# Relevance of a statement about monster models

I was reading this question and I have trouble understanding what the second statement really says. I know almost nothing about monster models and have never worked with them, so I should probably read some more introductory material but I still would like to understand that bit.

I copy below this second statement I'm speaking about. $$p(x)$$ is a not necessarily complete type and $$A$$ and $$B$$ are from what I understand submodels of this monster model (is $$B$$ supposed to be an elementary submodel?).

Let $$A\subset B$$ and $$p(x)\subset L(A)$$ suppose that for any solution $$a$$ of $$p(x)$$ the complete type over $$B$$, $$tp(a/B)$$ is isolated. Show that $$p(x)$$ is isolated.

Let me try to express this in a language which is more familiar to me. Let $$M$$ be the monster model, let $$M_A$$ be the same structure but where we add one constant per element of $$A$$ to the language and let $$\DeclareMathOperator{\Typ}{Typ}\Typ_{M_A}(n)$$ be the space of $$n$$-types of the theory of $$M_A$$. Then from what I understand, $$p(x)$$ is a closed set in $$\Typ_{M_A}(1)$$ and $$tp(a/B)$$ is a point of $$\Typ_{M_B}(1)$$. We have a function $$\Typ_{M_B}(1) → \Typ_{M_A}(1)$$ and the statement says that if all the points in the inverse image of $$p(x) ⊆ \Typ_{M_A}(1)$$ are isolated (open), then $$p(x)$$ is open itself. This follows from the fact that $$\Typ_{M_B}(1) → \Typ_{M_A}(1)$$ is surjective and continuous: it is then a quotient.

Is what I said above correct? If so, I don't understand what is the significance of this property. From what I understand, when working with a monster model, we place everything inside of it and it should stay "implicit" but ease the arguments. I have never worked with a monster model, so I don't really know what is going on here: what does the above property says about an embedding $$A⊆B$$ between two structures when we forget that there is a monster model?

## 1 Answer

There are a couple of issues raised in your question that I want to address, so this answer might seem a bit unfocused.

1. Some comments on terminology and notation:

• I prefer not to use the word "isolated" to describe a partial (possibly incomplete) type. I would rather call such a type "equivalent to a formula" or "finitely axiomatizable" or even "clopen". I prefer to reserve the word "isolated" for complete types (since when a complete type is equivalent to a formula, it is an isolated point of the Stone space). I used the word "isolated" in my answer to the question you linked to because I wanted to follow the OP's terminology - I probably should have also made this comment there.
• The standard notation for the Stone space of complete $$n$$-types relative to a theory $$T$$ is $$S_n(T)$$. The $$S$$ stands for Stone space. Often, we want to consider types with parameters coming from a subset $$A$$ of some model $$M$$ (types over $$A$$). Then we write $$S_n(A)$$ as shorthand for $$S_n(\text{Th}(M_A))$$. This is the same as what you denote by $$\text{Typ}_{M_A}(n)$$.
• Another common notational convention in model theory is to use single letters to denote tuples by default. So e.g. $$p(x) = \text{tp}(a/B)$$ could be an element of $$S_n(B)$$ if $$x = (x_1,\dots,x_n)$$ and $$a = (a_1,\dots,a_n)$$. This convention is not universal, but really de-clutters notation - and in many situations, there's really no meaningful difference between singletons and tuples.

2. Ok, now let's look at the statement (rewritten the way I would state it):

Let $$A\subseteq B$$ be sets, and let $$p(x)$$ be a partial type over $$A$$. Suppose that for any realization $$a$$ of $$p(x)$$, $$\text{tp}(a/B)$$ is isolated in $$S_n(B)$$. Show that $$p(x)$$ is equivalent to a formula over $$A$$.

You're right that this statement "takes place in the monster model". Here's the same statement, rewritten to make the monster model $$\mathbb{M}$$ explicit:

Let $$A\subseteq B$$ be subsets of $$\mathbb{M}$$, and let $$p(x)$$ be a partial type over $$A$$. Suppose that for any realization $$a\in \mathbb{M}$$ of $$p(x)$$, $$\text{tp}(a/B)$$ is isolated in $$S_n(B)$$. Show that $$p(x)$$ is equivalent to a formula over $$A$$.

If we drop the monster model convention entirely, here's the correct translation of the statement:

Let $$A\subseteq B$$ be subsets of a model $$M$$, and let $$p(x)$$ be a partial type over $$A$$. Suppose that for any realization $$a\in N$$ of $$p(x)$$ in an elementary extension $$M\preceq N$$, $$\text{tp}(a/B)$$ is isolated in $$S_n(B)$$. Show that $$p(x)$$ is equivalent to a formula over $$A$$.

Note in particular that $$A$$ and $$B$$ are just sets of parameters, not models. The advantage of the monster model convention here is just that we don't have to quantify over models and elementary extensions each time we have an alternation of quantifiers in the statement. We never have to move to an elementary extension, because everything you would find in an elementary extension is "already there" in the monster.

3. Your topological proof of the statement is correct. Note that you're implicitly using two facts in the proof: (1) every clopen set in the Stone space $$S_n(A)$$ is definable by a formula over $$A$$ (this is why it suffices to show that the closed set defined by $$p(x)$$ is also open), and (2) every surjective continuous map from a compact space to a Hausdorff space is a quotient map. These roughly correspond to the two main steps in the "concrete" proof I gave in the answer to the linked question: (1) The use of the compactness theorem to show that $$p(x)$$ is equivalent to a finite disjunction of formulas over $$B$$, and (2) the fact I cited that if a set definable by a formula over $$B$$ is invariant over $$A$$ then the set is actually definable by a formula over $$A$$.

4. As for "what is the significance of this property?" ... Well, to be honest, the statement in question doesn't actually seem that interesting or useful to me. One could view it as a generalization of the basic (and important) fact that if a partial type $$p(x)$$ is algebraic (has only finitely many realizations), then it is equivalent to a formula. [List the realizations of $$p(x)$$ as $$b_1,\dots,b_k$$ and let $$B = A\cup \{b_1,\dots,b_k\}$$. Then every realization of $$p(x)$$ has a complete type over $$B$$ which is isolated by one of the formulas $$x = b_i$$. Applying our property, we can conclude that $$p(x)$$ is equivalent to a formula over $$A$$.] Certainly isolated types are very important in model theory, and I could also imagine our property coming up as a step in a proof about atomic or prime models, or about Morley rank. If nothing else, it's a good exercise to get some practice thinking about Stone spaces.