Implicancias del Estudio

In this section we work in the theorySST, unless explicitly stated otherwise.

Definition 10.1 Let x ∈ A ∈ Sα. An α-pedigree for x over A is a sequence

~u=hun :n≤νiwhere ν∈ω and

(i) everyun is a stratified ultrafilter over A[ie, un∈β∞A]

(ii)u0vα; uν =x

(iii) (∀n<m≤ν)(un<αum)

(iv) (∀z∈un)(z<αun+1⇒un+1 ∈z), for all n< ν.

The ultrafilter u₀ is called the α-type ofx over A; we denote it tpα(x;A). We also use~u+:=hun: 0<n≤νi. Pedigreeandtypemean 0-pedigree and 0-type, resp. The definition as stated is clearly of the formPα_{(x,A) [where P(x,A) is P}0_(x,A)], but it is useful to note that

(iii) holds iffα<u₁ ∧ (∀n,m)(1≤n<m≤ν ⇒un<um), forν >0, and (iv) holds iff (∀z∈un)(z<un+1⇒un+1∈z), for all n< ν.

The condition (iv) perhaps becomes more meaningful when stated in terms of monads. Definition 10.2 For U∈_Sα we writexMαU iff (∀X∈U∩Sα)(x∈X).

The classMαU=T(U∩Sα) is theα-monadofU.

In general,MαU is a proper class. Also, it can be empty. Here is a list of some useful elementary facts about monads.

(1) If U is principal and generated by a, then a ∈ _Sα and MαU = {a} is a set. Conversely, ifMαU∩Sα 6=0, then U is principal. Hence U is nonprincipal if and only ifMαU∩Sα =0.

(2)αvβ impliesMαU⊇MβU.

(3) LetU,V ∈Sα. IfU∼V, thenMαU=MαV. IfMαU∩MαV6=0, thenU∼V. The condition (iv) in Definition10.1can be restated as:

Proposition 10.3 (SST) Everyx∈A∈_Sα has at most oneα-pedigree over A.

More generally: Letx∈A∈_Sα0,α0vα1. Let~u=hun :n≤νi be anα0-pedigree

forx overA, and let~v=hvm:m≤µi be anα1-pedigree forx overA. Then µ≤ν

andvm=u(ν−µ)+mfor allm≤µ.

Proof We begin by noticing that u_ν =x=v_µ.Let jbe the largest integer such that

j≤min{ν, µ} anduν−i =vµ−i holds for alli≤j. If j=µ, thenµ≤ν and we are done. If j= ν, then ν ≤µ. As u0 ∈Sα0 ⊆Sα1, we have u0 =uν−ν = vµ−ν, so

v_µ−ν ∈Sα1 andµ−ν =0 [if not, vµ−ν−1 <α1 vµ−ν, but this is impossible], hence

µ=ν, and again we are done.

It remains to show that the remaining casej< ν, j< µleads to a contradiction. We consideruν−j−1 andvµ−j−1. Let us assume thatuν−j−1vvµ−j−1. We letβvµ−j−1 and eu := vµ−j = uν−j, so that β < eu. By clause (iv) in the definition of pedi-

grees,euMβvµ−j−1 and alsoeuMβuν−j−1. By the property (3) of monads, this implies

uν−j−1 ∼vµ−j−1 and hence [Proposition9.1] uν−j−1 =vµ−j−1, a contradiction with the choice ofj.

The argument for the casevµ−j−1vuν−j−1 is analogous.

Proposition 10.4 (SST) LetA,B∈_Sα, x∈A∩B. If there is an α-pedigree forx

overA, then there is anα-pedigree for xoverB.

Proof Let~u=hu₀, . . . ,uνi be anα-pedigree forx overA. LetC:=A∩B. Claim. un∈β∞A/C, for alln≤ν.

Proof of Claim. We proceed by induction. The claim is clearly true for uν = x. Let ξ be the rank of un and η the rank of un+1. Assume that un+1 ∈βηA/C; then

un+1 ∈β<ξA/C, as ξ > η. For βun we have β<ξA/C ∈Sβ andun+1Mβun; so

β_<ξA/C∈un andun ∈β_ξA/C.

We now fix ρ, σ ∈ _Sα as in the assumptions of Proposition 9.6, and prove that

ρ◦~u=hρ(u0), . . . , ρ(uν)i is anα-pedigree forx overB.

Conditions (i) and (ii) from Definition10.1are clearly satisfied byρ◦~u. Asρ(un)vαun andun=σ(ρ(un))vαρ(un), we have unαρ(un); this implies (iii).

