Past exam of the model theory and Gödel's incompleteness theorems course of the University of Oxford / 2009 (C1.1, DMAC 2460, DMCO 2460, DMPC 2460)

This University of Oxford Part C paper covers Gödel's Incompleteness Theorems in section A and Model theory in section B, cross-listed for Mathematics, Mathematics and Computer Science, and Mathematics and Philosophy. ^[ref] The archive contains the complete editable five-page transcription. Bounded official-source searches found no advertised answers for this exact paper. A2(c) is answered with a counterexample to its literal semantic hypothesis and a proof of the intended canonical case. The vector-space comparison keeps infinitude of the underlying set distinct from infinite dimension.

Write $\square A$ for the arithmetical sentence asserting $S$-provability of $A$. If $S⊢G$, the first derivability condition gives $S⊢\square G$. The stipulated fixed-point equivalence also gives $S⊢\neg \square G$. This contradicts consistency. Conversely an inconsistent classical theory proves every sentence, including $G$. ThusOnly the direction from an actual proof to its formalized provability is used here. There is no inference from a theorem $\square G$ to a theorem $G$, and no assumption that every theorem of $S$ is true.

The Hilbert-Bernays-Löb derivability conditions are: if $S⊢A$ then $S⊢\square A$; $S⊢\square (A\to B)\to (\square A\to \square B)$; and $S⊢\square A\to \square \square A$. The first is an external rule about provability; the latter two are internal theorems. Standard numerical correctness and the stipulated ${\Sigma }_1$ form accompany the predicate in this question. These conditions do not include reflection $\square A\to A$.

Choose a fixed refutable sentence $\perp$ and identify ${\mathrm{Con}}_S$ with $\neg \square \perp$. Different fixed refutable choices have provably equivalent consistency sentences, because $S$ proves their equivalence to a contradiction. From $\perp \to G$, necessitation and distribution give $\square \perp \to \square G$. Since $G\to \neg \square G$, this yields $G\to {\mathrm{Con}}_S$.

For the converse, necessitate $G\to \neg \square G$ and distribute to get $\square G\to \square \neg \square G$. Positive introspection also gives $\square G\to \square \square G$. The two boxed contradictories imply $\square \perp$ by formalized propositional reasoning. Thus $\square G\to \square \perp$, and contraposition gives ${\mathrm{Con}}_S\to \neg \square G\to G$. ConsequentlyFinally the deduction theorem says that $S+G$ is inconsistent exactly when $S⊢\neg G$. The internal equivalence identifies $S+G$ and $S+{\mathrm{Con}}_S$ in their deductive consequences. Hence $S⊬\neg G$ exactly when $S+{\mathrm{Con}}_S$ is consistent. Consistency of $S$ alone need not ensure this stronger consistency condition.

For consistency of a negated-consistency extension, put $A=S+\neg {\mathrm{Con}}_S$. The arithmetized deduction theorem, together with double-negation elimination and formalized proof transformations, givesFrom the preceding proof, $S⊢{\square }_{S}G\to {\square }_S\perp$. Necessitating the equivalence between $G$ and ${\mathrm{Con}}_S$ and using distribution therefore gives $S⊢{\square }_S{\mathrm{Con}}_S\to {\square }_S\perp$. Its contrapositive isThis is an internal conditional statement. It does not assert that $S$ proves its own consistency or permits the consistency premise to be discarded.

For unrestricted consistency self-strengthening is inconsistent, again take $A=S+\neg {\mathrm{Con}}_S$. In $S^{++}$ the axiom ${\mathrm{Con}}_S$ and the preceding implication prove ${\mathrm{Con}}_A$. The additional rule then proves ${\mathrm{Con}}_{A+{\mathrm{Con}}_A}$.

However consistency of an extension implies consistency of its subtheory by the formalized inclusion of proofs: $S⊢{\mathrm{Con}}_A\to {\mathrm{Con}}_S$. Thus the ordinary theory $A+{\mathrm{Con}}_A$ contains $\neg {\mathrm{Con}}_S$ and derives ${\mathrm{Con}}_S$, so it has a finite contradiction proof. Checking that concrete proof in arithmetic gives $S⊢\neg {\mathrm{Con}}_{A+{\mathrm{Con}}_A}$. This theorem is also present in $S^{++}$, contradicting the result of the new rule. Therefore $S^{++}$ is inconsistent.

