Chris Nolan #15: I was wondering that too, thanks for asking. Testing my understanding of the answers then:
For any pair of objects \\(x, y\\) in the set of objects, exactly one of the following must hold:
1. \\(x\\) is comparable to \\(y\\), or
2. \\(x\\) is not comparable to \\(y\\).
Case 2 fails trichotomy per John's note @ #18. Stated alternately: if all \\((x, y)\\) are comparable, then the set of objects has a total order provided reflexivity and transitivity hold.
That then raises a question about antisymmetry. But I think I read somewhere that the concept of equality is non-trivial in category theory. So I'll leave that one for now!
It does, however, raise another minor question for me: why use \\(\le\\) to mean "is less than" instead of \\(<\\)? I can see that reflexivity gets troublesome if I use a conventional interpretation of "less than": \\(x\\) can't be less than \\(x\\) in any meaningful way. \\(x\\) "is less than or equal to" \\(x\\) does have intuitive meaning. But that gets us back into equality again!