Why do ultraproduct structures use a quotient as their universe?

Question

For an $L$-structure $\mathfrak{A}$ with universe $A$, if we have an index set $I$, with an ultrafilter $U$, we create an ultraproduct structure having as its universe $\Pi_I \;A_i/U$. This is the set of equivalence classes of $\Pi_I A_i$ modulo ~, where $a$ ~ $b$ iff $\{i \;|\; a(i) = b(i) \} \in U$. Why do we need, or why is it better to use the quotient, rather than just $\Pi_IA_i$ as the universe? It seems we could still define a model having this product as the universe - relations would still be those elements where the relation in $\mathfrak{A}$ holds over the ultrafilter, and functions and constants would also be defined as they are normally for ultraproducts. So why the need (or desire) to work with equivalence classes? They seem to just make things more confusing.

Answer 1

For the most part, model theorists are concerned with complete theories. This means, for a fixed language $\mathcal{L}$, an $\mathcal{L}-$theory $T$ is complete if for every $\varphi$ in this language, $T \vdash \varphi$ or $T \vdash \neg \varphi$. Now, there are exceptions to this rule. Sometimes you care about model companions, or quantifier elimination for a collection of theories (e.g. ACF), but again, for the most part, many model theorists care about complete theories.

Question 0: Why ultrapowers and not Cartesian products?

By the argument above, we care about the class (or category) of models of a complete theory $T$ (where our morphisms are elementary embeddings). Notice that if $T =$ $\mathbf{AFC_0}$, then $\mathbb{C} \models T$. However, $\mathbb{C}\times \mathbb{C} \not \models T$ since $(0,1)$ does not have a multiplicative inverse. Actually, it is somewhat rare for cartesian products of models (with pointwise interpretations) to be models of our original theory.

On the other hand, ultrapowers/ultraproducts fixes our problem. By Łoś's Theorem, $\{i:A_i\models\varphi(a_{i_1},...,a_{i_n})\}\in D$ iff $\prod_D A_i \models \varphi([a_1],...,[a_n])$. Furthermore, if $A_i \equiv A_j$ for each $i,j \in I$, we can conclude that $A_i \equiv \prod_DA_i$. In English, a first order sentence is true in the ultraproduct iff it is true in over a "large" subcollection of our models and if each $A_i$ is a model of $T$, then $\prod_D A_i$ is also a model of $T$.

Remark: In the category theory interpretation, if $A_i \in \mathcal{C}(T)$, then $\prod_DA_i \in \mathcal{C}(T)$. Furthermore, there is a natural elementary embedding from a model $A_i$ to an ultrapower $\prod_DA_i$ via the diagonal embedding (i.e. $d(a)=[a]$).

Question 1: How do we generate new models of a complete theory $T$ from old models of $T$?

The quick and fast answer is compactness and the upward and downward Löwenheim–Skolem theorems. However, these theorems just tell us "there is some model of cardinality $\kappa$ that realizes this collection of types". With more work, you can prove the omitting-types theorem which allows you to conclude "there is some model of cardinality $\kappa$ that omits this (small) collection of non-principal types".

On the other hand, ultraproducts and ultrapowers give you more control over saturation, realizing types, and actually elements. For instance, Let $A= (\mathbb{N};<)$ and let $|I|=\aleph_0$. Then, consider $\prod_DA_i$. In this case, I can actually give you an infinite descending chain in this structure (i.e. $(1,2,3,4,...),(0,1,2,3,...),(0,0,1,2,...)$).

Question 2: How exactly does this differ from controlled compactness?

Here is where I place my plug for the Keisler order. Classical "classification theory" deals with how one finds dividing lines among the collection of all complete theories. Why is arithmetic over the natural numbers so much harder than arithmetic over the reals? What do linear orders and the random graph have in common? The Keisler order is a lens in which we can under stand "how hard" it is to saturate all types over a model of a (countable) complete theory. This (pre-)order determines how "hard" this process is by considering "how strong" of an ultrafilter is necessary to achieve $\kappa-$saturation. Therefore, ultraproducts are also useful in finding dividing lines between first-order theories.