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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
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.
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.
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 | Main Page
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