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I can't really tell if ($Q$, $\leq$)$\cong$($Q \times Q$, $\leq _e$), where $\leq_e$ denotes the left lexicographic order. Neither have a last/first element, both are dense and have the same cardinality, so intuitively they are isomorphic. If so, how can I construct such an isomorphism? If not, why are they not isomorphic?

Generally speaking, what are the "criterions" for the existence of an isomorphism between two totally-ordered sets? As far as I have seen, if two sets are isomorphic then they preserve density (If one is dense then so is the other), a first/last element is mapped to a first/last element, open intervals are mapped to open intervals, and if each open interval has an inf/sup then so does its image. What other "criterions" are there?

Would appreciate any help.

Pedro Sánchez Terraf
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Dylan132
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  • Assuming by isomorphism you mean order-isomorphism, the only criteria is that a bijective mapping exists that preserves order. So you have to show $|A| = |B|$ and $x \leq y \leftrightarrow f(x) \leq f(y)$ – Dan Simon Jan 26 '16 at 15:25
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    The theory of dense linear orders without endpoints is $\aleph_0$-categoretical. So any two countable dense linear orders without endpoints are isomorphic. Probably, related. – Random Jack Jan 26 '16 at 15:31
  • [lease look here:http://math.stackexchange.com/questions/37151/showing-any-countable-dense-linear-ordering-is-isomorphic-to-a-subset-of-mat –  Jan 26 '16 at 15:31
  • @DanSimon That's the formal definition of an order-isomorphism, but I'm asking for virtues that are preserved through order-isomorphism (As to help me negate the existence of isomorphisms in certain cases). – Dylan132 Jan 26 '16 at 15:32
  • The elephant in the room: two isomorphic ordered sets have the same cardinality :) – BrianO Jan 26 '16 at 15:35

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