The error in the added rule is its unrestricted internal use on arbitrary arithmetically presented systems, including systems containing false consistency assertions. A true external consistency progression for a specified sound system is not this unrestricted inference rule.

For separation of number sets in an arithmetic theory, a formula separates $A$ from $B$ in $S$ if $n\in A$ implies $S⊢F(\overline{n})$, and $n\in B$ implies $S⊢\neg F(\overline{n})$. No demand is made for numbers outside their union. Consistency forces disjointness when such a separator exists.

Let $r$ be the code of the fixed-point sentence. If $S⊢R$, then $r$ is in the provable set, so separation gives $S⊢H(\overline{r})$. The fixed-point equivalence gives its negation, a contradiction. If $S⊢\neg R$, then $r$ is in the refutable set, so separation gives $S⊢\neg H(\overline{r})$, and the fixed-point equivalence gives $S⊢R$, again a contradiction. ThusThe argument is syntactic: it needs only the separating implications and consistency, not omega-consistency or general soundness.

For Rosser separation by a bounded stage comparison, use the numeralwise enumeration in arithmetic properties meant by enumeration in $S$: membership supplies some standard stage with a provable positive instance, while nonmembership makes every standard-stage instance refutable. These are stronger than merely attaching a ${\Sigma }_1$ syntactic label to an arbitrary formula. The finite-initial-segment axiom lets us combine finitely many numeralwise refutations into a bounded universal assertion.

If $n$ is provable, choose a standard stage $p$ with $S⊢P(\overline{n},\overline{p})$. Consistency means $n$ is not refutable. Therefore $S⊢\neg R(\overline{n},\overline{j})$ for each $j\le p$. Using the displayed finite-segment axiom and equality substitution, combine these proofs to get $S⊢\forall z\le \overline{p}\neg R(\overline{n},z)$. Existential introduction with stage $p$ proves the separator at $n$.

If $n$ is refutable, choose a standard stage $q$ with $S⊢R(\overline{n},\overline{q})$. Consistency implies that $n$ is not provable, so all standard instances of $P(\overline{n},\cdot )$ are refutable. In particular their finitely many instances through $q$ give $S⊢\forall z\le \overline{q}\neg P(\overline{n},z)$. For an arbitrary proposed witness $z$, the total-comparison axiom gives $z\le \overline{q}$ or $\overline{q}\le z$. In the first case its $P$ conjunct fails; in the second case the known $R$ instance at $q$ contradicts its bounded no-refutation conjunct. Generalize on $z$ to refute the existential separator. Hence the proposed first-proof-before-refutation formula separates the two sets.

The literal semantic hypothesis is insufficient: a semantic Gödel fixed point need not be canonical. A sentence can be false and provable in a consistent, unsound arithmetic theory, in which case both sides of the stipulated semantic biconditional can be false.

For a concrete counterexample, assume PA is consistent and set $S=\mathrm{PA}+\neg {\mathrm{Con}}_{\mathrm{PA}}$. Gödel's second incompleteness theorem makes this extension consistent: otherwise PA would prove its own consistency. It is still bounded-complete. Use ordinary numeralwise proof and refutation enumerators for this recursively axiomatized theory, and take $G=\neg {\mathrm{Con}}_{\mathrm{PA}}$. This $G$ is false in the standard natural numbers but is an axiom of $S$. Since it is provable, $\exists yP(┌G┐,y)$ is true; its negation is false. Thus the semantic fixed-point biconditional required in this part is true, with both sides false.

Choose $R$ by the usual canonical Rosser construction for these proof matrices. Its undecidability follows from the preceding parts, and it is a true ${\Pi }_1$ sentence: if it were false, its true ${\Sigma }_1$ negation would be provable in this bounded-complete theory, contradicting undecidability. Hence $G$ is false and $R$ true, so their proposed semantic equivalence is false. This counterexample meets the literal assumptions and does not exploit a missing source page.

Under the intended canonical ${\Pi }_1$ Gödel/Rosser convention, the expected answer does hold. Consistency plus bounded completeness implies ${\Pi }_1$ soundness: a false ${\Pi }_1$ theorem has a standard bounded counterexample whose negation the theory proves. If canonical $G$ were false, its truth fixed point would make it provable, contradicting this soundness; thus $G$ is true. Canonical $R$ is true by the preceding ${\Sigma }_1$-negation argument, so both canonical sentences are true and their equivalence is true. This last conclusion needs that canonical construction or an equivalent additional restriction; truth of a biconditional alone does not supply it.

Use the two formal consistency equivalences expressly supplied for this part. If $S⊢G\leftrightarrow R$, the derivability conditions give $S⊢\square \neg G\leftrightarrow \square \neg R$. Taking negations and applying those supplied equivalences yieldsOne-consistency implies ${\Sigma }_1$ soundness under bounded completeness. Hence $S+{\mathrm{Con}}_S$ is consistent: if not, $S$ would prove $\neg {\mathrm{Con}}_S$, a false ${\Sigma }_1$ sentence. But in that consistent extension the last equivalence and its consistency axiom prove its own consistency ${\mathrm{Con}}_{S+{\mathrm{Con}}_S}$. Gödel's second incompleteness theorem rules this out. ThereforeThis establishes the requested unprovability using the explicit formal hypotheses, without inferring internal provability from the semantic equivalence discussed in the preceding part. The hypotheses for the second theorem are the standard effective arithmetic presentation and adequate provability predicates stipulated here.

Gödel's second incompleteness theorem says that a consistent effectively presented arithmetic theory with an adequate proof predicate cannot prove its formal consistency statement. The background strength must support the diagonal lemma and the derivability conditions, as in the present arithmetic setting.

Löb's theorem saysIt is a theorem about derivability for every sentence $A$, not merely an external soundness assertion about the interpretation of $\square$. Taking $A=\perp$, a proof of consistency gives $S⊢\square \perp \to \perp$; Löb then gives $S⊢\perp$. Thus a consistent $S$ cannot prove consistency. This is how the second incompleteness theorem is the contradiction-sentence instance of Löb's more general result.

Suppose $S+A$ proves every specified reflection instance. Use only the instance with $X=\neg A$. The deduction theorem givesClassical propositional reasoning turns this into $S⊢\square \neg A\to \neg A$: if $\neg A$ already holds the implication is true, while otherwise $A$ holds and the displayed theorem supplies it. Löb's theorem now yields $S⊢\neg A$, making $S+A$ inconsistent. Therefore no sentence consistent with $S$ can supply all the local reflection instances in a one-sentence extension. The proof uses just one well-chosen instance, and never applies necessitation to a derivation with the extra assumption still open.

GL has all classical propositional tautologies, the normal distribution axiom $\square (A\to B)\to (\square A\to \square B)$ and the Löb axiom $\square (\square A\to A)\to \square A$, closed under uniform substitution, modus ponens and necessitation of theorems. We use its derived positive-introspection theorem $\square A\to \square \square A$, normal boxed reasoning, and the boxed replacement principle supplied in the question.

For completeness, positive introspection follows from these axioms. Put $C=A\wedge \square A$. Distribution gives $\square C\to \square A$, so propositional logic proves $A\to (\square C\to C)$. Necessitate and distribute to get $\square A\to \square (\square C\to C)$; Löb's axiom gives $\square A\to \square C$, and boxing $C\to \square A$ finishes $\square A\to \square \square A$.

For a GL fixed point by truth substitution, set $\top =(\perp \to \perp )$, $U=\square Y(\top )$ and $V=\square Y(U)$. From the tautology $U\to (\top \leftrightarrow U)$, boxed reasoning gives $\square U\to \square (\top \leftrightarrow U)$. The supplied replacement principle and distribution then giveSince $U$ is itself boxed, positive introspection gives $U\to \square U$, and hence $U\to V$.

