Difference between revisions of "Short recursively saturated models and boundedly saturated models"
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If an elementary cut of a countable recursively saturated model is tall, it is recursively saturated (and hence isomorphic to the model). If the elementary cut is short it is bounded recursively saturated, and hence short recursively saturated. | If an elementary cut of a countable recursively saturated model is tall, it is recursively saturated (and hence isomorphic to the model). If the elementary cut is short it is bounded recursively saturated, and hence short recursively saturated. | ||
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| + | Two countable short recursively saturated models $M$ and $N$ are isomorphic if and only if $SSy(M)=SSy(N)$, $Th(M)=Th(N$), and there is an element $a$ in the last gap of $M$ and $b$ in the last gap on $N$ such that $tp(a)=tp(b)$. | ||
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| + | === Automorphisms === | ||
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| + | Every automorphism of a countable recursively saturated model fixes elements in the last gap of the model. | ||
== Short Saturated Models == | == Short Saturated Models == | ||
Revision as of 16:03, 15 December 2014
Contents
Short Recursively Saturated Models
A model is short if there is an element in the model whose Skolem closure is cofinal in the model. That is, M is short iff there is a in M such that M=Sup(Scl(a)). If a model is not short, it is called tall.
A type p(v,b) is bounded if it contains the formula v<b.
A model M is bounded recursively saturated if for every b in M, every bounded recursive type p(v,b) is realized in M.
A model is short recursively saturated if it is short and bounded recursively saturated.
Every elementary cut of a recursively saturated model is boundedly saturated.
Countable Short Recursively Saturated Models
If an elementary cut of a countable recursively saturated model is tall, it is recursively saturated (and hence isomorphic to the model). If the elementary cut is short it is bounded recursively saturated, and hence short recursively saturated.
Two countable short recursively saturated models $M$ and $N$ are isomorphic if and only if $SSy(M)=SSy(N)$, $Th(M)=Th(N$), and there is an element $a$ in the last gap of $M$ and $b$ in the last gap on $N$ such that $tp(a)=tp(b)$.
Automorphisms
Every automorphism of a countable recursively saturated model fixes elements in the last gap of the model.
