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An elementary embedding $j:M\to N$ between first-order structures preserves and reflects every first-order formula with parameters: for every finite tuple $\bar{a}$ from $M$,In particular it is injective, by applying the condition to equality, and its image is an elementary substructure of $N$. An elementary inclusion is the defining map of an elementary extension.

For example, the rational order is an elementary substructure of the real order in the language containing only $<$, by quantifier elimination for dense linear orders without endpoints. Elementarity is stronger than preserving just the atomic relations. For embeddings between suitable transitive models of set theory, one can additionally ask which ordinals are moved; the least moved ordinal, when there is one, is the critical point of an elementary embedding. This ordinal notion is not part of the definition for arbitrary structures.

A counterexample can even use a nonalgebraic isolated type. Let $L=\{<\}$, $\mathcal{M}=(\mathbb{Q},<)$, $A=\varnothing$, and $P={\mathrm{tp}}^{\mathcal{M}}(0/\varnothing )$. The quantifier elimination for dense linear orders without endpoints shows that all elements have the same complete type over the empty set: a one-variable quantifier-free formula has no order information except $x=x$ and $x<x$.

Thus $x=x$ isolates $P$. Every $\mathcal{N}\equiv \mathcal{M}$ is a dense linear order without endpoints, and every element of $N$ realizes $P$. Such a model cannot omit it. Even a type with infinitely many realizations need not be omittable; isolation is the relevant obstruction.

Use the partial-isomorphism criterion for quantifier elimination: it is enough that every finite partial embedding between sufficiently saturated models have the one-point extension property.

Let $f:A\to B$ be a finite partial order isomorphism between models $M,N$ of the theory of dense linear order without endpoints, and take $c\in M\setminus A$. The finite set $A$ partitions $M$ into the points of $A$, the intervals between consecutive elements, and the two outer rays. The order relations between $c$ and $A$ identify one of those intervals or rays. The image $B$ determines the corresponding interval or ray in $N$. Density supplies a point there when it is bounded, and the absence of endpoints supplies one in either outer ray. Choose such a point $d$; then $f\cup \{(c,d)\}$ remains a partial order isomorphism.

The same argument works in the reverse direction. A back-and-forth method therefore extends finite partial order isomorphisms, making them partial elementary. The criterion proves quantifier elimination for dense linear orders without endpoints.

Use the finite-partial-isomorphism criterion for quantifier elimination. Let $M,N\models \mathrm{DLO}$ and let $h:A\to B$ be an isomorphism between finite suborders. For $a\in M\setminus A$, its position relative to $A$ is one of the finitely many open intervals determined by $A$, or one of the two exterior rays. The corresponding interval or ray determined by $B$ is nonempty because the orders are dense and have no endpoints. Choose $b$ there. Then $h\cup \{(a,b)\}$ remains a partial order isomorphism.

The same argument extends in the other direction. The back-and-forth method criterion therefore applies, proving quantifier elimination for dense linear orders without endpoints. Hence DLO eliminates quantifiers.

Yes: $Q^{+}$ is prime. In any model of its complete diagram, map each rational to the interpretation of its named constant. The diagram ensures this is an embedding, and quantifier elimination for dense linear orders without endpoints makes it elementary. Alternatively every finite tuple of $Q^{+}$ has its complete type isolated by the conjunction equating its coordinates to their named rational constants. Thus it is a countable atomic model, and the earlier prime-model equivalence applies. A prime model may omit many types; primeness is not saturation.

For cut types over the named rational order, choose an irrational real cut, for example $r=\sqrt{2}$, and formEvery finite subset is realized by a rational between its finite lower and upper bounds, so compactness makes it a consistent partial type; quantifier elimination for dense linear orders without endpoints makes these comparisons determine a complete type in the language naming all rationals. No rational realizes the cut, so $Q^{+}$ omits it.

Add a new constant realizing this type and apply compactness, followed by downward Löwenheim-Skolem in the countable expanded language. This gives a countable model of $T$ realizing it. An isomorphism in $L_{\mathbb{Q}}$ fixes all named rationals and preserves realization of the cut, so it cannot be isomorphic to $Q^{+}$. Thus $T$ is not countably categorical. Omission by itself would need this additional realization-and-countable-model argument to establish noncategoricity.

An atomic model is a model in which every finite tuple realizes an isolated complete type over the empty set. For a complete theory in a countable language, every countable atomic model is prime. The underlying argument is a back-and-forth method in the one-sided form needed for an embedding: enumerate the atomic model, and extend a finite partial elementary map one element at a time using an isolating formula for the enlarged tuple. The target model has the required witness because the already matched tuple has the same complete type.

Thus property (iii) implies property (iv). It need not imply pointwise definability or term-generation. The countable dense linear order without endpoints with a single named point, for example $(\mathbb{Q},<,0)$, is atomic: quantifier elimination for dense linear orders without endpoints isolates a tuple by its finite order-and-equality pattern relative to zero. But order structure automorphisms fixing zero move every nonzero rational. Such elements are not parameter-free definable, and no closed term names them. This supplies the failed implications (iii)$\nRightarrow$(ii) and (iii)$\nRightarrow$(i).

Yes. The pure order on the reals is a dense linear order without endpoints and has quantifier elimination for dense linear orders without endpoints. Over a finite parameter set $a_1<\cdots <a_m$, a complete one-variable type either specifies equality to one of these points or specifies one of the open intervals and exterior rays they determine.

Each such interval or ray contains a real number. Hence every complete one-variable type over a finite parameter set is realized. The successive-coordinate characterization gives saturation for all finite tuples. This uses finite parameter sets; it does not say that arbitrary types over a countably infinite set of parameters are realized.

For the pure orders, yes. Both structures satisfy the theory of dense linear orders without endpoints. By quantifier elimination for dense linear orders without endpoints, each formula is equivalent to a quantifier-free order formula. Inclusion of the rationals into the reals preserves and reflects all such formulas, hence is elementary:

With a common named unit, yes. In general an elementary substructure remains elementary when elements already in the substructure are named by new constants. A formula using the new constant is just an old-language formula with that common element as a parameter. Since $1\in \mathbb{Q}$, this applies directly.

With the real constant interpreted by $\pi$, no choice inside the rationals works for inclusion. A substructure embedding must preserve every constant. The inclusion would send a rational interpretation $q$ to that same real $q$, whereas the target constant is $\pi \notin \mathbb{Q}$. Thus the expanded rational structure is not even a substructure of the expanded real one.

This last failure concerns the specified inclusion. There is an abstract elementary embedding carrying a chosen rational $q$ to $\pi$: the increasing map $r\mapsto r+\pi -q$ has that property, and its image is a dense order without endpoints inside the reals. Quantifier elimination makes the embedding elementary, but its image is a translated copy of the rationals rather than the original rational subset.