Metin Ersin Arıcan

David Marker, Model Theory: An Introduction, Exercise 2.5.12

Metin Ersin Arıcan — Tue, 25 Jun 2024 00:00:00 GMT

Exercise 2.5.12. Let be an -formula. Show that the following are equivalent.

There is a universal formula such that .
If and are models of with , , and , then .

Proof. Let be a universal formula such that . Let and be models of with , , and . We have

Consider the following collection:

By compactness and because is closed under conjunctions, implies that , and hence for some .

To that end, let . We will show that . Consider the theory . For a contradiction, assume . By compactness, there exist and such that . This means , showing . Thus but , a contradiction.

Therefore is satisfiable. Let . Then, by (2), .

David Marker, Model Theory: An Introduction, Exercise 2.5.2

Metin Ersin Arıcan — Mon, 24 Jun 2024 00:00:00 GMT

Exercise 2.5.2. Suppose that has arbitrarily large finite models. Show that has an infinite model.

Proof. Let be the sentence

asserting that “there exist at least elements.” Consider the theory

Because has arbitrarily large finite models, every finite subset of is satisfiable. Therefore, by compactness, is also satisfiable. Clearly, any model of is an infinite model of .

David Marker, Model Theory: An Introduction, Exercise 2.5.10

Metin Ersin Arıcan — Mon, 24 Jun 2024 00:00:00 GMT

Exercise 2.5.10. Let be an -theory and be the set of all universal sentences such that . Show that if and only if there is with .

Proof. Consider the theory . Any model of is an extension of which is also a model of .

For a contradiction, assume that is unsatisfiable. By compactness, there exists a quantifier-free -formula with such that , implying . Therefore, , and hence . In particular, , which is a contradiction.

Suppose that for some . Every sentence in is true in , and universal sentences are preserved under substructures. Thus .