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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