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I've seen ultrapowers of $\mathbb{Z}$ and $\mathbb{R}$ discussed in this question, with the $\mathbb{R}$ case being described as the more straightforward case. I'm wondering what happens when you iterate the ultrapower construction (in this case just apply it twice).

Let $\mathfrak{U}$ and $\mathfrak{V}$ be non-principal ultrafilters of $\mathbb{N}$.

We get the hyperreal numbers if we take the ultrapower of $\mathbb{R}$ one time:

$$ \prod_\mathfrak{U} \mathbb{R} \;\; \text{is the ordered field of hyperreal numbers} $$

We can think of each element as a sequence of reals $\mathbb{N} \to \mathbb{R}$ with $f \le g$ holding of two sequences $f, g$ precisely when $\{ i : f(i) \le g(i) \} \in \mathfrak{U}$ and $+, *, \bigcirc^{-1}$ being defined componentwise. I don't have a strong intuition for how hyperreals behave, but that definition at least makes sense.

I'm curious what happens if we do this twice:

$$ \prod_\mathfrak{V} \prod_\mathfrak{U} \mathbb{R} \;\; \text{is some ordered field}$$

We can think of these things being doubly indexed sequences $\mathbb{N} \times \mathbb{N} \to \mathbb{R}$.

$ f \le_j g $ holds if and only if $\{ f(j, i) \le g(j, i) : i \in \mathfrak{U} \} $.

And $f \le g$ holds if and only if $\{ f \le_j g : j \in \mathfrak{V} \}$.

The behavior of $\le$ is completely determined by the indices where it holds.

Using ultrafilters as quantifiers like this but without the explicit $\forall$, we get the following.

$$ f \le g \;\; \text{if and only if} \;\; [\mathfrak{V} j][\mathfrak{U} i](f(j, i) \le g(j, i)) $$

I'm not sure how to prove it, but I don't think this is equivalent to the following for some ultrafilter $\mathfrak{W}$ on $\mathbb{N} \times \mathbb{N}$.

$$ f \le g \;\; \text{perhaps if and only if} \;\; [\mathfrak{W}\vec{p}](f(\vec{p}) \le g(\vec{p})) $$

We know by Łoś's Theorem that this thing is at least elementarily equivalent to our starting object $\mathbb{R}$ as an ordered field, or a real-closed field, or a ring or whatever.

Beyond that though I have no intuitions about its structure.

Greg Nisbet
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    Actually, a double ultrapower can just be reduced to a single ultrapower on the product of index sets. Indeed, if $U$ is an ultrafilter on index set $I$, $V$ is an ultrafilter on index set $J$, then you can defined the ultrafilter $U \times V$ on $I \times J$ by $A \in U \times V$ iff ${j \in J: p_I(A \cap (I \times {j})) \in U} \in V$, where $p_I: I \times J \rightarrow I$ is the projection onto the first coordinate. Then for any structure $M$, $(M^U)^V$ is canonically isomorphic to $M^{U \times V}$. – David Gao Jan 03 '24 at 05:25
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    This is an answer to the question, not just a comment. Per site policy, comments should only be used to clarify, not answer the question. See How do comments work for more information. Therefore I suggest that you post your answer as an answer. This brings extra visibility to the answer und puts the question off the unanswered list. Also notice that comments are not indexed by the full text search, don't have any revision history, can only be edited for 5 minutes and allow only limited markup. – Martin Brandenburg Jan 03 '24 at 09:18

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Per Martin’s suggestion, this is a slightly edited version of my comment turned into an answer.

Actually, a double ultrapower can just be reduced to a single ultrapower on the product of index sets. Indeed, if $U$ is an ultrafilter on index set $I$, $V$ is an ultrafilter on index set $J$, then you can define the ultrafilter $U \times V$ on $I \times J$ by $A \in U \times V$ iff $\{j \in J: \{i \in I: (i, j) \in A\} \in U\} \in V$. Then for any structure $M$, $(M^U)^V$ is canonically isomorphic to $M^{U \times V}$.

David Gao
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