Intuition behind quotienting by an ultrafilter -- why do we need all the conditions on ultrafilters?

Question

I'm trying to understand this answer by Noah Schweber, specifically the first paragraph about the truth value of existential second-order sentences being preserved by ultrapowers. It seems like an interesting fact. In order to understand that, I first need to understand what the construction $\prod_{i \in I} M_i / U$ means.

I can follow the definition and have a vague sense that elements in $\prod_{i \in I} M_i / U$ are the same if and only they are eventually the same, with $U$ providing our notion of eventualness, but I'm having a hard time picturing why $U$ needs to be an ultrafilter rather than merely being upwardly closed. I'm also struggling to see why we needed to rule out $U=2^I$ (why the filter needs to be proper) and rule out $U=\varnothing$ (filters can't be empty).

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Following the Wikipedia article, the initial definition of a product over an indexed family of structures seems straightforward to define.

I'll use $M_i$ to refer to both the structure $M_i$ and its universe, depending on the context.

Let $I$ be an index set. Let $M_i$ be a relational structure for each $i$. I am ignoring function and constant symbols for the time being. Let $F_i$ be the frame associated with $M_i$. Let $(r, n, \alpha) \in F_i$ hold if and only if $r$ is an n-ary predicate in the shared signature of all $M_I$ and the interpretation of $r$ in $M_i$ is $\alpha \subset M_i^n$.

$ \prod_{i \in I} M_i $ is straightforward to define.

$ v \in (\prod_{i \in I}M_i)$ holds if and only if $\forall k \in I \mathop. v_k \in M_k$ holds.

$(r, n, \alpha) \in (\prod_{i \in I} F_i)$ holds if and only if $\forall v \in r \mathop. \forall k \in I \mathop. v_k \in M_i$ holds.

Cool, so now $\prod_{i \in I}M_i$ is a structure.

I'm operating under the assumption that $\prod_{i \in I} M_i/U$ can be interpreted as $(\prod_{i \in I} M_i)/U$, i.e. there's a sensible way to interpret an ultrafilter as an equivalence relation $R$ on $\prod_{i \in I} M_i$ that respects the interpretations of all the relation symbols. (I would call $R$ a congruence relation if the symbols were function symbols, but I'm not sure that congruence relation is technically correct here.)

$U$ is an ultrafilter, meaning it is a maximal element in the lattice of proper filters on $I$ ordered by inclusion.

If $a$ and $b$ are two elements of $\prod_{i \in I}M_i$, let $J(a, b)$ be defined as the set of indices where they are the same.

$$ J(a, b) = \{ i : i \in I \land a_i = b_i \} $$

So now we define $R$, $R(a, b) \iff J(a, b) \in U$.

This makes sense, but it's not really clear where we used the maximalness assumption, or why we needed to rule out $U=2^I$, or why we needed to rule out $U=\varnothing$.

Answer 1

Any filter $F$ on $I$ determines a quotient of the product, usually written "$\prod_{i\in I}M_i/F$." This is called a reduced product, and if I recall correctly both Chang/Keisler and (big) Hodges have a lot of information about them.

(We do crucially need nonemptiness of $F$ here: the definition of the equivalence relation $R$ would break down if we tried to use $F=\emptyset$, since then $R$ would never hold - quotienting out by the emptyset doesn't make sense! Intuitively we need at least $I\in F$ to reflect the fact that each thing should be equal to itself.)

What makes ultrafilters particularly special in terms of the reduced products they yield (= "ultraproducts") is Los' theorem. If $U$ is an ultrafilter, then not only is $\prod_{i\in I}M_i/U$ a well-defined structure, it also takes on the "average behavior" of the $M_i$s in many ways: any "simple" (i.e. first-order) sentence satisfied by "most" (i.e. $U$-many) of the factor structures is satisfied by the whole $\prod_{i\in I}M_i/U$.

This doesn't happen in general! For example, if we take $F=\{I\}$ (not $F=I$), we just get $$\prod_{i\in I}M_i/F\cong \prod_{i\in I}M_i.$$ So for example even if all of the $M_i$s have exactly two elements (= satisfy "$\exists x,y(x\not=y\wedge\forall z(x=z\vee y=z))$") the structure $\prod_{i\in I}M_i/F$ won't have exactly two elements as long as $\vert I\vert>1$. Meanwhile, if we take a filter $F$ with $\emptyset\in F$ (not $F=\emptyset$), everything collapses to a single point: $\prod_{i\in I}M_i/F$ is a one-element structure.

If you go through the proof of Los' Theorem, you'll see in detail how the ultrafilter properties are needed to get the kind of transfer we want. It then turns out that in fact Los' Theorem can be strengthened to give "upwards transfer" for existential second-order sentences, but the point is that already non-ultra-filters don't even transfer very simple first-order properties.

Answer 2

In the referenced answer, we want

• If $\prod \mathcal{A}/\mathcal{U}\models\varphi$ and $\psi$ is arbitrary, then $\prod \mathcal{A}/\mathcal{U}\models\varphi\lor \psi$. The members of the index set $I$ that vote for $\varphi\lor\psi$ to be true can be an arbitrary superset of those who vote for $\varphi$ to be true. Hence we want $\mathcal U$ to be upward-closed
• If $\prod \mathcal{A}/\mathcal{U}\models\varphi$ and $\prod \mathcal{A}/\mathcal{U}\models\psi$, then $\prod \mathcal{A}/\mathcal{U}\models\varphi\land\psi$. The set of elements $I$ that vote for $\varphi\land\psi$ to be true are the intersection of the set voting for $\varphi$ and the set voting for $\psi$. Hence we want $\mathcal U$ to be closed under $\cap$.
• Exactly one of $\prod \mathcal{A}/\mathcal{U}\models\varphi$ or $\prod \mathcal{A}/\mathcal{U}\models\neg\varphi$ should hold. Hence for every subset of $I$, $\mathcal U$ should contain either that subset or its complement.

From the last, it also follows that $\mathcal U$ is neither empty nor the full power set. Finally, if $\mathcal U$ is principal, then the notion of "majority" is dictated by a single factor of the product, i.e., we end up precisely with that factor again.