Can you build a new model of a theory out of the set of all complete 1-types?

Question

Can you build a new model of a theory out of the set of all complete 1-types?

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Marker in Model Theory: An Introduction on page 117 says the following:

Complete types tell us what possible first-order properties an element can have in an elementary extension.

That's pretty cool, and it provides a nice motivation for types.

The first thing it makes me wonder is whether we can use the complete 1-types themselves as building material to make a new structure ... and if we do whether that structure will be the model of the theory we started out with.

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Here's my attempt to build such a thing.

I'm restricting attention to theories with no non-nullary functions, i.e. all function symbols are constant symbols.

Let $L$ be a language with only constants and relations.

Let $T$ be an $L$-theory.

Let $M \models T$.

Let $A$ be a subset of the domain of $M$.

I'm looking at $S^M_1(A)$, which I'll call $S^M(A)$ for brevity.

I'm going to try to build a new model $M'$.

Let the domain of $M'$ be $S^M(A)$.

For every constant in $L$ and fresh constant from $c_A$ (created implicitly by talking about parameters), we choose the type $p(x)$ consisting of precisely the formulas $\varphi(x)$ that hold in $M$ as the interpretation for the constant in question.

Let $R$ be a relation symbol in $T$.

Suppose $R$ is unary.

$M' \models R(p)$ holds if and only if, for every finite set of formulas $q(x)$ in $p(x)$, $M \models \exists x \mathop. q(x) \land R(x)$.

And in general, $M' \models R(p_1 \cdots p_n)$ if and only if, for every sequence of finite sets of formulas $q_1(x) \cdots q_n(x)$ where it holds for all $1 \le i \le n$ that $q_i(x) \subset p_i(x)$, it holds that $M \models \exists x_1 \cdots x_n \mathop. q_1(x_1) \land \cdots \land q_n(x_n) \land R(x_1 \cdots x_n)$.

I'm pretty sure that $M'$ gives me back a well-defined relational structure, but it isn't clear to me whether it's actually a model of $T$ or not.

Answer 1

No, looking at the $1$-types is not in general enough.

One way to see this is to observe that there are "interesting" structures whose automorphism group acts $1$-transitively (so a fortiori there is only one $1$-type). For example, every group is bi-interpretable (with parameters) with such a structure: given a group $\mathcal{G}=(G;*,e,{}^{-1})$ consider its torsor reduct $$\mathcal{T}_\mathcal{G}=(G; (a,b,c)\mapsto a*b^{-1}*c).$$ The torsor reduct is to the original group as affine space is to a vector space; basically, we "forget the origin" in a controlled way. Incidentally, with a bit more work you can show the same result for arbitrary rings - see this answer of Matt F. (which while phrased for the rationals in particular, works for all rings).

Answer 2

This is just to add to Noah's answer - I think we're really making the same point.

Your construction usually fails to produce a model of $T$. One reason is that in a model of $T$, if some type is realized, it often must be realized more than once.

Let's look at some examples.

1. Let $T$ be the theory of infinite sets. Let $A$ be a finite subset of some model $M$. Then $S^1(A)$ consists of the realized types $x=a$, together with one additional type which says $x\neq a$ for all $a\in A$. Your construction will produce a finite set of size $|A|+1$, which is not a model of $T$.

2. Let $T$ be the theory of algebraically closed fields of characteristic $0$, and let $A=\mathbb{Q}$. Then $S^1(A)$ contains one type for each irreducible monic polynomial in $\mathbb{Q}[x]$ (the type of an element algebraic over $\mathbb{Q}$ is determined by its minimal polynomial over $\mathbb{Q}$) together with the type of a transcendental element (which says $x$ is not a zero of any non-zero polynomial). Now your construction adjoins to $\mathbb{Q}$ a single element for each of these types. This fails to be an algebraically closed field for a number of reasons: For the algebraic types, an irreducible polynomial of degree $n$ should have $n$ roots, not just one. And for the transcendental type, once a field has a transcendental element over $\mathbb{Q}$, it will have infinitely many, e.g. if $t$ is transcendental, so is $t^2$ and $t+q$ for all $q\in \mathbb{Q}$.

3. Let $T$ be DLO (the theory of dense linear orders without endpoints), and let $A=\mathbb{N}\subseteq \mathbb{Q}$. Your construction will add one new element between each natural number and its successor, plus one element greater than each natural number and one less than each natural number. The result is a linear order which has endpoints and is not dense.