An ordinal (in von Neumann sense) can be thought as the "canonical" representative of the order type (i.e. of the class of all order-isomorphic sets) of a well-ordered set. This is very convenient, because it makes an ordinal to be a set (which allows it to be an element of a set in ZFC).

Is there a similar approach which allows to assign the "canonical" representative to the order type of every totally ordered set, or even every partially ordered set?

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    $\begingroup$ Actually, there's a generic technique that allows us to reify equivalence classes of equivalence relations on a proper class: we use Scott's trick, i.e. we represent an equivalence class by the set of all its members of lowest rank. This is used to define e.g. cardinals in a choice-free context. Unfortunately we do not get a canonical representative of the equivalence class per se. $\endgroup$ – Zhen Lin Nov 30 '11 at 23:31
  • $\begingroup$ @Zhen: Using Scott's trick only allows you to make equivalence classes into sets; however choosing from class-many sets may require more. This is exactly the question here. $\endgroup$ – Asaf Karagila Nov 30 '11 at 23:33
  • $\begingroup$ @Asaf: But do we need to? Perhaps, for some application, it is sufficient to have a set ‘name’ for each equivalence class, rather than needing an actual member for each. $\endgroup$ – Zhen Lin Nov 30 '11 at 23:35
  • $\begingroup$ @nikov: Perhaps you can be a bit more clear of what you mean by "canonical." To me, a basic interpretation is the following: there is a (absolute) formula which (in certain models of ZFC) assigns to each poset an OrderType, so posets are assigned the same OrderType iff they are order-equivalent. The answer for this notion of canonicality seems to be "yes" in $V=L$, but will likely be "no" in general. $\endgroup$ – user642796 Dec 1 '11 at 9:04

While I do not have an actual answer, the following observation may be useful:

Assuming Global Choice the answer is trivially yes. You can preform Scott's trick on order type equivalence classes, and choose the representative accordingly.

Assuming the axiom of choice, it is always possible to assign a set collection of order types. This set can be made as large as you want, however we are not guaranteed to have a class of representatives.

The ordinals are definable in ZF (without the axiom of choice at all) due to von Neumann's axioms of foundation and replacement schema (both attributed to Frankael and Skolem). This allows the unique definition of well-order types using transitive $\in$ ordered sets.

On the other hand, there is no way to define general partial orders in ZF or even ZFC in a way which would result in a definable representative to each class.

Much like Jech's proof that without the axiom of choice it is possible to have a model of ZF which has a set of cardinalities (defined as per Scott's trick) which has no definable choice of representatives, I would believe that such proof would be possible to achieve via forcing in ZFC.

It is immediate that such claim is likely to be independent of ZFC, since V=L implies Global Choice, which in turn implies that there is a canonical representative. However if it is possible to construct a model in which there is a class of order types without definable representatives (and I very much believe that it is possible) then it is independent of ZFC.

  • $\begingroup$ You can slightly simplify you presentation of the situation under Global Choice by just taking the minimal (according to a well-ordering of the universe) representative of every order-equivalence class. Of course, these methods may not lead to canonical representatives, since they depend heavily on the well-ordering used, and unless this well-ordering is canonical your choice of representatives will likely also fail to be canonical. $\endgroup$ – user642796 Dec 1 '11 at 8:37
  • $\begingroup$ @Arthur: Depending on the formulation of global choice, you may augment your language with a predicate which is the choice function. In that case the choice is indeed canonical. $\endgroup$ – Asaf Karagila Dec 1 '11 at 8:44

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