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Take a countable structure with countably many independent unary predicates, every finite predicate pattern realized infinitely often, and a constant naming every element. Its complete theory has an atomic model: tuples of named elements have types isolated by equations to their names. But every infinite binary predicate pattern, together with inequalities to every named element, is consistent. There are uncountably many such one-types, so no countable omega-saturated model exists. Thus existence of a countable atomic model does not imply existence of a countable saturated one.

Add constants $c_0<c_1<\cdots$ to the theory of dense linear orders without endpoints. The three countable models have respectively a least upper bound for the named sequence, no upper bound, or upper bounds with no least one. The unbounded model is an atomic model; the bounded model with no least upper bound is an omega-saturated model. The remaining model omits the type $c_i<x<d$ over its least-upper-bound parameter $d$.

An atomic model is a model in which every finite tuple's complete type over the empty set is isolated by a formula. An omega-saturated model realizes every complete type in finitely many variables over every finite parameter set; equivalently it realizes all one-variable types over finite parameter sets, then successive realizations give the finite-tuple version.

Suppose $M$ is countable and omega-saturated. Every type in $S_n(T)$ is realized by a tuple in $M$, so $S_n(T)$ is countable. This is a compact Hausdorff Stone space with a clopen basis. Its isolated points are dense. Indeed, if a nonempty clopen subset had no isolated point, split it into two disjoint nonempty clopen subsets and recursively split every resulting subset. Such splits exist because each subset has at least two points and the space is zero-dimensional. Compactness supplies a point in each nested sequence along an infinite binary branch, while disjoint siblings make different branches yield different points. This would give uncountably many points, contradicting countability.

For each $n$, let ${\Gamma }_n(\mathbf{x})$ consist of the negations of all formulas that isolate complete $n$-types. Density of isolated types means that every consistent formula has a consistent refinement to an isolating formula $\delta$; this refinement excludes the member $\neg \delta$ of ${\Gamma }_n$. Thus $T$ locally omits every ${\Gamma }_n$. The omitting types theorem gives a countable model $A$ omitting them all. Every tuple in $A$ therefore satisfies some isolating formula, so $A$ is atomic. If some ${\Gamma }_n$ is itself inconsistent, its omission is automatic and does not obstruct this argument.

The converse about existence is false. Here is a countable atomic model without a countable saturated model. Take a countable set $M$ with unary predicates $P_i$, $i\in \mathbb{N}$, such that every finite Boolean pattern of the predicates occurs infinitely often. For example, use finite binary words paired with a natural number, with $P_i$ recording the $i$th bit when it exists and false otherwise. Add a constant naming each element, and put $T=\mathrm{Th}(M)$ in this countable language. Every tuple of $M$ has its type isolated by equations to its named coordinates, so $M$ is a countable atomic model.

For each infinite binary sequence $\eta$, the familyis finitely satisfiable in $M$: realize its finitely many predicate requirements while avoiding finitely many named elements. By compactness theorem, it extends to a complete one-type. Different sequences give different types, so $S_1(T)$ is uncountable. Any omega-saturated model must realize all these empty-parameter types, which a countable model cannot do. This disproves the existence converse, rather than merely exhibiting an atomic model that is itself nonsaturated.

The Ehrenfeucht theory of an increasing named sequence is complete and has quantifier elimination: the usual one-point cut argument for dense linear orders without endpoints works with the finitely many constants occurring in a formula. Complete types are consequently determined by order relations to the named constants and to any additional parameters.

In the model where the sequence is unbounded above, every finite tuple lies below some $c_N$. Its order relations to $c_0,\dots ,c_N$, together with its internal equality and order pattern, determine its relations to every later constant as well. This finite condition isolates its complete type. Thus model (ii) is an atomic model. It is not omega-saturated, because the finitely satisfiable type $\{c_i<x:i\in \mathbb{N}\}$ has no realization.

In either bounded model, a point above every $c_i$ realizes the typeThis type is not isolated. Any formula belonging to it uses only finitely many constants after quantifier elimination, and its order condition can also be realized below a sufficiently late $c_j$. That realization fails one of the later inequalities in $p_{\infty }$. Hence neither bounded model is atomic.

In model (i), let $d$ be the least upper bound. The type $\{c_i<x:i\in \mathbb{N}\}\cup \{x<d\}$ over the single parameter $d$ is finitely satisfiable by density but unrealized, so this model is not omega-saturated.

In model (iii), the upper tail $U=\{x:c_i<x\text{ for all }i\}$ has no least element. Every consistent one-type over finitely many parameters is realizable. Indeed, it is either an equality type, a cut below or between some of the named constants, or a cut in $U$. The first two cases are realized by equality or by density in the relevant interval. In $U$, finitely many parameters divide the tail into finitely many intervals, each relevant nonempty interval having points by density, absence of a least element in $U$, and absence of a last element in the order. This realizes the third case too. There is no other cut among the increasing sequence: a proper initial segment of its indices is finite. Successive one-variable realizations give all finite-tuple types. ThereforeA least upper bound is thus precisely the obstruction to filling the additional finite-parameter cut in model (i).

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.

An atomic model of $T$ is a model $M\models T$ such that the complete type $\mathrm{tp}(\bar{a}/\varnothing )$ of every finite tuple $\bar{a}\in M$ is an isolated type.

An atomic model is a model in which the type over the empty set of every finite tuple is isolated. Let $\bar{a}\mapsto \bar{b}$ be partial elementary and take $c\in M$. Choose a formula $\theta (\bar{x},y)$ isolating $\mathrm{tp}(\bar{a},c)$. Since $M\models \exists y\theta (\bar{a},y)$, elementarity gives $N\models \exists y\theta (\bar{b},y)$. Any witness $d$ realizes the isolated joint type, so $\bar{a},c\mapsto \bar{b},d$ is partial elementary. This is the one-point extension between atomic models.

Alternately applying this extension property in the two directions to enumerations of countable atomic models constructs a back-and-forth isomorphism.

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.