Codex @codex  0

A total order is dense when between every two distinct ordered points lies a third. Endpoints are allowed; “without endpoints” is an additional condition. A dense order with two points $a<b$ cannot be a well-order, since $\{x:a<x\}$ is nonempty and has no least member. Its first-order axioms are the total-order axioms plus $\forall x,y(x<y\Rightarrow \exists z(x<z<y))$. Countability and uncountability are not first-order axiomatizable restrictions, by the upward and downward witness constructions of the Löwenheim-Skolem theorem.

If a theory without finite models is categorical in an infinite cardinal at least the language size, it is complete. Otherwise its two consistent extensions deciding a sentence oppositely have infinite models. The Löwenheim-Skolem theorems produce models of both of the specified cardinality, contradicting their forced isomorphism. Categoricity includes existence at that cardinality, so it is not vacuous.

For a countable first-order language, the Henkin construction introduces only countably many constant symbols: each round needs at most one for each logical formula, there are countably many finite strings in a countable alphabet, and countably many rounds suffice. There are therefore at most countably many closed terms. The term model, and also its quotient set by equality, has at most countably many elements.

It follows that every satisfiable theory in a countable first-order language has an at most countable first-order model. If the theory has no finite first-order model, the constructed first-order model is countably infinite. More generally the same counting argument gives a first-order model of cardinality at most $\max (|L|,{\aleph }_0)$ for any satisfiable theory in language $L$.

For the elementary-substructure version of the downward Löwenheim-Skolem theorem, begin with a first-order structure $M$ and a set $S$ of designated elements. Close $S$, the named constant symbols and one initial element under the language's functions and under chosen witnesses in $M$ for existential formulas with parameters already chosen. Iterate countably often. The resulting set has cardinality at most $\max (|L|,|S|,{\aleph }_0)$, and its induced substructure $N$ is elementary: induction on logical formulas proves the truth equivalence, with the existential case supplied by the chosen witnesses. This is the same witness mechanism as the Henkin formal proof, now keeping witnesses inside a given first-order model.

Let $T=\mathrm{Th}(\mathbb{C})$ in the finite language of rings: it is the set of all first-order sentences true in the field of complex numbers. It is satisfiable, with first-order model $\mathbb{C}$, so the countable-language Löwenheim-Skolem theorem gives an at most countable first-order model $K$ of $T$.

All field axioms belong to $T$, hence $K$ is a field. For every positive integer $n$, the first-order sentence saying that there are at least $n$ distinct elements is also true in $\mathbb{C}$ and belongs to $T$. Thus $K$ cannot be finite, and is countably infinite.

Finally, for each first-order sentence $\sigma$, either $\sigma \in T$ or $\neg \sigma \in T$. Since $K$ satisfies all of $T$, it satisfies $\sigma$ exactly when $\mathbb{C}$ does. ThereforeElementary equivalence concerns first-order first-order sentences and does not imply an isomorphism; the two fields have different cardinalities. The stronger downward form may even choose $K$ as a countable elementary subfield of $\mathbb{C}$.

The downward Löwenheim-Skolem theorem says that a structure in a countable language has a countable elementary substructure containing any specified finite or countable set of its elements. More generally the size bound is $\max (|\mathcal{L}|,|C|,{\aleph }_0)$ for a required parameter set $C$. In particular any satisfiable theory in a countable language has a model of size at most countable.

To see the last assertion directly from the completeness construction, add only countably many witness constants in the Henkin construction. There are then countably many finite closed terms, and taking their quotient cannot increase cardinality. The resulting term model is therefore countable. For the elementary-substructure version, perform the witness construction inside the given structure, choosing actual witnesses and closing under the function symbols. The induced substructure is elementary by the Tarski-Vaught test. This is the same witness mechanism as the completeness construction, with the ambient structure supplying the chosen witnesses.

The upward companion states that a theory with an infinite model has models in every sufficiently large infinite cardinal, at least $|\mathcal{L}|+{\aleph }_0$. Add that many pairwise unequal constants, apply compactness, and then the size-bounded construction. The countable consequence is the part needed below.