In order to prove (iv), we note thatρ(un)=ρ[un] for alln< ν, by the remark following Proposition 9.6. Given β such that un v β < un+1, and Y v β, Y ∈ ρ[un], there is some X ∈ un such thatρ[X] ⊆ Y. By Transfer into Sβ, we can assume X ∈Sβ. Henceun+1 ∈X [by (iv) for~u] and ρ(un+1) ∈ ρ[X] ⊆Y. This shows (iv) holds for

Proposition 10.5 (SST∗) Let~u be theα-pedigree forx overA∈_Sα.

Ifα vβ <x, then there is a unique n< ν such that un vβ <un+1. The sequence

~u β := hun,un+1, . . . ,uνi is then the β-pedigree for x over A. If x v β, we let

~uβ:=huνi=hxi; it is again the β-pedigree forx overA.

Proof The set ran~u={u₀, . . . ,uν} is a level set by Proposition8.6. Hencevran~u is a well-ordering, and there is a leastn such thatunvβ<un+1.

Corollary 10.6 (SST∗) Let A ∈ _Sα and α v β; if tpβ(x,A) = tpβ(y,A), then tpα(x,A)=tpα(y,A).

Proof Let~u = hu₀, . . . ,uνi and~v = hv0, . . . ,vµi be the α-pedigrees for x and y over A, resp., and let un v β < un+1, vm v β < um+1. Then un = tpβ(x,A) = tpβ(y,A) =vm. It is easy to check that the sequence hu0, . . . ,un =vm, . . . ,vµiis an

α-pedigree for yoverA. By Proposition10.3nowv0=u0.

Theorem 10.7 (SST∗) Ifx∈A∈_Sα, then an α-pedigree for xoverAexists. Proof Let Pα_{(x) be the statement “For every A ∈}

Sα, if x ∈ A, then there exists an α-pedigree for x over A.” By Proposition10.4, Pα(x) is equivalent to “For some

A∈_Sα such thatx∈A, there exists anα-pedigree forxoverA.” Our goal is to prove thatPα_{(x) holds for allα.}

Pα_{(x) holds for all α wx, because hu}

0i with u0 =x is then clearly an α-pedigree forx overA.

By Granularity, there is a least α for whichPα(x) holds. Let us assume 0 <α. We fix A ∈_S0 such that x ∈ A, and an α-pedigree~u= hun :n≤ νi for x overA, and obtain a contradiction by showing that there is a β-pedigree for x over A for some

β<α. If u0∈S0, then~u is also a 0-pedigree forx overA, and we let β=0. Otherwise we fixξ ∈S0 such that u0 ∈β_<ξA. Let W :=Wu0,β<ξA be the principal ultrafilter generated by u₀; we note Wu₀ =0. By Block Standardization, there is

u0 < u0 such that (∀X < α)(X ∈ u0 ⇔ X ∈ W); we let βu0. Then u0 ∈ Sβ is an ultrafilter overβ_<ξA, u0Mγu0 for all β vγ <α, and u0 is nonprincipal [because

u₀∈/ _Sβ]. By Proposition9.2,u0 ∼u for someu ∈β_ξA∩_Sβ. Then alsou_0Mγufor allβ vγ <α, and the sequencehui_a~uis aβ-pedigree forxoverA, a contradiction. Henceα0 andP0_{(x) holds. Proposition10.5now implies thatP}α_{(x) holds for all} levelsα.

Definition 10.8 An α-pedigree~u for x over A is a good α-pedigreeif, for u :=

tpα(x;A),~u+Mα(Σu) and

(j) f v_ug⇔f(~u+)v_α g(~u+) for allf,g∈_VΣu_∩

Sα.

An equivalent statement of the property (j) is that jα;x,A: VΣu ∩Sα → Sα[~u+], defined by j_α;x,A(f) = f(~u+), is an isomorphism of interpretations [U`t(V;Σu)]Sα

and (Sα[~u+],=,∈,vα) [see Theorem 3.7]. Of course, Sα[~u+] = Sα[~u], because

u∈_Sα. AlsoSα[~u]=Sα[ran~u], becauseun <α umiff rankun>rankum, and so the sequencehu₀, . . . ,uνiis∈-definable from the set{u0, . . . ,uν}.

Theorem 10.9 (SST∗) For anyx∈A∈Sα, theα-pedigree forx overA is good. Proof We fix an α0-pedigree~u=hu0, . . . ,uνi for x over A∈Sα0, and use Granu- larity to prove thatα0 is the least level αwα0 for which~uα isα-good. We write

jα for jα;x,A.

