let Qi = P(i, ℓi, γi) in L[Gi] where ℓi isthe least ℓsuch that the ⟨0#, C⟩, i −Σℓstables in C are bounded in i and γi = αjwhere j = S{⟨0#, C⟩, i −Σℓi stables in C}.

Let fk ↾i = S{f pk|p ∈gj for somej ∈I ∩i}. Then gi = Qi-generic determined by ⟨fk ↾i|k ∈ω⟩.

We will verify laterthat ⟨fk ↾i|k ∈ω⟩does indeed determine a Qi-generic over L[Gi].Lemma 9. Assume that the verifications claimed in the construction can be carriedout.

Then 1–4 hold.Proof. Everything is clear, with the possible exception of the first statement in 4:i ≤ℓi αi.

But note that in the construction of Gi∗+1, i = α(p0), αi = α(pi) wherep1 ≤p0 in P(i∗, ℓi + 1, αj) for some j, so we’re done by the definition of extension

8of conditions.⊣Lemma 10. By induction on i ∈I :(a)gi is well-defined and Qi-generic, where Qi = P(i, ℓi, γi) as interpreted byGi.

(b)Define p0 by:α(p0) = i, f p0k↾i = S{f pk|p ∈gi}, f p0k (i) = i if k < ℓi,f p0ℓi (i) = γi. Then p0 is a stability system.

(c) Lemma 6 holds for indiscernibles ≤i. (d) If p0 is defined as in (b) then p0 ∈Qi∗, as defined in the construction.Proof.

(a) This follows by induction unless ℓi > 1 and there is a final segment i0

By construction, pn1 meets ∆in L[Gi∗n] defined from Gi∗n and parameters in (in+1) S (least n indiscernibles ≥i∗n)where ∆is dense on P(i∗n, ℓi, γi) and pn1 is the condition in gi∗n with α(pn1) = αin. Byan inductive use of (c), in n, where p0 is defined asin (b).

So αi0

Sop0 obeys the requirements for a stability system. (c) The proof of Lemma 6 for indiscernibles ≤i only used the facts that p0 is astability system and 1–4 hold ≤i.

So we are done by Lemma 9 through i. (d) We must verify that αj

This follows from (c).⊣

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