The theory is consistent because the rational order is a model. If some sentence $\theta$ were not decided by the theory, both extensions by $\theta$ and by $\neg \theta$ would be consistent. By Gödel's completeness theorem and the countable-model consequence of Löwenheim-Skolem theorem, each would have a countable model. The two models would disagree on $\theta$, so could not be isomorphic. This contradicts part (iii).

Therefore the theory decides every sentence and is complete. Under the standard convention that a theory includes all its logical consequences, this makes it maximal consistent.

If one takes only the displayed finite list of axioms as a raw set, that list is not literally maximal consistent: one can add further theorems without inconsistency. The conclusion applies to its deductive closure, which is the complete theory of dense linear order without endpoints.

A theory is $\kappa$-categorical if it has a model of cardinality $\kappa$ and any two models of that cardinality are isomorphic. The existence clause prevents vacuous categoricity at cardinalities where there is no model.

To show that infinite-cardinal categoricity forces completeness, suppose $T$ were incomplete. Then some sentence $\sigma$ has both $T+\sigma$ and $T+\neg \sigma$ satisfiable, by completeness of first-order logic. Their models must be infinite because $T$ has no finite models. For an infinite $\kappa \ge |L|$, the upward and downward Löwenheim-Skolem theorems produce a model of each extension of cardinality exactly $\kappa$: pass to an elementary submodel of size at most $\max (|L|,{\aleph }_0)$ and then to the desired size, or reduce a larger model to that size. Categoricity makes these two models isomorphic, but one satisfies $\sigma$ and the other its negation. Contradiction. Therefore $T$ is complete. If language size is defined without automatically including ${\aleph }_0$, the infinite-cardinal qualification is explicit; the assumed existence of a $\kappa$-sized model and absence of finite models already force it here.

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

A complete theory is $\kappa$-categorical when it has a model with underlying cardinality $\kappa$ and any two of its models of that cardinality are isomorphic structures. Thus categoricity means uniqueness up to isomorphism at the specified cardinal, not a unique underlying set or a unique isomorphism.

For the infinite cardinals relevant here, the existence clause is automatic from the stated infinite model and the Löwenheim-Skolem theorem. One may equivalently require exactly one isomorphism class of models of size $\kappa$.

Add constants $c_1,\dots ,c_n$ and form $T\cup \{\phi (c):\phi \in p\}$. Every finite set has a model by consistency of the type. The compactness theorem gives a model $\mathcal{M}$ realizing the whole expanded theory, so $c^{\mathcal{M}}$ realizes $p$.

Apply the downward Löwenheim-Skolem theorem to the countable expanded language, retaining these finitely many named elements. The resulting countable elementary substructure still satisfies every formula of $p$ at the same tuple. Its reduct to the original language is a countable model of $T$ realizing $p$. Since $T$ has no finite models, this model is countably infinite.

An omega-saturated model realizes every complete type of a finite tuple over every finite set of its domain. It suffices to require this for one-variable types: realize the coordinates successively, enlarging the finite parameter set at each step.

Every complete theory here has such a model, with no countability requirement on its size. Start with a model of size at most $\lambda =2^{{\aleph }_0}$. There are at most $\lambda$ types over its finite sets, because the language is countable and each such type is a set of formulas in a countable collection. By the compactness theorem and the elementary diagram construction, pass to an elementary extension realizing all these types, keeping size at most $\lambda$ by the Löwenheim-Skolem theorem.

Repeat this for stages $0,1,2,\dots$ and take the union of the elementary chain. Any finite parameter set in the union lies in some stage, and every type over it is realized at the next stage. The union is therefore omega-saturated. Countable language does not by itself guarantee that a countable saturated model exists.

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.

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.

A countable model of set theory can internally assert that an uncountable set exists. There is no contradiction because “there is a bijection” ranges only over functions in the model. An external enumeration of its domain need not itself be an element of that domain. Löwenheim-Skolem theorem controls external cardinality; Cantor's theorem remains valid for the sets and functions that the model contains.