4
$\begingroup$

Edit: Following discussion with Tomasz below, I want to clarify that I mean for definable subsets here to range over arbitrary sorts; ie, $\mathfrak{C}$ can be multi-sorted, but I'm not considering it as having a distinguished "home sort" or using the term "definable subsets" to refer to subsets of a distinguished sort.

Otherwise, as he points out, one can consider the theory of infinite sets as a multi-sorted theory $T^*$, with sorts $(S_n)_{n\in\omega}$ and $n$-ary projection maps $\pi_n:S_1^n\to S_n$, where $S_1$ is taken to be the "home sort" and each $S_n$ is taken to be the sort for the subsets of $S_1$ of size at most $n$. Then any finite subset $\{x_1,\dots,x_n\}\subset S_1$ of the home sort has a canonoical parameter given by $\pi_{n}(x_1,\dots,x_n)$, but (as Tomasz argues) the pair of tuples $\{(a,b),(c,d)\}\subset S_1^2$ for distinct elements $a,b,c,d\in S_1$ has no canonical parameter.

However, this is not quite what I'm looking for, since $T^*$ only has elimination of $1$-imaginaries for subsets of the home sort. Indeed, for instance, the finite subset $\{\pi_2(a,b),\pi_2(c,d)\}\subset S_2$ has no canonical parameter, so $T^*$ does not eliminate $1$-imaginaries for subsets of arbitrary sorts.


Let $T$ be a complete first-order theory and fix a monster model $\mathfrak{C}\models T$. Recall that, for a $\mathfrak{C}$-definable collection of $n$-tuples $E\subseteq\mathfrak{C}^n$, a canonical parameter for $E$, denoted $\ulcorner E\urcorner$, is a finite tuple $\overline{e}\in\mathfrak{C}$ such that an automorphism of $\mathfrak{C}$ fixes $E$ setwise if and only if it fixes $\overline{e}$; this is unique up to interdefinability. Now, $T$ eliminates imaginaries if every $\mathfrak{C}$-definable set of $n$-tuples has a canonical parameter, and $T$ eliminates finite imaginaries if every finite set of $n$-tuples has a canonical parameter.

Example: If $T$ is the theory of infinite sets, then $T$ does not eliminate finite imaginaries, since for any $c_1\neq c_2\in\mathfrak{C}$ the set $\{c_1,c_2\}$ has no canonical parameter.

Example: If $T$ is totally ordered, then $T$ eliminates finite imaginaries, since an automorphism fixes a finite collection of $n$-tuples setwise if and only if it fixes it pointwise.

Now, I'm curious about how much elimination of imaginaries can fail for tuples as opposed to elements. Say that $T$ eliminates $1$-imaginaries if every $\mathfrak{C}$-definable subset of $\mathfrak{C}$ has a canonical parameter, and say that $T$ eliminates finite $1$-imaginaries if every finite subset of $\mathfrak{C}$ has a canonical parameter.

Question: Is there a theory that eliminates $1$-imaginaries but does not eliminate imaginaries? Is there a theory that eliminates finite $1$-imaginaries but does not eliminate finite imaginaries?

I suspect the answer is yes, but I'm struggling to come up with any examples.

$\endgroup$
8
  • 2
    $\begingroup$ I think that if you take a pure infintie set expanded by canonical parameters for finite subsets, then you trivially have elimination of finite 1-imaginaries, but a set of the form $\{(a,b),(c,d)\}$ has no canonical paremeter, since its real dcl will consist of $\{a,c\}$ and $\{b,d\}$, and the tuple $\{a,c\}\{b,d\}$ is not interdefinable with $\{(a,b),(c,d)\}$ (the latter has the same type as $\{(c,b),(a,d)\}$. More or less the same example should work for your other question. $\endgroup$
    – tomasz
    Commented Apr 15, 2021 at 0:19
  • 2
    $\begingroup$ Actually, exactly the same example, as in this case, the only definable sets in the home sort are finite or cofinite, so eliminating finite 1-imaginaries is the same as eliminating all imaginaries. It all gets a bit more tricky if you insist on a one-sorted structure, but just throwing in some sort of random encoding should do the trick. $\endgroup$
    – tomasz
    Commented Apr 15, 2021 at 0:24
  • 1
    $\begingroup$ @tomasz I'm sure I'm missing something, but – denoting by $e$ the canonical parameter for $\{a,b\}$ and $f$ the canonical parameter for $\{c,d\}$ – don't we then have the problem that the finite subset $\{e,f\}$ has no canonical parameter? $\endgroup$ Commented Apr 15, 2021 at 0:33
  • 1
    $\begingroup$ (indeed, note that, for any $a'\in\{b,c,d\}$, there is an automorphism fixing $\{e,f\}$ setwise and swapping $a$ and $a'$. so the only possible candidate for a canonical parameter of $\{e,f\}$ is the canonical parameter of $\{a,b,c,d\}$. but this element is fixed by the automorphism swapping $a$ and $c$ and fixing all other real elements, and this automorphism does fix $\{e,f\}$ setwise.) $\endgroup$ Commented Apr 15, 2021 at 0:57
  • 2
    $\begingroup$ Yes, that is what I am arguing (since then you can make the whole thing one-sorted). $\endgroup$
    – tomasz
    Commented Apr 15, 2021 at 19:51

1 Answer 1

1
$\begingroup$

Elimination of 1-imaginaries does not imply elimintaion of imaginaries in a one sorted theory. For example, every o-minimal structure eliminates 1-imaginaries, since definable subsets of the home sort are finite unions of intervals which are coded by the endpoints. However, an o-minimal structure need not eliminiate imaginaries, see e.g here - the example is $(\mathbf{Q}, <)$ with a 4-ary predicate for $x + y = w + z$

$\endgroup$
1
  • $\begingroup$ thank you so much, this resolves the question completely!! this is a great class of examples for theories eliminating $1$-imaginaries $\endgroup$ Commented Jul 5, 2021 at 23:48

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .