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I have a question about the following paper:

One-dimensional asymptotic classes of finite structures

by Macpherson and Steinhorn (link at Trans. AMS website).

Let $\mathbf{K}$ be a one-dimensional asymptotic class of finite structures.

The phrase "every infinite ultraproduct" appear in this paper several times. For example Lemma 2.5. This phrase is vague for me. What does that mean? Does it mean that the ultraproduct of evey infinite subclass of $\mathbf{K}$? or it means the ultraproduct of all members of $\mathbf{K}$ under different ultrafilters?

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  • $\begingroup$ "the ultraproduct of all members" does not mean anything. Given a set $X$ of sets, I think that "The ultraproducts of $X$" can indifferently mean all ultraproducts $\prod^\eta_{A\in X}A$, for $\eta$ ranging over ultrafilters on $X$, and all ultraproducts of all families $(A_i)_{i\in I}$ of elements of $X$ with respect to all set $I$ and ultrafilters $\eta$ on $I$. Here in this papers all structures are finite, and "infinite" just means we consider only those ultraproducts that are infinite, i.e., restrict to $\eta$ such that $\lim_{i\to\eta}|X_i|=\infty$. $\endgroup$
    – YCor
    Commented Aug 7, 2019 at 10:40

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At risk of triviality: An infinite ultraproduct (of members of $\mathcal{C}$) is a structure which is (a) infinite, and (b) an ultraproduct (of members of $\mathcal{C}$).

So, for example, Lemma 2.5 reads: "Let $\mathcal{C}$ be a class of finite $\mathcal{L}$-structures, and suppose that every infinite ultraproduct of members of $\mathcal{C}$ is strongly minimal. Then $\mathcal{C}$ is a $1$-dimensional asymptotic class."

This could be equivalently stated as follows: "Let $\mathcal{C}$ be a class of finite $\mathcal{L}$-structures. Suppose that for every family $\{A_i\}_{i \in I}$ of members of $\mathcal{C}$, and every ultrafilter $\mathcal{U}$ on $I$, if the ultraproduct $M = \prod_{i\in I} A_i\,/\, \mathcal{U}$ is infinite, then $M$ is strongly minimal. Then $\mathcal{C}$ is a $1$-dimensional asymptotic class."

Of course, if you take the ultraproduct of a family of finite structures by a principal ultrafilter, the result will be a finite structure. The point of writing "infinite ultraproduct" is to rule out such trivial cases.

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  • $\begingroup$ Thanks! So if $\mathcal{C}=\{A_i\}_{i\in I}$, "an infinite ultraproduct of members of $\mathcal{C}$" does Not mean only a member of the set $\left\{\prod_{i\in I} A_i /\mathcal{U} : \mathcal{U} \text{ is an ultrafilter on } I \right\}$ ? $\endgroup$
    – Mark Smith
    Commented Aug 7, 2019 at 14:07
  • $\begingroup$ @MarkSmith Right. "Infinite" here refers to the cardinality of the resulting structure, not to the cardinality of the set of factors in the ultraproduct. $\endgroup$
    – Alex Kruckman
    Commented Aug 7, 2019 at 17:52

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