If $M$ and $N$ are two $L$-structures, and $f: M \rightarrow N$ is an elementary extension, then given any subset $A$ of $M$, $f$ induces in a natural way a morphism $S^M_n(A) \rightarrow S^N_n(f(A))$ of type spaces which is in fact a homeomorphism of Stone spaces (in fact the underlying sets are equal: see the answer below).

What results exist about converse statements? If two type spaces are equal, when can you say it's because of the existence of an elementary embedding? What if you have a non-trivial homeomorphism, or some more general injective continuous map? Can you sometimes conclude that that it's induced by a suitable morphism between the structures?


The two type spaces that you are considering are equal: types over $A$ are the same thing whether we look at $A$ in $M$ or in $N$.

An element of $S_n^M(A)$ is a maximal consistent set of formulas with parameters in $A$ and free variables $x_1,...,x_n$ containing $T_M(A)$, where $T_M(A)$ is the set of sentences with parameters in $A$ that are true in $M$. By definition of an elementary extension, $T_N(f(A))=T_M(A)$: the same sentences are satisfied by $A$ and by $f(A)$. Hence also $S_n^M(A)=S_n^N(f(A))$.

  • $\begingroup$ Thanks for pointing this out. However, it does not answer the question. I have edited it to be more general to take into account non-trivial maps as well. $\endgroup$ – KristianJS May 11 '13 at 22:11
  • $\begingroup$ If I understand well, you want to look at type spaces as pure topological spaces and ask if maps between them translate to maps between the structures. This is hopeless, because the topological structure of type space does not carry enough information. For an easy counter-example, take the ring language. M is an ordinary ring and N is the same as M, but exchanging the roles of addition and multiplication. The type spaces are homeomorphic, but clearly the structures are completely different. $\endgroup$ – Pierre Simon May 12 '13 at 6:07
  • $\begingroup$ I would say in that example that you know something very interesting about the relationship between M and N actually. However, I appreciate the point that type spaces are perhaps quite bad at telling structures apart. Perhaps in certain specific cases though you can say something. I'll leave this question open for now. $\endgroup$ – KristianJS May 12 '13 at 13:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.