A complete theory is a consistent first-order theory deciding every sentence in its language:Equivalently, every two first-order models of $T$ are elementarily equivalent. Consistency excludes the vacuous situation in which a first-order theory has no first-order models and entails both alternatives.

For the additive first-order structure, letIt defines $n\mathbb{Z}$. These sets are pairwise distinct: if $m>n>0$, the integer $n$ belongs to $n\mathbb{Z}$ but not to $m\mathbb{Z}$. Thus there are infinitely many inequivalent one-variable first-order formulas modulo the complete theory of the additive integers. The Ryll-Nardzewski theorem implies that first-order theory is not aleph-zero-categorical.

For the multiplicative first-order structure define divisibility by $d(y,x):=\exists z(yz=x)$. For $k\ge 1$, the first-order formulaasserts the existence of at least $2k$ distinct integer divisors. The integer $2^{k-1}$ has exactly $2k$ such divisors, namely $\pm 2^j$ for $0\le j\le k-1$, so it satisfies ${\psi }_k$ but not any ${\psi }_{\ell }$ with $\ell >k$. Again there are infinitely many inequivalent one-variable first-order formulas. The Ryll-Nardzewski theorem therefore implies the first-order theory of the multiplicative integers is not aleph-zero-categorical.

A partial type $p(\mathbf{x})$ is a set of $L$-first-order formulas in a fixed tuple $\mathbf{x}=(x_1,\dots ,x_n)$ jointly consistent with $T$. A complete type, the usual meaning of an $n$-type here, is maximal among such consistent sets: for each first-order formula $\phi (\mathbf{x})$ it contains exactly one of $\phi ,\neg \phi$.

Add new constants $\mathbf{c}$ and considerEvery finite subset is satisfiable by consistency of the type. The compactness theorem gives a first-order model of the full set. Its $L$-reduct is a first-order model of $T$, and the interpretations of $\mathbf{c}$ realize $p$. This proof applies to partial as well as complete types.

For a complete theory in a countable language, the types realized in every first-order model are exactly the isolated types. An isolated type has a consistent first-order formula $\theta \in p$ such that $T\models \forall \mathbf{x}(\theta \to \phi )$ for every $\phi \in p$. Completeness forces $T\models \exists \mathbf{x}\theta$, so every first-order model realizes $p$; the converse is the omitting types theorem. Countability is a hypothesis of that converse, not of the preceding compactness realization argument. For a partial type the corresponding criterion is an isolating consistent first-order formula, together with its existence in every first-order model; it need not itself belong to the partial type.

If $T$ is not complete, no one complete $n$-type can be realized in every first-order model of $T$: a complete type decides sentences, and two completions of $T$ disagree on some sentence. Thus the completeness and countability hypotheses must be retained when using the isolated-type characterization.

An aleph-zero-categorical theory has, up to isomorphism, exactly one first-order model of cardinality ${\aleph }_0$.

The Ryll-Nardzewski theorem says that, for a complete theory $T$ in a countable first-order language, with infinite first-order models, the following conditions are equivalent: $T$ is aleph-zero-categorical; for every positive integer $n$, its type space $S_n(T)$ of complete types over the empty set is finite; and, for every $n$, there are only finitely many $L$-formulas in $n$ free variables modulo equivalence over $T$. Here $\phi$ and $\psi$ are equivalent over $T$ when $T\models \forall \mathbf{x}(\phi \leftrightarrow \psi )$.

For the additive zero singleton, $x+x=x$ is already a quantifier-free formula, so it is also an existential formula and a universal formula with an empty quantifier block. If a nonempty block is required, add a dummy $\exists y$ or $\forall y$; first-order domains are nonempty.

For the two definable multiplicative singletons, existential formulas areIn the integers, the equation $x^2=x$ restricts $x$ to zero or one. In the first line, one never satisfies the inequality, whereas zero does with $y=1$. In the second, zero never satisfies it, whereas one does with $y=0$. The respective universal formulas are $\forall y(xy=x)$ and $\forall y(xy=y)$, as already verified.

