# Dimension and model theory

Consider an elementary class $\mathcal{K}$. It is quite common in model theory that a structure $K$ in $\mathcal K$ comes with a closure operator $$\text{cl}: \mathcal{P}(K) \to \mathcal{P}(K),$$ which establishes a pregeometry on $K$.

Any pregeometry yields a notion of dimension, say:

$$\text{dim} (K) = \min \{|A|: A \subset K \text{ and } \text{cl}(A) = K\}$$ I am interested in some natural properties shared by dimensions induced by pregeometries.

What kind of properties I am looking for? An example is the following,

Suppose $K = \bigcup K_i$ (non redundant increasing chain) and that its dimesion is infinite, is it true that $\text{dim}(K) = \sum_i \text{dim}(K_i)$?

I already know that there are many results like "trivial geometries are modular" but this is not the kind of result I am looking for. I am looking for structural properties of dimension because I am interested in giving an axiomatic definition of dimension.

• It's not clear to me exactly what setup you have in mind for the axiomatic definition of dimension. Are you considering $\dim$ as a function from $\mathcal{K}$ to cardinals? It's often more useful in model theory to assign dimensions to subsets of a model (e.g. the monster model), rather than to models. – Alex Kruckman Dec 27 '17 at 20:46
• Yes, it is what I have in mind, I did not say not to bias your answers. Moreover, maybe a functor is too much, I don't care about what it should do on morphisms. – Ivan Di Liberti Dec 27 '17 at 20:46
• And regarding the highlighted question, are you assuming that the union is an increasing union of a chain of models? Otherwise, the answer is no, since any model (of a first-order theory in a countable language) can be written as a union of countable elementary submodels. – Alex Kruckman Dec 27 '17 at 20:48
• Absolutely yes. – Ivan Di Liberti Dec 27 '17 at 20:48
• And are you asking only about unions along chains indexed by $\omega$? Otherwise, the answer is no, since structures of dimension $\aleph_1$ can be written as (uncountable) unions of structures of dimension $\aleph_0$. – Alex Kruckman Dec 28 '17 at 2:56

Here's an example showing that in general, for pregeometries arising in model theory, you can't characterize the dimension of a union of a chain of models just in terms of the dimensions of the models. In other words, it matters how the models embed into each other.

Consider the theory of a single equivalence relation $E$ with infinitely many infinite classes, and define $\dim(M) = |M/E|$.

First union: Let $M_0$ be the unique countable model of this theory up to isomorphism. For every countable ordinal $\alpha$, let $M_{\alpha+1}$ be the elementary extension of $M_\alpha$ obtained by adding a single new equivalence class with countably many elements. For a limit ordinal $\lambda$, let $M_\lambda = \bigcup_{\alpha<\lambda} M_\alpha$. Then $\dim(M_\alpha) = \aleph_0$ for all $\alpha<\omega_1$, but $\dim(M_{\omega_1}) = \aleph_1$,

Second union: Let $N_0 = M_0$, and pick an equivalence class $C$. This time, for every $\alpha$, let $N_{\alpha+1}$ be the elementary extension of $N_\alpha$ obtained by adding a single new element to $C$. As before, take unions as limit stages. Then $\dim(N_\alpha) = \aleph_0$ for all $\alpha<\omega_1$, and also $\dim(N_{\omega_1}) = \aleph_0$.

Now I need to convince you that this dimension function arises in a standard way from a pregeometry studied in model theory. In stability theory, there's the notion of a regular type: a stationary type which is orthogonal to all of its forking extensions. The key point is that if $p(x)$ is a regular type (let's say over $\emptyset$ for simplicity), then forking dependence gives rise to a pregometry on the realizations of $p$ via the closure operator $\mathrm{cl}(A) = \{b\models p(x)\mid \text{tp}(b/A) \text{ forks over }\emptyset\}$.

In my example, the theory is stable, the unique $1$-type is a regular type, and forking is governed by the equivalence relation $E$, so we get a pregometry on the whole model with closure operator $\mathrm{cl}(A) = \bigcup_{a\in A} [a]_E$, where $[a]_E$ is the $E$-class of $a$. And the induced dimension function is $\dim(M) = |M/E|$.

