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!

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!