# Are $\mathbb{R}$+$\mathbb{R}$ and $\mathbb{R}$ isomorphic models of DLO?

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?

• I'm guessing you're talking about the cartesian product of $\mathbb{R}$. – offret Mar 2 '19 at 10:57
• @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 – Nicola Carissimi Mar 2 '19 at 11:04
• $\mathbb R$, $\mathbb R\setminus\{0\}$, $\mathbb R\setminus\{0,1\}$, $\mathbb R\setminus\{0,1,2\}$, $\mathbb R\setminus\mathbb N$, $\mathbb R\setminus\mathbb Z$, $\mathbb R\setminus\mathbb Q$ are all nonisomorphic linear orders which are elementarily equivalent to $\mathbb Q$. Your $\mathbb R+\mathbb R$ is isomorphic to $\mathbb R\setminus\{0\}$. – bof Mar 2 '19 at 13:55

$$\mathbb{R}$$ 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.