For the reverse direction, the displayed implication entails the theorem $V\to (\square U\to U)$. Necessitate it and distribute to obtain $\square V\to \square (\square U\to U)$. Since $V$ is boxed, $V\to \square V$ holds. The Löb axiom therefore yields $V\to \square U$, and the same displayed implication then yields $V\to U$. ConsequentlyAll occurrences of the unknown letter have been removed from $U$ by substitution of truth; any other parameter letters remain available. The proof does not assume the invalid reflection principle $\square U\to U$ as an unconditional axiom.

The fixed point of boxed negated provability gives, in fact, that GL proves $F\leftrightarrow \square \perp$. Here is a direct derivation. Since $F$ is provably equivalent to a boxed formula, positive introspection and necessitation of that equivalence give $F\to \square F$. Another introspection step gives $F\to \square \square F$. The stipulated fixed point also gives $F\to \square \neg \square F$. Combining these two boxed contradictories yields $F\to \square \perp$. Conversely $\square \perp \to \square \neg \square F$ follows by normal boxed explosion, and the fixed point then yields $\square \perp \to F$.

Arithmetical soundness of GL, proved by translating distribution into the second derivability condition, necessitation into the first, and Löb's axiom into its internal arithmetic theorem, now givesAny remaining propositional parameters may be assigned arbitrary arithmetic sentences; this equivalence is uniform in those assignments. If PA is one-consistent, it is consistent and ${\Sigma }_1$ sound. Its standard provability predicate for a contradiction is then a false ${\Sigma }_1$ sentence and cannot be a theorem. Hence $F^{\ast }$ is not provable in PA. Moreover its negation is equivalent to PA's consistency statement and is unprovable by the second incompleteness theorem. Thus this fixed point is, under the stated hypothesis, undecidable rather than an unboxed reflection principle.

An $L$-embedding $\pi :A\to B$ is an injective map preserving the interpretations of all constants and functions and preserving and reflecting every relation. Explicitly $\pi (c^A)=c^B$, $\pi (f^A(\bar{a}))=f^B(\pi \bar{a})$, and $A\models R(\bar{a})$ if and only if $B\models R(\pi \bar{a})$. Injectivity handles logical equality. Equivalently it is an isomorphism onto an $L$-substructure of $B$, or a map preserving the truth of all quantifier-free formulas. Preserving relations in only one direction would instead give a weaker homomorphism notion, not an embedding.

The relation $A\subseteq B$ means that the domain of $A$ is a subset of the domain of $B$ and the inclusion map is an $L$-embedding. Thus it contains all named constants, is closed under every named function, and uses the restrictions of the ambient relations.

The stronger relation $A\preccurlyeq B$ means that for every first-order formula $\varphi (\bar{x})$ and tuple $\bar{a}$ from $A$,This includes formulas with quantifiers ranging over the respective domains. Elementary inclusion implies ordinary substructure inclusion, but the latter need only preserve quantifier-free truth. In particular an existential witness available in $B$ need not lie in an arbitrary substructure.

The Tarski-Vaught test says that a substructure $A\subseteq B$ is elementary exactly when for every formula $\varphi (x,\bar{y})$ and tuple $\bar{a}\in A$, if $B\models \exists x\varphi (x,\bar{a})$, some $b\in A$ satisfies $B\models \varphi (b,\bar{a})$. Thus every existential assertion with parameters from $A$ that has a witness in $B$ already has a witness in $A$. The matrix may contain quantifiers; restricting it to quantifier-free formulas would not state the full test.

Finite satisfiability means that every finite subset $T_0\subseteq T$ has an $L$-structure satisfying all its sentences. Different finite subsets may have different witnessing structures. The first-order compactness theorem says this condition is equivalent to satisfiability of the entire set, but finite satisfiability itself is defined by the finite-subset condition.

Satisfiability means that there exists one nonempty-domain $L$-structure $M$ such that $M\models \varphi$ for every $\varphi \in T$. It is this single simultaneous model that is required. By the soundness and completeness theorems, satisfiability is equivalent to syntactic consistency in classical first-order logic.

