Ultraproducts - Model Theory - Cambridge III notes

5 Ultraproducts

Definition 5.1. Let ${(A_{j})}_{j \in J}$ be a family of sets ( $J \neq \emptyset$ ), with $A_{j} \neq \emptyset$ for all $j \in J$ . Take $J$ to be an ultrafilter on $J$ , and define the following equivalence relation $\sim$ on $\prod_{j \in J} A_{j}$ :

{(a_{j})}_{j \in J} \sim {(b_{j})}_{j \in J} iff {j \in J : a_{j} = b_{j} \in U} .

Proposition 5.2. The relation $\sim$ defined above is an equivalence relation on $\prod_{j \in J} A_{j}$ .

Proof. Symmetric / reflexive is obvious.

Transitivity: let ${(a_{j})}_{j \in J}, {(b_{j})}_{j \in J}, {(c_{j})}_{j \in J} \in \prod_{j \in J} A_{j}$ , and suppose ${(a_{j})}_{j \in J} \sim {(b_{j})}_{j \in J}$ and ${(b_{j})}_{j \in J} \sim {(c_{j})}_{j \in J}$ .

Let

\begin{array}{l} F_{a b} & = {j \in J : a_{j} = b_{j}} \\ F_{b c} & = {j \in J : b_{j} = c_{j}} \\ F_{a c} & = {j \in J : a_{j} = c_{j}} \end{array}

Note $F_{a b}, F_{b c} \in U$ . Also,

\underset{\in U}{\underset{⏟}{F_{a b} \cap F_{b c}}} = {j \in J : a_{j} = b_{j} = c_{j}} \subseteq F_{a c}

hence $F_{a c} \in U$ , i.e. ${(a_{j})}_{j \in J} \sim {(c_{j})}_{j \in J}$ . □

Definition 5.3. Let ${(A_{j})}_{j \in J}$ be a non-empty family of non-empty sets and $U$ an ultrafilter on $J$ .

Write $\prod_{j \in J} A_{j} ∕ U$ to be $\prod_{j \in J} A_{j} ∕ \sim$ (where $\sim$ is defined as in Definition 5.1).
${[{(a_{j})}_{j \in J}]}_{U}$ is the equivalence class of ${(a_{j})}_{j \in J}$ with respect to $\sim$ .
Let $B_{j} \subseteq A_{j}$ for every $j \in J$ . Then
${[{(B_{j})}_{j \in J}]}_{U} = {[{(a_{j})}_{j \in J}] \in \prod_{j \in J} A_{j} ∕ U : {j \in J : a_{j} \in B_{j}} \in U} .$

Is the third item well-defined?

Proposition 5.4. Assuming that:

${(A_{j})}_{j \in J}$ , ${(B_{j})}_{j \in J}$ satisfy $B_{j} \subseteq A_{j}$ for all $j$
${(a_{j})}_{j \in J}, {(b_{j})}_{j \in J} \in \prod_{j \in J} A_{j}$ satisfying ${(a_{j})}_{j \in J} \sim {(b_{j})}_{j \in J}$

Then

{j \in J : a_{j} \in B_{j}} \in U

if and only if

{j \in J : b_{j} \in B_{j}} \in U

Proof. Know ${(a_{j})}_{j \in J} \sim {(b_{j})}_{j \in J}$ . Define

\begin{array}{l} U & = {j \in J : a_{j} = b_{j}} \in U \\ V & = {j \in J : a_{j} \in B_{j}} \\ W & = {j \in J : b_{j} \in B_{j}} \end{array}

Note $U \cap V \subseteq W$ and $U \cap W \subseteq V$ .

If $V \in U$ then $U \cap V \in U$ so $W \in U$ . Similarly, if $W \in U$ then $V \in U$ . □

So $[{(B_{j})}_{j \in J}]$ is well-defined.

Proposition 5.5. Assuming that:

${(A_{j})}_{j \in J}$ , $U$ , $\sim$ as usual.
$B_{j}, C_{j} \subseteq A_{j}$

Then

(1) $[{(B_{j})}_{j \in J}] \cap [{(C_{j})}_{j \in J}] = [{(B_{j} \cap C_{j})}_{j \in J}]$ .
(2) $[{(B_{j})}_{j \in J}] \cup [{(C_{j})}_{j \in J}] = [{(B_{j} \cup C_{j})}_{j \in J}]$ .
(3) $[{(B_{j})}_{j \in J}] ∖ [{(C_{j})}_{j \in J}] = [{(B_{j} ∖ C_{j})}_{j \in J}]$ .

Proof. Example Sheet 1. □

Definition 5.6. Let ${(A_{j})}_{j \in J}$ , $U$ (ultrafilter on $J$ ), $\sim$ as before, and let $n \in ℕ$ .

