# Theory of bijection with no finite cycle has trivial associated geometry

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!

• 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. Commented Feb 19, 2021 at 13:53
• @AlexKruckman thank you for that clarification! I concluded that a little too quickly. Commented Feb 20, 2021 at 0:10

• 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.