Ifα wx, then all is trivial: ~uα =huνi whereuν =x=tpα(x;A)=:u, ~u+ =0,

Σu ={{0}} is the principal ultrafilter over{0}, Sα[~u+] =Sα, vα is the identity on Sα, and for any f = {(0,a)} ∈ Sα, jα(f) = f(~u+) = f(0) = a; evidently, jα preservesvα.

Letαwα0be the least level such that~uαisα-good. Thenα<x, there is a unique

n< ν such that un vα <un+1, and~uα =hun,un+1, . . . ,uνi. If αα0, we are done; otherwise, we obtain a contradiction by showing that~uβ isβ-good, for some

β<α. There are two cases to consider. Case 1. un<α.

Then, for everyun vβ <α,~uα =~uβ is also the β-pedigree for x overA and

u:=tpβ(x;A)=tpα(x;A)=un.

By the inductive assumption, (~u α)+MαΣu, hence also (~u β)+MβΣu, and

f v_u g⇔ f(~u+) v_α g(~u+) for allf,g∈ _VΣu_∩

Sβ. It remains to show that the last statement holds withvα replaced by vβ.

It suffices to prove that, for f ∈ Sβ, f((~u β)+) vα 0 implies f((~u β)+) vβ 0. [Indeed, from (zvα0⇒zvβ 0) it follows that, for all y, zvα y⇔zvα0 ∨ z v

y⇔zv_β 0 ∨ zvy⇔zv_β y.]

Butf((~u β)+)v_α 0 means [apply j−_α1] thatU`t(V;u) f =kΣu(c), for some c∈

Case 2. unα.

Then n > 0 and we consider all β w α0 such that un−1 v β < un. We have

~u β = hun−1,un, . . . ,uνi = hun−1i a ~u α, (~u β)+ = ~u α, and u := tpβ(x;A) = un−1. We note that unMγu for all β v γ < α; by the inductive assumption, also (~uα)+MαΣun.

We first prove that (~uβ)+MβΣu.

By Definitions 9.8 and 6.2, Σu = Σu(Σu0), where u0 ranges over the domain of

u. Let X ∈ _Sβ, X ⊆ domΣu = S

u0_∈uhu0i a domΣu0, and X ∈ Σu; then

Y := {u0 _{∈ domu : (X)} hu0_i ∈ Σu0} ∈ u. As Y ∈ _Sβ and u_nMβu, we get u_n ∈ Y, ie, (X)huni ∈ Σun. As un ∈ Sα, (X)huni ∈ Sα, and so (~u α) + _{∈ (X)} huni, ie, (~uβ)+=hunia(~uα)+∈X. This proves (~uβ)+MβΣu.

By the inductive assumption we have the isomorphism jα of [U`t(V;un)]Sα and

(Sα[(~uα)+],=,∈,vα).

From Sβ[un] 4 Sα we get that jα, restricted to Sβ[un], is an isomorphism of [U`t(V;un)]Sβ[un] _{and (}

Sβ[un][(~u α)+],=,∈,vα) =(Sβ[(~u β)+],=,∈,vα) [the ultrapower is defined becauseun∈Sβ[un]].

We also have the isomorphismjof [U`(V;u)]Sβ _{and (S}

β[un],=,∈)4(Sα,=,∈) given byj(f)=f(un); especially,j(Iddomu)=un [Theorem3.7].

jinduces an isomorphism [also denotedj] of

[U`t(V;j−1(un))][U`(V;u)]

Sβ

and [U`t(V;un)]Sβ[un].

(*)

Observing that [U`(V;u)]Sβ =(Vu∩Sβ, . . .) andIddomu ∈Sβ, and usingj−1(un) =

Iddomu = hu0 : u0 ∈ domui, we get j−1(un) = hu0 : u0 ∈ domui = huhu0_i : u0 ∈

domui[recall the definition of u, the TOU associated tou]. If T(1) ={0} ∪ {hu0_{i :u}0 _{∈domu} denotes the u-subtree of T}

u of level 1, we have furtherhuhu0_i : u0 ∈ domui = h(u/T(1))(u0) : u0 ∈ domui. [For the last step, recall

thatΣT(1) is identified with domu(0)=domu via the maphu0i 7→u0.]

We can thus write u/T(1) for j−1(un) in (*). Furthermore, by the Factoring The- orem, ΩT(1)_,u [restricted to _Sβ] is an isomorphism of (U`(_V;Σu),v_u;Id

T(1))

Sβ _and [[U`t(V;u/T(1))]U`(V;u)]Sβ =[U`t(V;u/T(1))][U`(V;u)]

Sβ

.

