Difference between ultraproduct and ultrapower (with example)

Question

An ultraproduct's domain is the quotient set of the cartesian product of all of the domains of its factors. However, a constraint on this construction says all the factors of the ultraproduct have the same signature.
From what I understand, the signature of a structure tells us the functions and relations that are defined on said structure. Now, if all the factors of the ultraproduct have the same signature, then that means that they all have the same functions and relations. These functions and relations in turn define their domains and codomains, therefore specifying the whole structure. Thus, all the factors of an ultraproduct are the same, making it an ultrapower.
In order to clear this up in my mind, I have tried to work through a very simple example.
$|A|=\{2,3\} \ \mathrm{and} \ f^A(2)=3 \ \mathrm{and} \ f^A(3)=2$
$|B|=\{3,4\} \ \mathrm{and} \ f^B(4)=3 \ \mathrm{and} \ f^B(3)=4$
From here, I think $I=\{0,1\}$ and $U=\{0,\{0,1\}\}$.
Then the domain of the ultraproduct should be: $$\{\{(2,3),(2,4)\},\{(3,3),(3,4)\},\{(2,3)\},\{(2,4)\},\{(3,3)\},\{(3,4)\}\}$$ Here I do not know how to continue because I am unsure whether I have done it right. Is it possible to complete the example?

Answer 1

In your example $A\times B=\{\langle 2,3\rangle,\langle 2,4\rangle,\langle 3,3\rangle,\langle 3,4\rangle\}$. Since $U$ is the principal ultrafilter at $0$, elements $\langle a,b\rangle$ and $\langle a',b'\rangle$ in $A\times B$ are equivalent mod $U$ if and only if $a=a'$, so $(A\times B)/U$ has just two elements, the equivalence classes

$$\{\langle 2,3\rangle,\langle 2,4\rangle\}$$

and

$$\{\langle 3,3\rangle,\langle 3,4\rangle\}\;.$$

Now

$$\begin{align*} f^{A\times B}(\langle 2,3\rangle)&=\langle 3,4\rangle\;,\\ f^{A\times B}(\langle 2,4\rangle)&=\langle 3,3\rangle\;,\\ f^{A\times B}(\langle 3,3\rangle)&=\langle 2,4\rangle\;,\text{ and}\\ f^{A\times B}(\langle 3,4\rangle)&=\langle 2,3\rangle\;, \end{align*}$$

so

$$f^{(A\times B)/U}(\{\langle 2,3\rangle,\langle 2,4\rangle\})=\{\langle 3,4\rangle,\langle 3,3\rangle\}$$

and

$$f^{(A\times B)/U}(\{\langle 3,3\rangle,\langle 3,4\rangle\})=\{\langle 2,4\rangle,\langle 2,3\rangle\}\;.$$

This means that the ultrapower $(A\times B)/U$ is isomorphic to $A$, the correspondence of underlying sets being

$$\begin{align*} &2\leftrightarrow\{\langle 2,3\rangle,\langle 2,4\rangle\}\\ &3\leftrightarrow\{\langle 3,3\rangle,\langle 3,4\rangle\}\;. \end{align*}$$

This is exactly what one would expect, since $U$ is the principal ultrafilter over $0$, the index corresponding to $A$. If you use the principal ultrafilter over $1$, you’ll get an ultraproduct isomorphic to $B$.

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A signature doesn’t tell you what functions and relations you have: it tells you what kinds of functions and relations you have. Every linear order, for instance, can be considered a structure with a single binary relation (satisfying certain axioms). Thus, you can form an ultraproduct of an arbitrary set of linear orders, no two of which even have the same cardinality.