Under the deductively closed convention, an $L$-theory is a set of $L$-sentences containing all its first-order logical consequences: if $T\models \varphi$, then $\varphi \in T$. By completeness this is equivalent to closure under provability. An axiom set is often called a theory too, with its deductive closure understood. Neither convention by itself requires completeness or satisfiability; an inconsistent deductively closed theory contains every sentence.

A complete satisfiable theory has a model and decides every $L$-sentence: for each $\sigma$, exactly one of $\sigma$ and $\neg \sigma$ is a consequence. Under the closed convention, exactly one belongs to $T$. Satisfiability excludes the possibility of both. Equivalently all models of the theory are elementarily equivalent, even though they need not be isomorphic or have equal cardinality.

The complete theory $\mathrm{Th}(A)$ is the set of all $L$-sentences true in $A$. The existential theory of a structure ${\mathrm{Th}}_{\exists }(A)$ consists of those true sentences of the form $\exists \bar{x}\theta (\bar{x})$ with quantifier-free $\theta$. The universal theory ${\mathrm{Th}}_{\forall }(A)$ uses $\forall \bar{x}\theta (\bar{x})$ with quantifier-free matrix. Allow equivalent formulas and empty quantifier blocks, or insert dummy variables in nonempty domains. These are sentences in the original language, distinct from diagrams, which add constants naming Elements of the structure.

If a model $B$ of $T$ contains $A$ as a substructure, every existential sentence true in $A$ remains true in $B$: use the same witness tuple and preservation of quantifier-free truth. Hence $B$ satisfies the required joint theory.

For the existential-theory embedding criterion, conversely suppose that joint theory has a model. Expand the language by a constant $c_a$ for each $a\in A$, and add the atomic and negated atomic diagram $D(A)$ to $T$. Any finite collection of its literals is a quantifier-free conjunction $\theta (\bar{c})$ true of the corresponding tuple in $A$. Replacing those named constants by variables and existentially closing gives a sentence in ${\mathrm{Th}}_{\exists }(A)$. A model of the assumed joint theory can supply its witnesses, and hence an expansion satisfying those finitely many diagram requirements and any finite subset of $T$.

Compactness therefore gives a model of $T+D(A)$. Its named Elements define an embedding of $A$ into its $L$-reduct: inequalities ensure injectivity, and all function and relation literals ensure preservation and reflection. Replace the ambient model by an isomorphic copy identifying this image with $A$ itself. Thus a model of $T$ containing $A$ exists. The diagram step explains how parameter-free existential information nevertheless yields an embedding of the whole structure.

If $B\models T$ and $B\subseteq A^{\ast }\equiv A$, every universal sentence of $A$ holds in $A^{\ast }$ and is preserved by passage to the substructure $B$. Thus $B$ witnesses satisfiability of the joint theory.

For universal theory and substructures of elementarily equivalent models, conversely let $B$ satisfy $T$ together with the universal theory of $A$. Every existential sentence true in $B$ must be true in $A$: if an existential $\sigma$ were false in $A$, its universal negation would belong to ${\mathrm{Th}}_{\forall }(A)$ and hold in $B$, a contradiction. Hence ${\mathrm{Th}}_{\exists }(B)\subseteq \mathrm{Th}(A)$. In particular $\mathrm{Th}(A)\cup {\mathrm{Th}}_{\exists }(B)$ is satisfiable, with model $A$. Apply the preceding embedding result to the theory $\mathrm{Th}(A)$ and structure $B$. It supplies a structure $A^{\ast }\models \mathrm{Th}(A)$ containing $B$. Therefore $A^{\ast }\equiv A$ and $B\subseteq A^{\ast }$, as required. Elementary equivalence here does not assert that $A$ itself contains a copy of $B$.

The natural-number successor chain embeds into the integer successor chain by ordinary inclusion. Thus $\mathbb{Z}$ witnesses the existential joint satisfiability in the first assertion. The two chains are not elementarily equivalent: every integer has a predecessor, whereas zero has none in the natural-number chain.

