Proving that the theory is $\aleph_{0}-$categoric, giving and example and going to a particular case Consider the language $\mathcal{L} = \{<,U\}$, where $<$ is a binary relation $U$ is a unary relation. Let $T$ a $\mathcal{L}-$theory given by:
$$\begin{array}{l r}
    \forall_{x}[\neg(x < x)] & \forall_{x,y}\{\neg [(x < y) \land (y < x)]\}\\
    \forall_{x,y,z}\{[(x < y)\land(y < z)] \longrightarrow (x < z)\} & \forall_{x,y}[(x < y)\lor(x = y)\lor(y < x)]\\
    \exists_{w}\{U(w) \land \forall_{x}[(w < x) \lor (w = x)]\} & \exists_{w}\{\neg U(w) \land \forall_{x}[(x < w) \lor (x = w)]\}\\ 
\end{array}$$
$$\forall_{x,y}\{ (x < y) \rightarrow \exists_{z_{1},z_{2}}[U(z_{1}) \land \neg U(z_{2}) \land (x < z_{1} < y) \land (x < z_{2} < y)]\}$$
1. Describe a model of $T$ which its domain is a subset of $R$ with the usual order.
2. Show that $T$ is $\aleph_{0}-$categoric (Use back and forth argument).
3. Use the last part to show that $\mathcal{M}_{1} \equiv \mathcal{M}_{2}$ but, $\mathcal{M}_{1} \not\cong \mathcal{M}_{2}$, where:
$$\begin{array}{l r}
        \mathcal{M}_{1} = \left([-\sqrt{2},1];<;U^{\mathcal{M}_{1}} = \mathbb{I} \cap [-\sqrt{2},1]\right) & \mathcal{M}_{2} = \left([0,\sqrt{2}];<;U^{\mathcal{M}_{2}} = \mathbb{Q} \cap [0,\sqrt{2}]\right)  
    \end{array}$$
I do not understand very well the staments of the theory I would like that we build a group solution in order to work on it.
 A: *

*The theory can be described succinctly as the theory of a dense linear order with endpoints with a dense subset $U$ whose complement is also dense, and such that the least element is in $U$ and the greatest isn't. They give two examples of models in part three, where the domain is a closed interval reals and $U$ represents either the rationals or irrationals.

*The back and forth argument is almost identical to the standard argument for dense linear orders. The only difference is that if an element of one model satisfies $U$, the element from the other model that you pair it with must also satisfy $U$ (and similarly for elements not satisfying $U$). There is no issue making this work since both $U$ and its complement are dense.

*The categoricity result (and the fact that there are clearly no finite models) implies completeness, so all models are elementarily equivalent. The fact that they are not isomorphic follows from the fact that the rationals are countable but the irrationals aren't.

