Does the isomorphism type of this specific ultraproduct depend on the ultrafilter?

Question

Preliminaries. I was wondering how much this specific example depends on the choice of ultrafilter. Let $\mathcal U$ be a free (i.e., non-principal) ultrafilter over $\mathbb N$ and consider the structures $$\mathcal A_n = \langle\{ 0, 1, 2, \dots, n\}, \leq \rangle.$$ Then the ultraproduct $$\left. \mathcal A = \left(\prod_{n \in \mathbb N}\mathcal A_n \right) \middle/ \mathcal U \right.$$ is totally-ordered, discrete and has a minimal and a maximal element, all by the Łoś Ultraproduct Theorem. We can even write down concrete representatives of the smallest (bottom) and biggest (top) elements $$\bot = [\langle 0, 0, 0, \dots\rangle]_{\mathcal U}\\ \top = [\langle 0, 1, 2, \dots\rangle]_{\mathcal U}$$ respectively, independent of $\mathcal U$. In fact, the successor $\bot_+$ of $\bot$ also has an explicit representative, namely $\bot_+=[\langle 0, 1, 1, \dots\rangle]_{\mathcal U}$. Continuing like this, we get explicit representatives of all finite successors of the bottom element.

Similarly, the predecessor $\top_-$ of $\top$ is $[\langle 0, 0, 1, 2, \dots \rangle]_\mathcal U$ (because it is the predecessor at all but finitely many indices). So we also get explicit representatives of all finite predecessors of $\top$.

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Question. The above makes it seem reasonable to ask whether the ultraproduct $\mathcal A$ is isomorphic to the structure $$ \mathcal B = \mathcal N \sqcup \mathcal N^{\text{op}},$$ where $\mathcal N$ is the natural numbers with the standard order, $\mathcal N^{\text{op}}$ is the natural numbers with the opposite of the standard order, and $\sqcup$ denotes the disjoint union. Is this true? If not, does the isomorphism type of $\mathcal A$ depend on the ultrafilter $\mathcal U$?

Answer 1

First, let me make the pedantic point that if $\mathcal{U}$ is a principal ultrafilter (generated by $n\in \mathbb{N}$), then the ultraproduct will be isomorphic to $\mathcal{A}_n$. Ok, so let's agree to consider only non-principal ultrafilters from now on.

Here's what I can tell you about the situation:

• Every ultraproduct of a family of structures (in a countable language) by a countably incomplete ultrafilter is $\aleph_1$-saturated (see Theorem 6.1.1 in Model Theory by Chang and Keisler, or Theorem 5.3 in this expository paper by Kyle Gannon). Every non-principal ultrafilter on $\mathbb{N}$ is countably incomplete, so every ultraproduct of the kind you're considering is $\aleph_1$-saturated.

• In particular, if you write $\mathcal{N}$ for the initial segment isomorphic to $\mathbb{N}$ and $\mathcal{N}^*$ for the end segment isomorphic to $\mathbb{N}^{\text{op}}$, then the type $\{x > n\mid n\in \mathcal{N}\}\cup \{x < n \mid n\in \mathcal{N}^*\}$ is realized in the ultraproduct. So your ultraproduct is not isomorphic to $\mathcal{B}$. Actually, we don't need to appeal to the theorem to see that this type is realized: it's realized by $[\langle 0, 1, 1, 2, 2, 3, 3, \dots\rangle]_\mathcal{U}$.

• Every ultraproduct of the $\mathcal{A}_n$ by a non-principal ultrafilter will be an infinite discrete linear order with endpoints by Łoś's theorem. The first-order theory $T$ of infinite discrete linear orders with endpoints is complete, so any two ultraproducts of the $\mathcal{A}_n$ will be elementarily equivalent. That is, the theory of the ultraproducts doesn't depend on the ultrafilter, but the isomorphism type might.

• Any model of $T$ looks like $\mathcal{N} \sqcup \bigsqcup_{i\in \mathscr{L}} \mathcal{Z} \sqcup \mathcal{N}^{\text{op}}$, where $\mathcal{N}$ is the order type of the natural numbers, $\mathcal{N}^{\text{op}}$ is its opposite, each $\mathcal{Z}$ is the order type of the integers, and these copies of $\mathcal{Z}$ are ordered like an arbitrary linear order $\mathscr{L}$. This last claim is not too hard to prove just by thinking about what the axioms of $T$ say - the completeness of $T$ is a little tricker, but it follows from an Ehrenfeucht-Fraïssé game argument. So your question now comes down to: What order types $\mathscr{L}$ can arise from this ultraproduct?

• Every ultraproduct of a countable family of finite structures of unbounded size by a non-principal ultrafilter has cardinality $2^{\aleph_0}$. (For example, see this answer.) This gives another reason why your ultraproduct is not isomorphic to $\mathcal{B}$.

• Any two saturated (i.e. $\kappa$-saturated where $\kappa = |M|$) structures of the same cardinality which are elementarily equivalent are isomorphic. So if we assume that continuum hypothesis, then for any non-principal ultrafilters $\mathcal{U}_1$ and $\mathcal{U}_2$ on $\mathbb{N}$, we have $$\prod_{n\in \mathbb{N}} \mathcal{A}_n/\mathcal{U}_1\cong \prod_{n\in \mathbb{N}} \mathcal{A}_n/\mathcal{U}_2$$ because both structures are $\aleph_1$-saturated of cardinality $2^{\aleph_0} = \aleph_1$, and they are both models of the complete theory $T$ of discrete linear orders without endpoints. Thus, under CH, the choice of ultrafilter doesn't matter. The structure $\prod_{n\in \mathbb{N}} \mathcal{A}_n/\mathcal{U}_1$ is always isomorphic to $\mathcal{N} \sqcup \bigsqcup_{i\in \overline{\mathscr{L}}} \mathcal{Z} \sqcup \mathcal{N}^{\text{op}}$, where $\overline{\mathscr{L}}$ is the unique saturated linear order of size $2^{\aleph_0}$ (which exists if we assume CH).

• On the other hand, if CH fails, we still know that the ultraproduct is isomorphic to $\mathcal{N} \sqcup \bigsqcup_{i\in \mathscr{L}} \mathcal{Z} \sqcup \mathcal{N}^{\text{op}}$, for some $\aleph_1$-saturated linear order $\mathscr{L}$ of size $2^{\aleph_0}$, but exact order type $\mathscr{L}$ of the copies of $\mathcal{Z}$ may depend on your choice of ultrafilter.