Well, maybe you don't like this kind of pregeometry, and you only want to consider the kind you meet more often in model theory, namely pregeometries induced by the $\text{acl}$ operator. That's fine, but then the dimensions are only interesting for models that are at most the size of the language (so only countable models if the language is countable).

Indeed, suppose $T$ is a theory such that $\text{acl}$ induces a pregometry on every model of $T$, and let $M\models T$ with $|M| > |L|$. Since $|\text{acl}(A)| = \max(|A|,|L|)$ for all $A\subseteq M$, any basis for $M$ must have cardinality $|M|$, and $\dim(M) = |M|$.

Added in edit: You might also decide that you're only interested in closure operators with the property that when $A\subseteq M\prec N$, $\text{cl}(A)$ in $M$ equals $\text{cl}(A)$ in $N$, i.e. closures don't grow in elementary extensions. This is the case for $\text{cl} = \text{acl}$, and it would salvage the proof in your answer that $\dim$ takes unions of chains to sums, since if $N$ is a proper elementary extension of $M$, the closure of a basis for $M$ is contained in $M$, and we need at least one new element to form a basis for $N$. But we actually don't get anything beyond $\text{acl}$ under this assumption.

Indeed, suppose $\text{cl}$ satisfies the condition above, and look at $A\subseteq M$. Embed $M$ in a large monster model $\mathbb{M}$. Then $\text{cl}_M(A) = \text{cl}_{\mathbb{M}}(A)$. In fact, for any $A\subseteq N\prec \mathbb{M}$, we have $\text{cl}_N(A) = \text{cl}_{\mathbb{M}}(A)$, so $\text{cl}_{\mathbb{M}}(A)\subseteq N$. But $\bigcap\{N\mid A\subseteq N\prec \mathbb{M}\} = \text{acl}(A)$, so $\text{cl}(A)\subseteq \text{acl}(A)$.

If you're interested in axiomatizing dimension functions, you might want to look this paper, which gives a number of equivalent axiom systems for infinite matroids. In particular, look at the axioms in terms of rank functions. Their rank functions take values in $\mathbb{N}\cup \{\infty\}$, but you might as well be in this situation if you're thinking about $\text{acl}$ pregometries ($\dim(M) = \infty$ means $\dim(M) = |M|$).

• Thanks! This is very interesting! Can you give a look to my proof to spot the mistake? – Ivan Di Liberti Dec 28 '17 at 19:06

The following is valid when $A\subseteq M\prec N$, $\text{cl}(A)$ in $M$ equals $\text{cl}(A)$ in $N$, i.e. closures don't grow in elementary extensions.

The answer to my questions looks to me to be yes.

Let $K = \bigcup K_i$ We know that $K_i$ has a basis $A_i$.

Step 1. We can suppose that $A_i \subset A_{i+1}$. In fact, if it is not the case, consider the $\text{cl}(A_i)$ inside $K_{i+1}$. We through in $A_i$ elements untill it gets a basis $A_{i+1}^*$ of $K_i$ and it must be the case that $|A_{i+1}^*| = |A_{i+1}|$ because of exachange property. By transfinite induction we can replace $\{A_i\}$ so that they are an increasing chain.

Step 2. $\bigcup A_i$ is a basis of $K$. In fact $$K = \bigcup K_i = \bigcup \text{cl} (A_i) \subset \text{cl} (\bigcup A_i)$$ because of monotonicity.

Step 3. $|\bigcup A_i| = \sum |A_i|$ as soon as they are infinite.

Is this correct? Am I really using that $A_i \subset A_{i+1}$?

• Oh, you posted this while I was typing my answer. I think your argument works, except in Step 3, where we don't know that $A_i$ is properly contained in $A_{i+1}$. – Alex Kruckman Dec 28 '17 at 19:06
• I've added a paragraph to my answer. – Alex Kruckman Dec 28 '17 at 19:12

Marcel van de Vel's "Theory of Convex Structures" has material on dimension theory. Currently I do not have access to the book and its been a very long time since I read it. Here is a copy of the table of contents. The second and third from the last items may be of interest.