For the universal assertion, construct an elementary extension of the natural-number chain containing a two-sided successor chain. To its elementary diagram add distinct constants $d_j$ for $j\in \mathbb{Z}$ with $s(d_j)=d_{j+1}$ and require them to differ from all named natural numbers. Every finite subset is realized in the natural-number chain: place the finitely many required consecutive blocks sufficiently far to the right, away from the finitely many forbidden named Elements. Compactness yields an elementary extension $A^{\ast }$ with the required two-sided chain. That chain is a substructure isomorphic to the integer successor structure and hence is a model of its complete theory.

Consequently both joint theories in the two assertions are satisfiable, though the integer chain cannot embed into the natural-number chain itself: repeated predecessors would give an infinite descending sequence of natural numbers. The second assertion correctly uses an elementarily equivalent ambient structure rather than the original one.

Existential sentences are preserved under extensions: witnesses remain present and quantifier-free truth is unchanged by an embedding. Therefore any existential axiomatization implies the specified extension-closure property.

For existential axiomatization from extension closure, let $E$ be all existential consequences of $T$, and suppose $A\models E$. We claim that $T\cup {\mathrm{Th}}_{\forall }(A)$ is satisfiable. Otherwise compactness supplies finitely many universal sentences ${\theta }_1,\dots ,{\theta }_m$ true in $A$ such that $T\models \neg ({\theta }_1\wedge \cdots \wedge {\theta }_m)$. The finite conjunction is universal after renaming bound variables, so its negation is existential. It belongs to $E$ but is false in $A$, a contradiction.

The preceding universal-theory embedding result now yields $B\models T$ and $A^{\ast }\equiv A$ with $B\subseteq A^{\ast }$. By the assumed extension closure, $A^{\ast }\models T$; elementary equivalence implies $A\models T$. Thus $E$ axiomatizes exactly the models of $T$, proving existential axiomatizability is equivalent to closure under extensions. This includes the inconsistent case, whose empty model class is axiomatized by an existential contradiction.

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.

Let $F_n$ be the formulas whose free variables are among $x_1,\dots ,x_n$. Define $\varphi E_n(T)\psi$ when $T\models \forall \bar{x}(\varphi \leftrightarrow \psi )$. The n-variable Lindenbaum algebra quotient is a Boolean algebra, with meet, join and complement induced by conjunction, disjunction and negation; its zero and one are contradiction and truth classes. Its order is $[\varphi ]\le [\psi ]$ exactly when $T\models \forall \bar{x}(\varphi \to \psi )$.

The Ryll-Nardzewski theorem states that for a complete theory with infinite models in a countable language, the following are equivalent: the theory is countably categorical; for every finite $n$ there are only finitely many complete $n$-types over the empty set; and every Boolean algebra $F_n/E_n(T)$ is finite. Equivalently each such type is isolated. The countable-language and completeness hypotheses are essential parts of this formulation.

For each fixed finite tuple length $n$, use the finite conjunctionIt asserts that no nontrivial scalar combination vanishes, exactly linear independence. It is an $L$-formula because the field is finite and every scalar multiplication is named. Scalar coefficients are indices of function symbols, not variables ranging over a separate field sort. There is no single finite formula with arbitrarily many free variables; the assertion is a formula for each prescribed finite $n$.

Use the vector-space axioms: addition is an abelian group operation with identity zero; ${\lambda }_1(x)=x$; ${\lambda }_a{\lambda }_b(x)={\lambda }_{ab}(x)$; ${\lambda }_a(x+y)={\lambda }_a(x)+{\lambda }_a(y)$; and ${\lambda }_{a+b}(x)={\lambda }_a(x)+{\lambda }_b(x)$, with ${\lambda }_0(x)=0$. Group inverses may be expressed by ${\lambda }_{-1}$, so no extra inverse symbol is needed.

Add the dimension-unbounded schemeOver a finite field, a finite-dimensional space has finitely many Elements, while every infinite-dimensional space is infinite. Hence this scheme axiomatizes exactly the infinite underlying vector spaces requested. In this finite-field case the vector-space axiom list itself is finite; the infinitude of the additional dimension scheme is substantive.

No. Universal sentences are preserved under substructures. Every infinite $K$-vector space has its zero subspace as an $L$-substructure, since zero is named and every scalar multiplication preserves it. This one-element structure is not infinite and is not a model of $T$. Thus the model class is not closed under substructures and has no universal axiomatization, even relative to the vector-space axioms.

