05:28
@LeakyNun That's why I said labelled linear orders. This is a very common notion in logic. As I said, an ordinary linear order is simply a binary predicate on a set, and there is internally nothing to distinguish one item from another unless they have different 'neighbourhoods'. For example, in a typical introductory course one would encounter the fact that any two countable dense linear orders without endpoints are isomorphic.
However, it's totally different once you add labels, because then when we say "isomorphic" we mean that the isomorphism must respect labels too.
@LeakyNun Actually I didn't define L to just have countable labelled linear orders, because the axioms actually hold for all labelled linear orders. But in ZFC my idea of L does not exist as a set, even though it exists as a definable collection. So I should indeed restrict L to countable labelled linear orders modulo isomorphism, with + meaning concatenation.
@Mathmore: See my last comment; I need to be more careful if I want my claims to be provable in ZFC. =)
@LeakyNun You are right that there is no way in general to write linear orders down, even if they are countable. After all, every real number in [0,1) would correspond to some countable linear order with the labels taken from the binary expansion. We sure can't write all of them down. I just gave you a few as examples.
@LeakyNun Order does not necessarily distinguish items, as you can see in the typical example of countable dense linear orders without endpoints. In some cases, you can distinguish items based on their neighbourhoods. I can give you a few examples to help your intuition.
(0,0,0,...): Every item is uniquely identified by the number of items before it. So any two items are distinguishable.
(...,0,0,0,...): There is no way to distinguish any two items because there is an isomorphism that 'shifts' the whole thing.
(...,0,0,1,0,0,...): This is exactly like the previous one except a single item is labelled 1. Now each item is uniquely identified by the directed distance from it to the 1-labelled item. So any two items are distinguishable.
(0,0,0,...)+(...,1,1,1,...): Given any two items x,y, if x,y are both in the first part then they are distinguishable. If exactly one of x,y is in the first part, then they are distinguishable too, because one has finitely many before it while the other has infinitely many before it. But if x,y are both in the second part then they cannot be distinguished.
(...,0,1,0,1,0,...): Here the items are alternately labelled 0 and 1. It should be clear that any two items can be distinguished iff they have the same label.
Rationals Q with integers labelled 0 and non-integers labelled 1: Trivially items with different labels can be distinguished. If you understood the previous examples, you can see that items with the same label cannot be distinguished. (Exercise for you!)
