# The theory of dense linear orders without end-points is not $2^\omega$-categorical

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 $$\mathbb{R}$$?

• Hint: Which of those two orders are Dedekind complete? – Alessandro Codenotti Dec 30 '18 at 23:01

## 2 Answers

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.

(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 $$\cal M\equiv M'\equiv N'\equiv N$$