Minimal types in simple theories(3)

We prove that if M0 is a model of a simple theory, and p(x) is a complete type of Cantor-Bendixon rank 1 over M0, then p is stationary and regular. As a consequence we obtain another proof that any countable model M0 of a countable complete simple theory T

CB-rank1.Thenpisstationary,namelyhasauniquenonforkingextensionoveranysetcontainingM0.

Proof.Wewillassumethatp(x)isnonstationaryandgetacontradiction.ClaimI.ThereisamodelMcontainingM0,anddistinctnonforkingextensionsp (x),p (x)∈S(M)ofp(x)suchthatp (x)isanheirofp.

Proof.Thisisclear,asanyheirofpisanonforkingextensionandweareassumingthatphasdistinctnonforkingextensionsoversomemodel.

ClaimII.ThereisamodelMofTandanonforkingextensionp (x)∈S(M)ofpwhichisnotacoheirofp(namelyisnot nitelysatis ableinM0).¯isProof.IfnotthenanynonforkingextensionofpoverthemonstermodelM¯ nitelysatis able nitelysatis ableinM0.ButthenumberoftypesoverM|M|inM0isboundedby220,sowehaveaboundednumberofnonforkingextensionsofp.Thisimpliesthatpisstationary.(BytheIndependencetheoremforexample:see[3].)

ByClaimIthereisa nitetupleaanddistinctnonforkingextensionsp1(x),q(x)∈S(M0a)ofp(x)suchthatp1isanheirofp.

ByClaimII,letcbea nitetuple(c0,)andr(x)anonforkingexten-sionofpoverM0cwhichisnot nitelysatis ableinM0.

Thereisnoharminextendingc,sowemayassumethatc0realizesp,namely,tp(c0/M0)=p.Also,asbyautomorphismwemayreplacecbyanyrealizationoftp(c/M0),wemayassumethatc0realizesthetypep1(overM0∪{a})mentionedabove.Notethatthentp(a/M0c0)is nitelysatis ableinM0sohasacompleteextensionoverM0cwhichwhichis nitelysatis ableinM0.Thus,byautomorphismagainwemayassumethattp(a/M0c)is nitelysatis ableinM0,namelythattp(c/M0a)isanheiroftp(c/M0).Letussummarisethesituationsofar.Wehavetuplesaandc=(c0,..,cn)¯andcompletetypesq(x)∈S(M0a)andr(x)∈S(M0c)suchthatinM

(i)tp(c0/M0)=p(x),

(ii)tp(c/M0a)isanheiroftp(c/M0),soinparticularcaswellasc0isinde-pendentfromaoverM0.

(iii)q(x)isanonforkingextensionofp(x),andq(x)=tp(c0/M0a).

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