Total order/Talk

The unique smallest unbounded connected total-ordered set is the real numbers.

What does "connected" mean here? -- JanHidders

Maybe connected in the order topology? This should be explained, I agree.

I think the statement about the rationals being the smallest dense total-ordered set is false. For instance Q\[0,1] is dense and smaller. --AxelBoldt

I'm not an expert in this area, but what definition of "smaller" are you using? I would say in this case that A <= B iff there is an order homomorphism from A to B. In that case obviously Q\[0,1] <= Q, but also Q <= Q\[0,1] by, for example, f : Q -> Q\[0,1] such that

f(x) = x if x < 0 and

f(x) = x + 2 if x >= 0.

--JanHidders

I think A <= B is supposed to mean that A is order-isomorphic with a subset of B. However, I don't think it's very clear, and would have no objection if someone decided to remove the whole paragraph. Note that Q\[0,1] is order-isomorphic to Q.
While I'm here, I'd like to suggest that we try to use "totally ordered" rather than "total-ordered". This is far more common in my experience, and is consistent with our use of "partially ordered" (rather than "partial-ordered").
Zundark, 2001-08-17

I must be missing something. Zundark, isn't my definition equivalent with yours? My f is an order-isomorphism from Q to Q\[0,2), which is a subset of Q\[0,1]. No!?

Btw., I agree with "totally ordered" replacing "total-ordered". In the computer-science literature I know, this is also the more common term. --JanHidders

Sorry, I wasn't being very clear. I wasn't disagreeing with you, I was just responding to Axel. If by "order homomorphism" you meant x < y implies f(x) < f(y), then your definition is the same as mine.
Zundark, 2001-08-17

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I agree with everything that was said above. I guess one could justify the statement about the rationals after all. I also like totally ordered better. --AxelBoldt