Suppose $T$ is a countable complete theory, with monster model $\mathbb{C}$. A definable set $D := \phi(\mathbb{C}, \overline a)$ is strongly minimal if given any other formula $\psi(x, \overline b)$ with $\overline b\in \mathbb{C}$, then $D\cap \psi(\mathbb{C}, \overline b)$ is either finite or cofinite.

One of the important facts about strongly minimal sets is that the closure operator $A\rightarrow acl(A)\cap D$ turns $D$ into a pregeometry. My question is about proving the exchange principle: if $A\subseteq D$; $b, c\in D$; $b\in acl(A\cup \{c\})\setminus acl(A)$, then $c\in acl(A\cup \{b\})$.

Most proofs I find start as follows. Suppose not, and find $b, c$ as above but with $c\not\in acl(A\cup \{b\})$. Now there is a step where we include $A$ and $\overline a$ as constants in the language. Including $A$ is fine, but including $\overline a$ makes me uneasy. It could be the case that $c\in acl(A\cup\{b\} \cup \{\overline a\})$, which would undermine the rest of the proof.

If anyone knows of a good proof of exchange in this setting, I would much appreciate it.


1 Answer 1


You're right to be worried about this point in the proof. In fact, what you're trying to prove is false: you have the wrong definition of the closure operator!

In a strongly minimal set $D$ defined by a formula with parameters $\varphi(x,\overline{a})$, the closure operator should be $X \mapsto \text{acl}_{\overline{a}}(X) = \text{acl}(X\cup\overline{a})\cap D$.

Here's a counterexample to exchange for your definition of closure, showing that including the parameters is necessary:

Let $L = \{E,f\}$, where $E$ is a binary relation and $f$ is a unary function. Let $T$ be the theory asserting that $E$ is an equivalence relation with infinitely many infinite classes, and $f$ chooses a representative for each $E$-class. That is, $\forall x\, (x\, E\, f(x))$, and $\forall x\, \forall y\, (x\, E\, y) \rightarrow (f(x) = f(y))$.

$T$ is a complete theory with quantifier-elimination. Let $\mathbb{C}$ be its monster model. For any $a\in \mathbb{C}$, the $E$-class of $a$, defined by $x\, E\, a$, is strongly minimal. Call it $D_a$. Note that $\text{acl}(\emptyset) = \emptyset$, since every element of $\mathbb{C}$ has infinitely many conjugates under automorphisms of $\mathbb{C}$ (permuting the equivalence classes).

Choose any $b\in D_a$ such that $f(b)\neq b$, and let $c = f(b)$. Now $c\in (\text{acl}(b)\cap D_a)\setminus (\text{acl}(\emptyset)\cap D_a)$, but $b\notin \text{acl}(c)\cap D_a$, since $b$ has infinitely many conjugates under automorphisms of $\mathbb{C}$ fixing $c$ (take any permutation of $D_a\setminus \{c\}$).

What went wrong, of course, is that $c$ should really be in the closure of $\emptyset$. Indeed, following the correct definition of closure, $c\in \text{acl}(\emptyset \cup \{a\}) \cap D_a$. It's the unique element of $\mathbb{C}$ satisfying the formula $(x\, E\, a) \land (f(x) = x)$.

It's easier to think about / work with strongly minimal sets defined over $\emptyset$, so we often add the parameters $\overline{a}$ to the language at the start of the discussion. Note that if there are constants for $\overline{a}$ in the language, then $\text{acl}(X\cup\overline{a})\cap D = \text{acl}(X)\cap D$.

The popular textbooks by Marker and Hodges are both unfortunately sloppy on this point. Hodges only proves exchange over base sets $A$ containing the parameters $\overline{a}$, but he doesn't explicitly write down the definition of closure in strongly minimal sets defined with parameters. The book by Tent & Ziegler, as usual, gets it right.

  • $\begingroup$ Thanks! One follow up question. If $D$ is strongly minimal and there are several different formulas with possibly different parameters defining $D$, is there a way to make a canonical choice? $\endgroup$
    – Andy
    Commented Nov 23, 2015 at 5:36
  • $\begingroup$ Yes, if you're willing to work in $M^\text{eq}$. In $M^\text{eq}$, definable sets have canonical parameters, up to interdefinability. Given $\varphi(x,y)$, the formula $E_\varphi(y,z)\colon \forall x\, \varphi(x,y) \leftrightarrow \varphi(x,z)$ is a definable equivalence relation, where $b$ and $c$ are equivalent just in case $\varphi(x,b)$ and $\varphi(x,c)$ define the same set. Now the equivalence class $[a]$, as an element of $M^\text{eq}$ has the property that an automorphism of $M$ fixes the set $D = \varphi(M,a)$ if and only if it fixes $[a]$. $\endgroup$ Commented Nov 23, 2015 at 5:47
  • $\begingroup$ A consequence is that if $\psi(x,b)$ defines the same set $D$, then any automorphism of $M$ that fixes $b$ fixes $D$, and hence fixes $[a]$. It follows that $[a]\in \text{dcl}^{\text{eq}}(b)$ (I should note that I'm calling the monster model $M$). So $[a]$ is a minimal for $D$ in a sense: it's definable from any other parameter (even using a different formula). Now $[b]$ is also a canonical parameter (using a different formula $\psi$), but by the argument above $[a]$ and $[b]$ are interdefinable. So canonical parameters are minimal for $\text{dcl}$ and unique up to interdefinability. $\endgroup$ Commented Nov 23, 2015 at 5:51
  • $\begingroup$ The upshot for strongly minimal sets is that if $[a]$ is the canonical parameter for $D$ and $\overline{b}$ is some other parameter for $D$, then $\text{acl}_{[a]}(X) \subseteq \text{acl}_{\overline{b}}(X)$, since $X[a]\subseteq \text{acl}(X\overline{b})$. So $\text{acl}_{[a]}$ is the minimal closure relation of the form $\text{acl}_A$ such that $D$ is defined over $A$ (i.e. such that we're sure to get a pregeometry). $\endgroup$ Commented Nov 23, 2015 at 5:54
  • $\begingroup$ If you're not comfortable working with $M^{\text{eq}}$, another way of describing this canonical closure operator on $D$ is: $b\in \text{cl}(X)$ if and only if $b$ has only finitely many images under automorphisms of $M$ fixing $D$ set-wise and fixing $X$ point-wise. $\endgroup$ Commented Nov 23, 2015 at 18:02

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