indiscernible sequence?
Morley sequence?
Ramsey theorem?
Erdos-Rado theorem?
Ehrenfeucht-Fraïssé games (back-and-forth games)
Hrushovski construction?
generic predicate?
An imaginary element of a (finitary) first-order theory is an equivalence class $X/E$ of a (finitary first-order) definable set $X$ quotiented by a (finitary first-order) definable equivalence relation $E$. A hyperimaginary is an equivalence class $X/E$ of a type-definable $X$ quotiented by a type-definable equivalence relation $E$.
The prototypical example of a hyperimaginary is an equivalence class of the quotient of the hyperreal numbers by the equivalence relation “$x$ is infinitesimally close to $y$”, which is definable by the countable conjunction $\left( \left| x - y \right| < \frac{1}{n} \right)_{n \in \mathbb{N}}$.
Let $X/E$ be a hyperimaginary. $T$ is said to eliminate the hyperimaginary $X/E$ (c.f. elimination of imaginaries) if $X/E$ is interdefinable with a sequence of imaginaries, i.e. if every type-definable equivalence relation is, in fact, a conjunction of definable equivalence relations.
Stable theories? eliminate
hyperimaginaries. This is related to a result of Hrushovski’s that in a stable theory, (pro-definable) groupoids are pro-(definable groupoids).
In the same way that the automorphism group $\operatorname{Aut}(M)$ of a model acts on imaginaries, $\operatorname{Aut}(M)$ acts on hyperimaginaries.
Any hyperimaginary $X/E$ in the sense of the previous paragraph can be extended to a quotient of the entire model $M^{\alpha}$ (in the (possibly infinitary) sort $\alpha$ of $E$) by a type-definable equivalence relation $E'$ obtained by preserving $E$ on $X$ and extending it to be the discrete equivalence relation on $M^{\alpha} \backslash X$. In symbols, $E' \overset{\operatorname{df}}{=} (E(x,y) \wedge (x,y) \in X \times X) \vee (x = y)$. (This is a special case of the fact that finite disjunctions of type-definable sets are again type-definable. It is not true that arbitrary disjunctions of type-definable sets are again type-definable, since $\operatorname{Aut}(\mathbb{M})$-invariant subsets of $\mathbb{M}$ the monster model have this form.)
The process of taking points of hyperimaginaries in models ($M \mapsto M(X/E)$) does not commute with ultraproducts. For example, if $E$ is the equivalence relation in the reals of being infinitesimally close together, it’s easy to see that $\mathbb{R}^{\mathcal{U}}/E \not \simeq \left(\mathbb{R}/E \right)^{\mathcal{U}}$, since $\mathbb{R}/E \simeq \mathbb{R}$ (after all, the real numbers are Hausdorff) so that the right hand side is the nonstandard reals $^* \mathbb{R}$ again (so every number has a cloud of infinitesimals around it) while the left hand side has no points which are infinitesimally close together.
This is actually an excellent example of why we shouldn’t expect in general the process of taking points of hyperimaginaries in a model to commute with ultraproducts: on one hand, if $E$ is a type-definable equivalence relation on $X$, and $\mathcal{U}$ an ultrafilter, then computing $\left(\prod_{i \in I} M_i/\mathcal{U} \right)(X/E)$ yields $E$-equivalence classes of germs $[m_i]_{\mathcal{U}}$. That is, for every condition $\varphi(x_1, x_2)$ in $E$ that needs to be checked to conclude that two tuples are $E$-equivalent, every pair of $E$-equivalent germs $[m_i]_{\mathcal{U}}$ and $[n_i]_{\mathcal{U}}$ agree on $\varphi$ on some subset $P \subseteq I$ for $P$ in $\mathcal{U}$.But these $P$s depend on $\varphi$. They aren’t uniform, and there’s possibly no set $P'$ in the ultrafilter (the intersection of all the $P$‘s might be empty) on which one could decide if $[m_i]$ and $[n_i]$ are $E$-equivalent for every condition $\varphi \in E$.
On the other hand, one could try computing $\left( \prod_{i \in I} M_i(X/E) \right) /\mathcal{U}$. One ends up instead with germs of hyperimaginaries, as in $[[m_i]_E]_{\mathcal{U}}$, but this is a tighter equivalence relation: $([x_i]_E)_i$ and $([y_i]_E)_i)$ are $\mathcal{U}$-equivalent if and only if there is a single $P \in \mathcal{U}$, uniform for every $i \in P$ and for every $\varphi \in E$ so that $M_i \models \varphi(x_i, y_i)$. This means that the commutation of these two procedures should fail precisely when the $P$‘s on the left hand side intersect to something outside of the ultrafilter.
Indeed, in the example we first mentioned, consider the sequences $(1, 2, 3, \dots)$ and $(1 + 1/2, 2 + 1/3, 3 + 1/4, \dots)$. If we take them mod-$\mathcal{U}$, then mod-$E$, then they are equivalent as hyperimaginaries in $^*\mathbb{R}$, since for each condition “x is less than $1/n$ away from y”, these two sequences agree cofinitely many times (and every ultrafilter contains the cofinite filter). But these cofinite sets of indices are getting smaller and smaller and intersect to $\emptyset$. And if we take them mod-$E$, then mod-$\mathcal{U}$, they remain distinct.
Frank Wagner, Simple Theories (2000)
Byunghan Kim, Simplicity Theory (2014)
Enrique Casanovas, Simple theories and hyperimaginaries (2011)
Last revised on June 20, 2017 at 15:34:54. See the history of this page for a list of all contributions to it.