The composition Ψ :=jα◦j◦ΩT(1)_,u is an isomorphism of (U`(_V;Σu),v_u;Id

T(1))

Sβ

and (Sβ[(~u β)+],=,∈,vα); it remains to show that Ψ = jβ;x,A, and that, for

Let f ∈ _VΣu_∩

Sβ; we show that Ψ(f) =f((~u β)+). First, ΩT(1)_,u(f) = f/T(1) =

hf_hu0_i : u0 ∈ domui, where u-almost everywhere domf_hu0_i ∈ Σu_hu0_i and f_hu0_i(t) =

f(hu0i _at). Hence [U`(V;u)]Sβ _“domf/T(1) _{∈ Σu/T}(1)_{”, so, second, j(f/T}1_{) =} (f/T(1))(un) = fhuni ∈ Sβ[un] and Sβ[un] “domfhuni ∈ Σun”, hence also [as

Sβ[un] 4 Sα], Sα “domfhuni ∈ Σun”. Finally, Ψ(f) = jα(fhuni) = fhuni((~u

α)+)=f(hunia(~uα)+)=f((~uβ)+).

This argument also shows that if f =Σu kΣu(c), then Ψ(f) =c ∈Sβ, and iff 6=Σu

k_Σu(c) for any c, then Ψ(f) ∈/ _Sβ. That is, Ψ preserves the level 0: f v_u k_Σu(0) if and only ifΨ(f)vβ 0. Moreover, since the last equivalence is true for allβ0 such that

βvβ0<α, we havef v_uk_Σu(0) if and only ifΨ(f)v_α0.

Ifk_Σu(0) <u f,g, then f vug iff Ψ(f) vα Ψ(g) iff Ψ(f) vα 0 ∨ Ψ(f) vΨ(g) iff

Ψ(f)vβ 0 ∨ Ψ(f)vΨ(g) iffΨ(f)vβ Ψ(g).

Theorem 10.10 (SST∗) Let ~u be the α-pedigree for x over A and let ~v be the α-pedigree for y over B, where y = f(x) and A,B,f ∈ Sα, f: A → B. Let

u:=tpα(x;A)andv:=tpα(y;B) Thenv=f(u) andran~v=f[ran~u].

If ϕ: u→ v is the morphism canonically induced by f, then the following diagram commutes. [U`t(V;v)]Sα [U`t(V;u)]Sα (Sα[~v],=,∈,vα) (Sα[~u],=,∈,vα) ϕ∗Sα ⊆ jα;y,B j_α;x,A - - 6 6

Proof We fix α0-pedigrees~u for x overA and~v for y overB, where y =f(x) and

A,B,f ∈_Sα0, and prove that α0 is the least levelαwα0such that the assertions hold for~uα and~vα.

If x v α, then also y vα, u :=tpα(x;A) = x, v :=tpα(y;B) = y, v = f(u),

Σu = Σv ={{0}}, ϕ(0) =0, ϕ∗ = Id, andSα[~u] =Sα[~v]= Sα. The diagram commutes trivially.

By Granularity, there is a least α wα0 such that the assertion is true for~u α and

~vα. We assume thatα=α0 and obtain a contradiction. Let unvα<un+1. Case 1. un<α.

We consider anyβ wα0such thatun vβ <α. Thenu:=tpβ(x;A)=tpα(x;A)=un and f(u) v β, so ~u β = ~u α and~v β = ~v α are β-pedigrees and v :=

tpβ(y;B) = tpα(y;B). The diagram commutes for α, by the inductive assumption. We have to prove that it commutes withβ in place of α. Asj_β;x,A(f)=f((~uβ)+)=

f((~u α)+) =jα;x,A(f), we have jβ;x,A = jα;x,A Sβ; similarly jβ;y,B = jα;y,B Sβ. As ran(~v β) = f[ran(~u β)] and f ∈ _Sβ, we have Sβ[~v β] = Sβ[ran(~v

β)] ⊆_Sβ[ran(~u β)] =Sβ[~u β]. The morphism ϕ: u → v is ∈-definable from

A,B,f ∈Sβ, and so it belongs toSβ. It follows, by Transfer betweenSα andSβ, that

ϕ∗ Sβ is a morphism of [U`t(V;u)]Sβ into [U`t(V;v)]Sβ. The commutativity of the

diagram is of course preserved. Case 2. unα.

Then n > 0 and we consider any β w α0 such that un−1 v β < un, so u := tpβ(a;A)=un−1. Letv0:=tpα(y;B).

Subcase A:v0∈Sβ.

Claim. Thenf(u)=v0.

Proof of Claim.

According to Definition9.3off(u), one considersw:=f_<ξ[u] for the appropriateξ. If Y ∈ w∩_Sβ, then f−_<ξ1[Y] ∈u∩_Sβ, hence un ∈f

−1

<ξ[Y] and v0 = f(un) ∈Y. We conclude thatwis a principal ultrafilter generated byv0, andf(u)=m(w)=v0. Returning to the proof of Subcase A, we have f(u) = v0 as well as f(un) = v0,

~v α =~v β, and v := v₀ =tpβ(y;B). As in Case 1, ran(~v β) =f[ran(~u β)],

Sβ[~vβ] ⊆Sβ[~uβ], andjβ;y,B is the restriction ofjα;y,B toSβ. In the definition of the induced canonical morphismϕ: u→v [proof of Proposition9.9] this is the case (i); ie, for u-almost all u0, ϕ(hu0_{i)=0 andϕ}

hu0_i=ϕu 0

, where ϕu0_{: u}0 _{→ v are the}

induced canonical morphisms. Considerg∈VΣv∩Sβ. On the one hand,

jβ;y,B(g)=jα;y,B(g)=jα;x,A((ϕun)∗(g)) (*)

by the inductive assumption.

On the other hand, ϕ∗(g)(hu0i _a t) = g(ϕ(hu0i _a t)) = g(ϕu0(t)) = (ϕu0)∗(g)(t)

u-almost everywhere, so [proof of Theorem 10.9] ΩT(1)_,u(ϕ∗(g))(u0) = (ϕu

0

)∗(g),

j(ΩT(1)_,u(ϕ∗(g)))=(ϕun)∗(g), and

j_β;x,A(ϕ∗(g))=jα;x,A((ϕun)∗(g)). (**)

Comparing (*) and (**) gives j_β;y,B(g) = j_β;x,A(ϕ∗(g)), ie, the commutativity of the diagram forβ.

We letv:=f(u); then v∈_Sβ∩β∞B.

Claim. v0Mβv.

Proof of Claim. u ∈ β_ξA, v ∈β_ηB, for suitable ξ, η ∈ _Sβ. Let w = f_<ξ[u]; from

unMβu we get v0 = f(un)Mβw [Proposition 3.5], so w is nonprincipal [v0 ∈/ Sβ],

w∼f(u) by definition of the latter, and, consequently,v0Mβv.

It follows that~vβ=hvi_a~vα. As in Case 1 now, ran(~vβ)=f[ran(~uβ)] and

Sβ[~vβ]⊆Sβ[~uβ]. Consider g ∈ _VΣv_∩

Sβ. On the one hand, jβ;y,B(g) = jα;y,B(jy(ΩT(1)_,v(g)), where

jy is the j from the proof of Theorem 10.9, Case 2, for y,v in place of x,u, and

ΩT(1)_,v(g)(v0)=g_hv0_i for v0 ∈v; then jy(Ω_T(1)_,v(g))=g_hv0i and

jβ;y,B(g)=jα;y,B(ghv0i)=ghv0i((~vα)

+_)=g(~v

α).

(*)

On the other hand, ϕ∗(g)(hu0i _a t) = g(ϕ(hu0i _a t)) = g(hf(u0)i _a ϕu0(t)) =

g_hf(u0_)i(ϕu 0

(t)) = (ϕu0₎∗_(g

hf(u0_)i)(t) u-almost everywhere, so Ω_T(1)_,u(ϕ∗(g))(u0) = (ϕu0)∗(g_hf(u0_)i) u-almost everywhere, and then, by definition of j = jx and Trans-

fer, jx(ΩT1_,u(ϕ∗(g))) = (ϕun)∗(g_hf(u

n)i). But f(un) = v0 and ϕ

un _{is the naturally}

induced morphism of un to v0. Hence [note that ghv0i ∈ Sα, so we can use the inductive assumption to justify the second equality]

j_β;x,A(ϕ∗(g))=j_α;x,A((ϕun₎∗_(g

hv0i))=jα;y,B(ghv0i)=ghv0i((~vα)

+_)=g(~v

α).

(**)

By comparing (*) and (**) we get jβ;y,B(g) =jβ;x,A(ϕ∗(g)), the commutativity of the diagram forβ.

The theory SST∗ is used in this paper mainly as a technical tool. Our main interest is in two related theories that are introduced next.

We writexMαU as shorthand for:

“There exists a good α-pedigree~u=hun :n≤νi forx over some A∈Sα such that

U=u0”.

We note thatU is then anα-type ofx.

The theorySST[ is SSTplus (∀α)(Bα), where

(Bα) (∀x,y)(∀U,F∈_Sα)[(ranF⊆DomU ∧ xMαU ∧ x=F(y))⇒ (∃V ∈Sα)(U=F(V) ∧ yMαV)].

Later, we consider one more axiom:

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