No. If a finite conjunction $\varphi$ axiomatized the infinite spaces, it would follow from the vector-space axioms together with all the $D_n$. Compactness would make it follow from finitely many of those axioms and dimension requirements. Let $N$ be at least every dimension mentioned there. The finite-dimensional space $K^N$ satisfies all vector-space axioms and those finitely many $D_n$, hence would satisfy $\varphi$. But it is a finite set and so cannot be a model of the purported axiomatization. This contradiction proves failure of finite axiomatizability, including finite axiomatizability relative to vector spaces.

For quantifier elimination for infinite underlying vector spaces, we first justify that formulas cannot make finer distinctions than their linear-equation data. In any infinite underlying vector space, eliminate an existential variable from a disjunction of conjunctions of linear equations and inequations. If an equation has a nonzero coefficient of that variable, solve for it and substitute. If none does, the inequations forbid only finitely many values, leaving an available value in the infinite domain; the remaining equations concern only the other variables. Terms are linear combinations, so induction gives quantifier elimination for $T$ and completeness, since closed terms reduce to zero.

For one variable there are two possible complete types, zero and nonzero. Both are isolated, and every definable set is their union. ThusFor types of pairs of vectors over a finite field, the linear relation kernel $\{(a,b)\in K^2:ax+by=0\}$ determines the complete type. If $q=|K|$, the possibilities are one rank-zero pair, $q+1$ rank-one directions, and one rank-two independent type. Concretely these are $(0,0)$; $x\ne 0$ with $y={\lambda }_a(x)$ for each $a\in K$; $x=0,y\ne 0$; and the independent pairs. Their isolating formulas are finite because $K$ is finite. HenceEvery class is the union of a selection of these atoms. Basis extension also shows that pairs with the same relation kernel are conjugate by a linear automorphism within any infinite-dimensional model.

Every vector space has a basis and is classified up to linear isomorphism by its dimension. Over a finite field, a vector is a finite linear combination of basis Elements. If the basis has infinite cardinality $\lambda$, the set of finite such combinations has cardinality $\lambda$: there are at least $\lambda$ basis vectors and at most a countable union of finite powers of $\lambda$, each of size $\lambda$. Thus the underlying cardinality of an infinite space equals its dimension.

Any two models of cardinality $\kappa \ge {\aleph }_0$ therefore have bases of cardinality $\kappa$. A bijection between their bases extends uniquely to an $L$-isomorphism. A space with that basis size exists, giving the required existence clause as well. Hence $T$ is categorical in every infinite cardinal.

The cardinality of a vector space is central: here “infinite” must continue to refer to the underlying set. Over a countably infinite field, every nonzero finite-dimensional space is already infinite. Thus the model class now consists of all nonzero $K$-vector spaces, not only infinite-dimensional ones.

Since independence of a pair is not definable over an infinite field, for (i), independence of a single vector is still $x\ne 0$, but independence of tuples of length at least two is not definable by one formula in this one-sorted language. Quantifier elimination follows from the linear-equation argument in the preceding part for all these infinite domains. Any candidate two-variable formula therefore involves only finitely many linear equations. Take an independent pair, and compare it with $(x,{\lambda }_{a}x)$, where $x\ne 0$ and $a\in K$ avoids the finitely many ratios making one of those nontrivial equations vanish. Both pairs satisfy the same listed atomic formulas, contradicting a proposed definition of independence. Longer tuple independence would define pair independence by existentially completing a pair inside an infinite-dimensional model, giving the same obstruction.

For (ii), the vector-space axiom schema together with $\exists x(x\ne 0)$ axiomatizes the infinite underlying spaces. For (iii), there is again no universal axiomatization, because the zero subspace is a substructure.

For finite axiomatizability with infinitely many scalar symbols, in (iv) there is no finite axiomatization in the full language: any finite axiom set mentions only finitely many scalar symbols. In a nonzero vector space alter an unused nonzero ${\lambda }_a$ to send zero to a nonzero vector, retaining all mentioned interpretations. The finite sentences keep their truth values, but a vector-space axiom is now false. Relative to the already imposed vector-space axioms, the extra condition is finitely axiomatized by the single nonzero-vector sentence. These absolute and relative answers must be distinguished.

