I am working through the following exercise in Tent and Ziegler's A Course in Model Theory.

Exercise 5.7.2. Show that the theory of an infinite set equipped with a bijection without finite cycles is strongly minimal and that the associated geometry is trivial.

I think I've managed to show that this theory has QE and hence (having no relation symbol) can only define finite or cofinite sets, and hence is strongly minimal. What I don't understand is what is meant by "the associated geometry is trivial". I am guessing this refers to the pregeometry $(M, \text{cl})$, where for each $A\subseteq M$, $\text{cl}(A)=\text{acl}^M(A)\cap M$. But what does it mean for it to be trivial and how do we prove it? Thank you!

  • 1
    $\begingroup$ Just to clear up a possible misconception: You write "this theory has QE and hence (having no relation symbol) can only define finite or cofinite sets". It is certainly not true that every theory with QE in a language with no relation symbols is strongly minimal. For example, consider the structure $(\mathbb{N};0,1,f)$ where $f$ is a unary function symbol sending every even number to $0$ and every odd number to $1$. The complete theory of this structure has QE, but the formula $f(x) = 0$ defines an infinite and co-infinite set. $\endgroup$ Feb 19, 2021 at 13:53
  • $\begingroup$ @AlexKruckman thank you for that clarification! I concluded that a little too quickly. $\endgroup$
    – ikrto
    Feb 20, 2021 at 0:10

1 Answer 1

  • Whenever we have a strongly minimal theory, we get a pregeometry on any model $M$ of that theory: $(M,\text{acl})$. There's no reason to write $\text{acl}^M(A)\cap M$, because $\text{acl}(A) = \text{acl}^M(A)\subseteq M$. Maybe what you were thinking of is that if $D$ is a strongly minimal set definable in a model $M$, then the pregeometry is $(D,\text{cl})$, where for $A\subseteq D$, $\text{cl}(A) = \text{acl}^M(A)\cap D$.

  • The meaning of "associated geometry" is explained in Section C.1 (bottom of p. 205). If $(X,\text{cl})$ is a pregeometry, let $X^\bullet = X\setminus \text{cl}(\emptyset)$. Define an equivalence relation $\sim$ on $X^\bullet$ by $x\sim y$ iff $\text{cl}(x) = \text{cl}(y)$. Finally, let $X' = X^\bullet/{\sim}$, and define $\text{cl}'$ on $X'$ by $\text{cl}'(A/{\sim}) = \text{cl}(A)^\bullet/{\sim}$. That is, $\text{cl}'$ of a set $B$ of equivalence classes contains the equivalence class of every point which is in the $\text{cl}$-closure of a set of representatives for $B$ but not in $\text{cl}(\emptyset)$. Then $(X',\text{cl}')$ is a geometry, the geometry associated to $(X,\text{cl})$.

  • The meaning of "trivial" is explained on p. 80 in passing: a geometry is trivial if $\text{cl}(A) = A$ for all sets $A$.

  • A word about terminology: Tent and Ziegler provide another definition of "trivial" in the context of pregeometries. On p. 207, we read that a pregeometry is trivial if $\text{cl}(A\cup B) = \text{cl}(A) \cup \text{cl}(B)$ for all sets $A$ and $B$. Some people (including me) prefer call such a pregeometry disintegrated to distinguish it from the more trivial meaning of trivial given above. But the reason for the terminology is that a pregometry is trivial (in the disintegrated sense) if and only if its associated geometry is trivial (in the truly trivial sense). This is a good exercise to strengthen your understanding of the associated geometry construction.

  • As for how to solve the exercise from Tent and Ziegler: Now that you know what the words mean, I think it's better to try to solve it yourself. If you get stuck, drop me a comment.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .