End extensions and numbers of countable models

by Shelah. [Sh:66]
J Symbolic Logic, 1978
The answer to the question from page 562 (the end). is negative; have known a solution but not sure if have Not record it. For any countable model M with countable vocabulary with predicates only. Not including < and E . First we choose a function F from Q the rationals onto M such that the pre-image of any element is dense Second we define a model N Universe. The rationals <. Is interpreted. As the rational order E is interpreted as the equivalence relation xEy iff F(x)=F(y) For any predicate P of the vocabulary of M is interpreted as it's pre-image by F No Th(N) is a countable fo theory with the same number of countable models up to isomorphism as Th(M) So we are done giving a negative answer to the question

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