For each $j \in J$ , suppose $B_{j} \subseteq A_{j}^{n}$ . Define

\begin{array}{l} [{(B_{j})}_{j \in J}] & = {([{(a_{j}^{1})}_{j \in J}], \dots, [{(a_{j}^{n})}_{j \in J}]) \in {(\prod_{j \in J} A_{j} ∕ U)}^{n} : {j \in J : (a_{j}^{1}, \dots, a_{j}^{n}) \in B_{j}} \in U} \\ \subseteq {(\prod_{j \in J} A_{j} ∕ U)}^{n} \end{array}

Note. If $n = 0$ , then we get that

\begin{array}{l} B_{j} & = {()} \\ [{(B_{j})}_{j \in J}] & = {()} \end{array}

Definition 5.7. Let $n > 0$ , $k \in {1, \dots, n}$ . Define:

$π_{k} (x_{1}, \dots, x_{n}) = (x_{1}, \dots, x_{k - 1}, x_{k + 1}, \dots, x_{n})$ .
For $X$ a set of $n$ -tuples,

$π_{k} (X) = {(x_{1}, \dots, x_{k - 1}, x_{k + 1}, \dots, x_{n}) : (x_{1}, \dots, x_{n}) \in X} .$

Proposition 5.8. Assuming that:

${(A_{j})}_{j \in J}$ , $U$ , $\sim$ as usual
$n \in ℕ$
for each $j$ we have $B_{j}, C_{j} \subseteq A_{j}^{n}$

Then

(1) $[{(B_{j})}_{j \in J}] \cap [{(C_{j})}_{j \in J}] = [{(B_{j} \cap C_{j})}_{j \in J}]$
(2) $[{(B_{j})}_{j \in J}] \cup [{(C_{j})}_{j \in J}] = [{(B_{j} \cup C_{j})}_{j \in J}]$
(3) $[{(B_{j})}_{j \in J}] ∖ [{(C_{j})}_{j \in J}] = [{(B_{j} ∖ C_{j})}_{j \in J}]$
(4) If $n > 0$ , $k \in {1, \dots, n}$ then
$π_{k} ([{(B_{j})}_{j \in J}]) = [{(π_{k} (B_{j}))}_{j \in J}] .$

Proof. (1) - (3): Straightforward. See Example Sheet.

(4): Let $n > 0$ , $k \in {1, \dots, n}$ . Let

α = ([{(a_{j}^{1})}_{j \in J}], \dots, [{(a_{j}^{k - 1})}_{j \in J}], [{(j_{j}^{k + 1})}_{j \in J}], \dots, [{(a_{j}^{n})}_{j \in J}]) \in π_{k} ([{(B_{j})}_{j \in J}]) .

So we have some $[{(a_{j}^{k})}_{j \in J}]$ such that

([{(a_{j}^{1})}_{j \in J}], \dots, [{(a_{j}^{n})}_{j \in J}]) \in [{(B_{j})}_{j \in J}] .

So $U = {j \in J : (a_{j}^{1}, \dots, a_{j}^{n}) \in B_{j}} \in U$ .

Consider

V = {j \in J : (a_{j}^{1}, \dots, a_{j}^{k - 1}, a_{j}^{k + 1}, \dots, a_{n}) \in π (B_{j})} \supseteq U .

So $α \in [{(π_{k} (B_{j}))}_{j \in J}]$ , i.e. $π_{k} [{(B_{j})}_{j \in J}] \subseteq [{(π_{k} (B_{j}))}_{j \in J}]$ .

Showing $[{(π_{k} (B_{j}))}_{j \in J}] \subseteq π ([{(B_{j})}_{j \in J}])$ is similar (do it as an exercise). □

Definition 5.9. Suppose $A_{j} = A$ for each $j \in J$ . Then we call $\prod_{j \in J} A_{j} ∕ U$ an ultrapower of $A$ , and write $A^{U}$ .

If $B \subseteq A$ , we write $[B]$ for $[{(B)}_{j \in J}]$ .

Theorem 5.10. Assuming that:

$A^{U}$ be as above
$J = ℕ$
${C \subseteq ℕ : ℕ ∖ C is finite} \subseteq U$
$n \in ℕ$
for each $m \in ℕ$ , let $B_{m} \subseteq A^{n}$ satisfying:
- (1) $[B_{m}] \neq \emptyset \forall m \in ℕ$
- (2) $[B_{k}] \subseteq [B_{m}]$ for all $m, k \in ℕ$ with $m \leq k$

Then

⋂_{m \in ℕ} [B_{m}] \neq \emptyset

Proof. Omitted. For $n = 1$ , this is a potential presentation topic. □