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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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    @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.2011-12-01

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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.

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    @Arthur: Depe$n$di$n$g 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.2011-12-01