I'm trying to get a better understanding of ultraproducts and their typical uses.
To that end, I'm wondering what happens if we fix an ultrafilter $U \in 2^{2^{\mathbb{N}}} $ and look at $\prod_{i \in \mathbb{N}}$ where $S_i$ is the symmetric group associated with $i \in \mathbb{N}$.
The result is a group, and by the intution behind Łoś's theorem (proofwiki), it should satisfy the properties that "most" (with heavy scare quotes) symmetric groups satisfy.
However, being a symmetric group in the first place is not a first-order property, so I'm curious what group you actually end up with (and whether it's sensitive to the choice of $U$).
Suppose $I$ is a poset.
$F \in 2^{2^I}$ is a filter if and only if
- $F$ is nonempty.
- It holds that $\forall x \in F \mathop. \forall y \in F \mathop. \exists z \in F \mathop. z \le x \land z \le y$.
- It holds that $\forall x \in F \mathop. \forall y \in I \mathop. x \le y \to y \in F$.
$F$ is a proper filter if and only if $F \neq 2^I$.
$U$ is an ultrafilter if and only if
- $U$ is a filter on $I$.
- $U$ is not equal to $2^I$.
- Every proper filter $F$ extending $U$ is equal to $U$.
An ultrafilter is principal if and only if it is the upward closure of a single element of $I$.
Let $(I, \le)$ be $(\mathbb{N}, \le^\mathbb{N})$.
Let $U$ be a fixed non-principal ultrafilter on $I$.
Let $S_k$ be the $n$th symmetric group as usual.
I'm curious about $\prod_{i \in I} S_k / U$ (which I'll be writing as $\prod^U_{i \in I} S_k$ for convenience.
If we restrict our attention to algebraic theories, then we can define the truth of a sentence in $\prod^U_{i \in I} S_k$ as follows.
$\prod^U_{i \in I} S_k \models t_1 = t_2$ if and only if if the set of indices for which the $i$th component of $t_1$ and $t_2$ are equal is in $U$. By the nonemptiness of $U$ and the upward-closedness property of $U$, it holds that $I \in U$.
The interpretation of $f(\vec{a})$ is the equivalence class associated with $\langle f(a_0), f(a_1), \cdots \rangle$ where $a_i$ is the $i$th component of the canonical representative of the equivalence class $\vec{a}$.
I'm pretty sure $\prod^U_{i \in I} S_k$ and that this follows from the fact that $\prod_{i \in I} S_k $ is a group and modding out by the equivalence relation associated with $U$ is a group homomorphism.
$\prod_{i \in I} S_k $ is a Cartesian product of models of an algebraic theory consisting entirely of $\forall$-sentences, so it is again a model of that theory (the theory of groups, not the theory of symmetric groups).
I'm using the non-principalness of $U$ to insist that there isn't a proper subset of indices that completely determines the behavior of $\prod^U_{i \in I} S_k $. Although if I'm being honest, I'm mostly insisting on this because I've heard "non-principal ultrafilter" before.
As an object, $\prod^U_{i \in I} S_k$ is a bit odd.
I'm pretty sure it's infinite, but all of its factors are finite, so we can't prove its infiniteness by constructing an infinite sequence of examples that differ at every index.
Let $G$ be a finite symmetric group, let $g_k$ be $(k \mathop{\text{mod}} |G|)$-th element of the group, where the elements are assigned an index in some arbitrary fashion.
Let $b_i$ be equal to $(S_0)_i, (S_1)_i, (S_2)_i, \cdots$.
I'm pretty sure that $b_0, b_1, b_2, \cdots$ contains infinitely many distinct elements of $\prod^U_{i \in I} S_k$.
I think you can show this by collecting the index sets between identical elements and saying that they intersect to the empty set when there are infinitely many of them ... and therefore that $U$ would have to equal $2^I$ (which contradicts the definition of an ultrafilter) ... but I'm really not sure.
