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It seems best to prove this by counter example. Both $\mathbb{R}$ and $I := \mathbb{R} \backslash \mathbb{Q}$ under the usual order $<$ are models of the theory of dense linear orders without end-points and I think they are not isomorphic (if I understand the definition correctly, that means there is no order preserving bijection between $\mathbb{R}$ and $I$). I couldn't manage to prove this.

My thoughts so far: suppose there exists such an isomorphism $\beta: \mathbb{R} \to I$, then no irrational elements can be mapped to $\beta(\mathbb{Q})$, which messes up the order maybe because $\mathbb{Q}$ is dense in $\mathbb{R}$?

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    $\begingroup$ Hint: Which of those two orders are Dedekind complete? $\endgroup$ – Alessandro Codenotti Dec 30 '18 at 23:01
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In the irrationals, take a sequence decreasing to $0$, then look at where those points map in the reals. Those real points will be a bounded decreasing sequence, therefore they will have a real limit $x$. Where does $x$ map in the irrationals ... call it $y$. Then $y<0$, but then there are irrationals between $y$ and $0$ and that should give you a contradiction since $x$ was the limit of the decreasing real sequence.

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(I decided to put it as an answer in the end)

A different $I$ that I believe is easier to work with is $I=\Bbb Q ∪[0,1]$, assume that there is isomorphism $φ:I\to \Bbb R$, then $φ(1.5)<φ(2)$, because it is order preserving, but $[1.5,2]$ in $I$ is countable but $[φ(1.5),φ(2)]$ in $\Bbb R$ is uncountable, which leads to contradiction.


Also, although it is not $2^\omega$-categorical, all models of cardinality $\kappa$ that satisfy those properties are elementary equivalent: let $\cal M,N$ be 2 models of that theory of size $\kappa$, then, by (downward) Löwenheim–Skolem theorem, there exists $\cal M',N'$ elementary substructure of $\cal M,N$ respectively of size $\omega$. Because the theory is $\omega$-categorical, $\cal M',N'$ are isomorphic, which implies that they are elementary equivalent, so $\cal M\equiv M'\equiv N'\equiv N$

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