Neither multiplicative singleton has a quantifier-free formula. In this constant-free, multiplication-only language, every term in the sole free variable $x$ is $x^k$ for a positive integer $k$. Every atomic formula $x^r=x^s$ is true at both zero and one. Every Boolean combination consequently has the same truth value at those two points, and cannot separate their singletons. This is the fact that zero and one have the same quantifier-free type in the multiplicative integers.

Thus additive zero admits all three forms; multiplicative zero and one admit existential and universal forms, but no quantifier-free form. All remaining cases were excluded from definability itself by structure automorphisms.

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.

Skolem's paradox is the apparent conflict between a countable model of a sufficiently strong set theory and that model's theorem asserting the existence of uncountable sets. Downward Löwenheim-Skolem theorem gives a countable elementary model whenever an appropriate infinite model exists; internally the same model still proves, for example, that its real-number set is uncountable.

The two uses of “countable” quantify over different collections of functions. Externally the whole model, and therefore the elements it regards as members of its real-number set, can be enumerated. Internally a set is countable only if an enumerating function exists among the sets of the model. The external enumeration need not be an element of the model.

Thus there is no contradiction: the model's assertion is that no function it contains is the required enumeration. Cantor's theorem remains valid inside the model. The model need not be transitive or have the actual standard natural numbers for this explanation; the essential issue is the restriction of internal quantifiers to its own domain.

For a structure in a countable language and any subset $C$ of its domain, there is an elementary substructure containing $C$ of cardinality at most $\max (|C|,{\aleph }_0)$. In particular a finite or countable $C$ lies in a countable elementary substructure. This is the downward Löwenheim-Skolem theorem.

Construct it directly inside the given structure $A$. Start with $C_0$ containing $C$, the interpretations of the constants, and one element if needed to ensure a nonempty domain. Given $C_n$, for every formula $\exists x\phi (x,a)$ with its finite parameter tuple from $C_n$ that is true in $A$, choose one actual witness in $A$ and place it in $C_{n+1}$. Retain $C_n$ as well. There are at most $\max (|C|,{\aleph }_0)$ choices at each stage. Let $B={\bigcup }_{n<\omega }{C}_n$.

The construction closes $B$ under every function of the language: for a tuple in $C_n$, the formula $x=f(a)$ forces its unique value into the next stage. Thus $B$ carries the induced substructure, and it has the asserted cardinal bound.

Prove elementarity by structural induction. Atomic formulas agree by the induced-substructure construction; logical connectives preserve agreement. For an existential formula with parameters in $B$, if it is true in $A$, all its parameters occur in some $C_n$, so the chosen witness lies in $C_{n+1}\subseteq B$. The induction hypothesis makes its matrix true in $B$. In the other direction a witness in $B$ is also a witness in $A$, again by induction. This is the Tarski-Vaught test, with its witness argument proved explicitly. No completeness or compactness theorem theorem enters the construction.

Expand the language by a constant $c_a$ for every element of the structure. Its diagram of a structure consists of all atomic and negated atomic sentences true in this expanded structure. Its elementary diagram, also called the complete diagram, consists of all first-order sentences true in the expansion.

For a model $M$ of the diagram, the map $a\mapsto c_a^M$ is injective because the diagram includes all required inequalities. It preserves functions and constants by the corresponding equations, and it preserves and reflects relations by the positive and negative atomic sentences. Thus it is a structure embedding onto a substructure of the reduct of $M$. Conversely, a structure containing an isomorphic copy of the original as a substructure can interpret the new constants at that copy and satisfy its diagram.

For the elementary diagram the same argument works for every formula, not just atomic ones. The map is therefore an elementary embedding. Conversely any elementary embedding gives an expansion satisfying the elementary diagram. These are characterizations up to the named copy; a model of either diagram may have additional unnamed elements.

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.