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I'm asking here if the following construction is of any interest. I can not find any reference for that kind of thing, so either the subject is completely trivial, either I just don't have the correct terminology to search more about it.

Take a first-order $\mathcal L$-theory $T$ (say consistent). Recall that the $n$-th Stone's space is the topological space $S_n(T)$ of all complete $n$-types of any model of $T$ with $(\{p \mid \varphi(\underline x) \in p\})_{\varphi \in \mathcal L-\text{formula}}$ as a basis of open sets.

Then I start the construction I talked about : taking a model $\mathcal M \models T$, we have a natural application $$f : M^n \to S_n(T),\, \underline m \mapsto \mathrm{typ}^{\mathcal M}(\underline m)$$ where $\mathrm{typ}^{\mathcal M}(\underline m) = \{ \varphi(\underline x) \mid \mathcal M \models \varphi(\underline m)\}$ is the type of the element $\underline m \in M^n$. This application $f$ provides a topology for $M^n$, the initial topology with respect to $f$ (that is the smallest topology on $M^n$ making $f$ continuous).

Can this topology on the models be used for anything ? For example, it seems that for $n=1$, the topology on $M$ is such that the group of model-theoric automorphisms of $\mathcal M$ is exactly the automorphisms's group of the object $M \stackrel f \to S_n(T)$ in the category $\mathbf{Top} / S_n(T)$.

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  • $\begingroup$ Note that the basic open sets in your topology would be $f^{-1}([\phi]) = \{\bar m \colon f(\bar m) \in [\phi]\} = \{\bar m \colon tp(\bar m) \in [\phi]\} = \{\bar m \colon \phi \in tp(\bar m)\} = \{\bar m \colon \phi(\bar m)\}$. Thus the basic opens sets are exactly definable subsets. $\endgroup$ – Levon Haykazyan Apr 29 '13 at 10:35
  • $\begingroup$ @LevonHaykazyan : I noticed it. Actually, every open subset is either definable or union of definables (the basic open sets are closed under intersection). $\endgroup$ – Pece Apr 29 '13 at 11:54
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This is a good observation. I also noticed this at some point, and I hoped that something interesting would come of it... But I still don't know of any reasons to consider this topology on a model that can't be rephrased in terms of a topological argument on the Stone space.

One reason for this is that the Stone space is a "better version" of the topological space associated to a model - it is compact (all consistent types correspond to a point) and Hausdorff (any two elements satisfying the same type are collapsed to a single point). I haven't thought this through carefully, but I'm fairly certain that $S_n$ is the Stone-Čech compactification of the induced topology on $M^n$.

One possible connection with the outside word is that for an algebraically closed field $K$, one can view the induced topology on $K^n$ as an alternative to the Zariski topology - in this context it's usually called the "logic topology". In fact, for $K\models ACF$, $S_n(K)$ is in bijection with $Spec(K[x_1,\dots,x_n])$ as a scheme! But they're not homeomorphic - the scheme theoretic toplogy on Spec is the Zariski toplogy, not the logic topology.

The difference is that in the Zariski topology, a closed set is determined by polynomial equations, not inequations, and in the logic topology we don't distinguish between the two - they're both formulas. One can't generalize the Zariski topology to a general model theoretic setting, because there may be no clear way of specifying which formulas are "positive" and "negative". It's interesting to think about examples of this.

One last point - your observation about automorphisms is incorrect. Take, for example, $M$ to be a structure in the language $\mathcal{L} = \{E\}$, where $E$ is an equivalence relation with two infinite classes. Then there is only one $1$-type, so $S_1$ is just a point, and $M$ has the trivial topology $\{\emptyset, M\}$. But not every bijection $M\rightarrow M$ is an automorphism.

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  • $\begingroup$ Thanks for the anwser. I'm not familiar with schemes, so I will read a little about it before understanding your example. But knowing the classic Zariski topology (say, on $\mathbb C^n$), I intuitively understand why the logic topology is finer than the Zariski one. $\endgroup$ – Pece May 6 '13 at 9:50

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