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I know that the theory of dense linear orders without endpoints is $\aleph_0$-categorical, and looking for two uncountable non isomorphic models of same cardinality, I found many examples, but nothing about what I initially thought: simply two copies of $\mathbb{R}$, one following the other (i.e. $2$ $\times$ $\mathbb{R}$ with lexicographical order) and $\mathbb{R}$ itself, are they isomorphic models of DLO?

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  • $\begingroup$ I'm guessing you're talking about the cartesian product of $\mathbb{R}$. $\endgroup$ – offret Mar 2 '19 at 10:57
  • $\begingroup$ @offret Umh, maybe i don’t catch your comment... I’m a bit sure i mean two real lines with order given by (1,r)>(0,s) for all r,s in \mathbb{R} and (i,r)>(i,s) iff r>s for i=0,1 as the first model of DLO; and just \mathbb{R} as the second model $\endgroup$ – Nicola Carissimi Mar 2 '19 at 11:04
  • $\begingroup$ $\mathbb R$, $\mathbb R\setminus\{0\}$, $\mathbb R\setminus\{0,1\}$, $\mathbb R\setminus\{0,1,2\}$, $\mathbb R\setminus\mathbb N$, $\mathbb R\setminus\mathbb Z$, $\mathbb R\setminus\mathbb Q$ are all nonisomorphic linear orders which are elementarily equivalent to $\mathbb Q$. Your $\mathbb R+\mathbb R$ is isomorphic to $\mathbb R\setminus\{0\}$. $\endgroup$ – bof Mar 2 '19 at 13:55
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$\mathbb{R}$ and $\mathbb{R} + \mathbb{R}$ are not isomorphic as linear orders. Suppose $f \colon \mathbb{R} + \mathbb{R} \to \mathbb{R}$ is order preserving. Let $A$ be the first copy of $\mathbb{R}$ in $\mathbb{R} + \mathbb{R}$ and $B$ be the second copy. Then $f(A)$ is bounded above in $\mathbb{R}$ by every element of $f(B)$, but $\sup f(A)$ can't be in $f(A)$ or $f(B)$, so $f$ isn't surjective.

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