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Take the standard successor structure and its disjoint union with one two-sided integer successor chain:Both interpret the named zero in the natural-number component. The classification of models of the successor theory describes every model as this distinguished one-sided chain together with an arbitrary collection of two-sided chains. Its axioms are injectivity of $s$, zero outside its range, a predecessor for every nonzero element, and absence of finite cycles. These axioms are complete; equivalently the theory has the successor quantifier-elimination property explained in the concept proposal.

Consequently both displayed structures are models of the standard complete theory. They are not isomorphic: any isomorphism must fix zero and hence every element $s^n(0)$. The first structure has no other elements; the second has the entire extra integer chain. Being in that additional component is an isomorphism invariant even though being outside all standard iterates of zero is not expressible by one first-order formula.

Consider the partial typeEach finite fragment is realized by a sufficiently large standard natural number, so it is consistent. It is omitted in the standard successor structure, since each of its elements equals one of the named iterates.

There is exactly one complete type extending this partial type. By classification of models of the successor theory and its quantifier-elimination property, a parameter-free one-variable formula is equivalent to a Boolean combination of finitely many equalities $x=s^n(0)$ and sentences whose truth is fixed by the theory. For a point on an extra integer chain every displayed equality is false, so all such points have the same complete type. Thusis the required omitted complete type. More formally take all formulas true at such a point; $p_{\infty }$ axiomatizes that type modulo the theory, and its deductive closure is complete.

The distinguished chain generated by zero is countable, and every other component in the classification of models of the successor theory is a countable copy of the integer successor chain. A model of cardinality ${\aleph }_1$ therefore has exactly ${\aleph }_1$ additional components: a finite or countable collection would be countable, while the union of $\kappa$ countable components has size $\max (\kappa ,{\aleph }_0)$.

Given two such models, match their zero chains in the forced way and choose a bijection between their additional components. On each component choose an origin and match corresponding integer iterates; this commutes with successor. The union of these maps is an isomorphism of structures. Existence is given by a disjoint union with ${\aleph }_1$ integer chains. HenceThe same argument works at every uncountable cardinal. Countable models can still differ by the number of extra components, as part (ii) shows.

Suppose $T$ is an omega-categorical theory. By the Ryll-Nardzewski theorem, every finite-arity complete type over the empty set is isolated. Let $p(x,a)$ be a complete type over a finite tuple in the unique countable model $M$, and take an elementary extension realizing it at $b$. The joint type of $(b,a)$ is isolated by a formula $\theta (x,y)$.

The complete type of $a$ contains $\exists x\theta (x,y)$, since this formula is true of $a$ in the extension. Hence it is true of $a$ in $M$, and some $b^{\prime }\in M$ satisfies $\theta (b^{\prime },a)$. Isolation of the joint type implies that $b^{\prime }$ realizes $p$. Thus the countable model is an omega-saturated model.

The converse fails. The complete successor theory from question 4 has a countable modelwith countably infinitely many two-sided successor components. It is omega-saturated. Over finitely many parameters, a complete one-variable type either fixes a finite successor/predecessor translate of a parameter or zero, or asks for a point in no component represented by those parameters and zero. The first kind is realized in the named component, and the second in an unused integer component. The quantifier-elimination description of classification of models of the successor theory shows there are no other cases.

Nevertheless this theory is not omega-categorical: $M$ and its standard model $\mathcal{N}$ are countable and not isomorphic. Thus existence of a countable saturated model is strictly weaker than countable categoricity.

The complete theory of the standard successor structure is axiomatized by injectivity of $s$, $0$ outside its range, a predecessor for every nonzero element, and the sentences $s^n(x)\ne x$ for each positive integer $n$. Its models have a single distinguished one-sided chain generated by zero and any collection of two-sided integer chains. The classification of models of the successor theory proves completeness and explains why the theory is categorical at every uncountable cardinal but not at the countable cardinal.