Ultraproduct Structures - Model Theory

6 Ultraproduct Structures

Definition 6.1. Let $M_{j} = (M_{j}, I_{j})$ be $L$ -structures for each $j \in J$ . Let $U$ be an ultraproduct on $J$ . Define an interpretation $I_{U}$ of $L$ on $\prod_{j \in J} M_{j} ∕ U$ . Let $S \in L$ .

If $S$ is an $n$ -ary relation:
$I_{U} (S) = [{(I_{j} (S))}_{j \in J}] \subseteq {(\prod_{j \in J} M_{j} ∕ U)}^{n} .$
If $S$ is a constant:
$I_{U} (S) = [{(I_{j} (S))}_{j \in J}] \in \prod_{j \in J} M ∕ U .$
Functions are a bit less clear. However, we can always turn a function into a relation by looking at its graph (i.e. $f : M^{n} \to M$ has graph $R_{f} = {(\bar{x}, y) \in M^{n + 1} : f (\bar{x}) = y}$ ).

So if $S$ is a function: for each $j \in J$ , define the graph of $I_{j} (S)$ as
$G_{j} (S) = {(a_{1}, \dots, a_{n}, b) \in M_{j}^{n + 1} : I_{j} (S) (a_{1}, \dots, a_{n}) = b} .$

Then $[{(G_{j} (S))}_{j \in J}]$ is the graph of a function
${(\prod_{j \in J} M_{j} ∕ U)}^{n} \to \prod_{j \in J} M_{j} ∕ U .$

(Checking this is left as an exercise). Now define $I_{U} (S)$ to be the function corresponding to $[{(G_{j} (S))}_{j \in J}]$ .

Example 6.2. $L = {+, R}$ (where $+$ is a function and $R$ is a unary relation). Let $C_{n} = (C_{n}, I_{n})$ with $C_{n} = ℤ ∕ n ℤ$ , with addition modulo $n$ , and let $I_{n} (R) = {x \in C_{n} : \exists y \in C_{n}, 2 y = x}$ .

Consider $C = (\prod_{n \in ℕ_{> 0}} C_{n} ∕ U, I_{U})$ .

What does the set $I_{U} (R)$ look like?

If $\gcd (n, 2) = 1$ , then $I_{n} (R) = C_{n}$ .

If $\gcd (n, 2) = 2$ , then $I_{n} (R) \neq C_{n}$ (for example, $1 \notin I_{n} (R)$ ).

If $U = {A \subseteq ℕ : 3 \in A}$ , then $C ≅ C_{3}$ , so $I_{U} (R) = C$ .
Homework: Can you think of two non-principal ultrafilters, with one of them having $I_{U} (R) = C$ , and the other having $I_{U} (R) \neq C$ .

Options to think about:

(a) Every $U \in U$ contains an even number.
(b) Every $U \in U$ contains an odd number.
(c) There is a $U \in U$ all even.
(d) There is a $U \in U$ all odd.

Consider $b = (1, 1, 1, \dots) \in \prod_{n \in ℕ} C_{n}$ .

Suppose (a), so for every $K \in U$ we have $i \in K$ even, i.e. $b_{i} \notin I_{R} (C_{1})$ , so $I_{R} (C) \neq C$ . Note (a) is equivalent to (c).

By similar reasoning, (b) implies $I_{R} (C) = C$ (and also (b) is equivalent to (c)).