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Use the countable-language hypothesis of the omitting types theorem, which is the usual convention for this prime-model result. Let $\mathcal{M}$ be a prime model of $T$ and $a\in M$.

First, $T$ is complete up to logical consequence: every model $\mathcal{N}\models T$ receives an elementary embedding of $\mathcal{M}$, so it satisfies exactly the same sentences as $\mathcal{M}$. We may therefore replace $T$ by $\mathrm{Th}(\mathcal{M})$ without changing its model class or the prime model property.

Suppose $p={\mathrm{tp}}^{\mathcal{M}}(a/\varnothing )$ were not isolated. The omitting types theorem, applied to this single complete type, would produce $\mathcal{N}\models T$ omitting $p$. By primality choose an elementary embedding $j:\mathcal{M}\to \mathcal{N}$. For every $\phi (x)\in p$,Then $j(a)$ realizes $p$, contradicting omission. ConsequentlyThe same proof applies to every finite tuple, giving an atomic model. Countability of the language is a hypothesis of this argument, not a restriction on parameter sets in the separate parts (a) and (d). The PDF does not state that standing hypothesis explicitly; it must be included when invoking this version of the theorem.

A prime model of a complete theory $T$ is a model $M\models T$ for which there is an elementary embedding $M\to N$ into every model $N\models T$.

Enumerate the countable atomic model as $M=\{a_0,a_1,\dots \}$ and fix any $N\models T$. We recursively construct finite partial elementary maps $f_n$ from the first $n$ elements of $M$ into $N$.

Suppose $f_n(\bar{a})=\bar{b}$. Since $M$ is atomic, choose a formula $\theta (\bar{x},y)$ isolating $\mathrm{tp}(\bar{a},a_n)$. Its existential consequence $\exists y\theta (\bar{x},y)$ belongs to $\mathrm{tp}(\bar{a})$, so partial elementarity givesChoose a witness $b_n$. Because $\theta$ isolates the complete joint type, extending $f_n$ by $a_n\mapsto b_n$ remains partial elementary. The union of the recursive maps is an elementary embedding $M\to N$. Thus every countable atomic model is a prime model, proving the countable atomic model is prime result.

Assume $M$ is a prime model. By the downward Lowenheim-Skolem theorem, $T$ has a countable model, and the elementary embedding of $M$ into it makes $M$ countable. If a tuple $\bar{a}\in M$ had a nonisolated type, the omitting types theorem would give a countable model of $T$ omitting that type. An elementary embedding of $M$ into this model would realize it, a contradiction. Thus $M$ is atomic.

Conversely, let $M$ be countable and atomic, enumerate it as $(a_i{)}_{i<\omega }$, and let $N\models T$. Construct an elementary embedding recursively. Suppose $a_0,\dots ,a_{n-1}$ have been mapped to $\bar{b}$. Let $\psi (\bar{x})$ isolate the type of $(a_0,\dots ,a_{n-1})$ and let $\theta (\bar{x},y)$ isolate the type of $(a_0,\dots ,a_n)$. Since the latter extends the former and is realized in $M$, completeness givesThe tuple $\bar{b}$ realizes $\psi$, so a suitable image of $a_n$ exists in $N$. The union of the finite partial elementary maps is an elementary embedding $M\to N$. Hence $M$ is prime.

A prime model of a complete theory $T$ is a first-order model admitting an elementary embedding of first-order structures into every first-order model of $T$. Let $\mathcal{A}\models T$ be canonical and $\mathcal{B}\models T$. For each closed term $t$, defineEvery element of $A$ has a name, so this defines the desired map on all of $A$. If two first-order terms agree in $\mathcal{A}$, their logical equality first-order sentence holds in one first-order model of the complete theory and therefore in every first-order model of it; they agree in $\mathcal{B}$. This proves well-definedness. If they are distinct in $\mathcal{A}$, their inequality holds throughout the complete theory as well, proving injectivity.

For every logical formula $\varphi (\vec{x})$ and named tuple ${\vec{t}}^{\mathcal{A}}$, the substituted first-order sentence $\varphi (\vec{t})$ has the same truth value in the two first-order models by completeness of $T$. ConsequentlyThus $e$ is elementary. Since $\mathcal{B}$ was arbitrary, every such canonical first-order structure is prime. This formal proof also handles closed-term generation when function symbols are present; under the constant-name convention one simply uses constants as the first-order terms.

A model is prime if it has an elementary embedding into every model of $T$. A model is atomic if every finite tuple realizes an isolated complete type over the empty set.

If $A$ is prime, the downward Löwenheim-Skolem theorem gives a countable model $M$ of $T$, and an elementary embedding $A\to M$ makes $A$ countable. If a tuple of $A$ realized a nonisolated type, the omitting types theorem would give a model of $T$ omitting that type. The prime embedding into it would preserve the tuple's full type and realize the omitted type, a contradiction. Thus a prime model is countable and atomic.

Conversely enumerate a countable atomic model as $a_0,a_1,\dots$, and take any $B\models T$. Build finite partial elementary maps into $B$. Suppose the current tuple $\bar{a}$ maps to $\bar{b}$. Choose a formula $\theta (\bar{x},y)$ isolating the type of $(\bar{a},a_n)$. The assertion $\exists y\theta (\bar{a},y)$ holds in $A$, so its counterpart holds at $\bar{b}$ in $B$ by partial elementarity. Choose a witness $b_n$. Since $\theta$ isolates the entire extended tuple type, the extended map is elementary on its finite domain; its equality formulas ensure distinct Elements remain distinct. The union is a total elementary embedding $A\to B$. The empty initial tuple works because $T$ is complete. Hence

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.

There are $2^{{\aleph }_0}$ distinct irrational cuts of the named rational order. Each defines a consistent complete one-type as above, already over the empty additional parameter set, since the rational constants are part of the language. Different cuts cannot be realized by the same element: a named rational between two cuts gives contradictory required inequalities.

An ${\aleph }_0$-saturated model must realize every one of these types, so it has at least $2^{{\aleph }_0}$ Elements. Consequently there is no countable ${\aleph }_0$-saturated model of $T$. This does not contradict existence of a saturated model from the preceding construction, whose size was unrestricted, or existence of the countable prime model, whose requirement is different. No continuum-hypothesis assumption is involved.

A prime model admits an elementary embedding into every model of its complete theory. In the countable setting a prime model is atomic. If a tuple in the prime model had a nonisolated complete type, the omitting types theorem would give a model omitting that type; an elementary embedding of the prime model into it would realize the omitted type, a contradiction.

Combining the four parts, all positive implications are exactly those generated byThe structure $(\mathbb{N},<,0)$ refutes (ii)$\Rightarrow$(i). The structure $(\mathbb{Q},<,0)$ is a countable atomic and hence prime model of its complete theory, while failing both (i) and (ii); it refutes every remaining reverse arrow, including (iv)$\Rightarrow$(i) and (iv)$\Rightarrow$(ii). The counterexamples therefore cover all ordered pairs of distinct properties not forced by the boxed chain.