For (v), the one-variable algebra still has two atoms and four Elements. In two variables the isolated atoms are $(0,0)$ and the countably many rank-one directions, represented by $x\ne 0,y={\lambda }_{a}x$ and by $x=0,y\ne 0$. There is also a nonisolated complete type of an independent pair. A formula using finitely many linear equations either picks finitely many isolated atoms, or contains that independent type and all but finitely many isolated atoms. Therefore $F_2/E_2(T)$ is the finite-cofinite Boolean algebra on a countably infinite set of atoms, with the independent type the extra nonisolated point in its Stone space. In particular this algebra is countably infinite, rather than the full power set of its atoms.

For (vi), dimension can be $1,2,3,\dots$ or ${\aleph }_0$ while the domain is countable, producing pairwise nonisomorphic countable models. Thus $T$ is not countably categorical. For an uncountable $\kappa$, a vector space over this countable field has cardinality $\kappa$ exactly when its dimension is $\kappa$, since in general a nonzero space has size $\max (|K|,\dim V)$ when that maximum is infinite. Basis bijections then give categoricity in every uncountable cardinal. Quantifier elimination also shows that these different positive dimensions are elementarily equivalent; independence of an arbitrarily long tuple is not being silently treated as definable over this infinite field.

A complete $n$-type over $T$ is a maximal set $p(\bar{x})$ of formulas in $n$ variables consistent with $T$; it decides each such formula or its negation. A partial type drops maximality. A tuple realizes the type if it satisfies every member. A complete type is principal, or isolated, if some formula $\theta (\bar{x})$ consistent with $T$ implies every member of the type modulo $T$; then $\theta$ itself belongs to that type.

The omitting types theorem says that in a countable first-order language, a consistent theory has a countable model omitting a specified nonprincipal type; more generally a countable family of nonprincipal types can be omitted simultaneously. In the present complete-theory setting it applies directly to complete nonisolated types. “Omitted” means that no tuple in the model realizes all formulas of the type, even though each finite subset is compatible with the theory.

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

An ${\aleph }_0$-saturated model realizes every complete type in finitely many variables over every finite parameter subset of itself. It suffices equivalently to require one-variable types and then realize finite tuples successively. The definition puts no countability restriction on the size of the model.

For any model $M$, collect all complete types over its finite parameter sets. Add to its elementary diagram, for each such type, a new tuple of constants satisfying all that type's formulas. Any finite selection of these requirements is realizable in $M$: each finite type fragment has its existential closure true there, since it is consistent with the complete theory of the corresponding named finite tuple. Distinct new tuples have no additional mutual restrictions. Compactness gives an elementary extension realizing all these types simultaneously.

Start from any model $M_0\models T$ and apply this construction repeatedly to obtain $M_0\preccurlyeq M_1\preccurlyeq \cdots$, with each successor realizing all types over finite subsets of its predecessor. The elementary-chain theorem makes $M_{\omega }={\bigcup }_n{M}_n$ a model of $T$ with every stage elementary in it. A finite parameter set $A\subseteq M_{\omega }$ lies in some $M_n$. Every type over $A$ consistent with $M_{\omega }$ is consistent with $M_n$, since the expanded complete theories at those parameters agree by elementarity. It is realized in $M_{n+1}$, hence in the union. Therefore an ${\aleph }_0$-saturated model always exists, possibly with uncountable cardinality.

For cut types over the named rational order, choose an irrational real cut, for example $r=\sqrt{2}$, and formEvery finite subset is realized by a rational between its finite lower and upper bounds, so compactness makes it a consistent partial type; quantifier elimination for dense linear orders without endpoints makes these comparisons determine a complete type in the language naming all rationals. No rational realizes the cut, so $Q^{+}$ omits it.

Add a new constant realizing this type and apply compactness, followed by downward Löwenheim-Skolem in the countable expanded language. This gives a countable model of $T$ realizing it. An isomorphism in $L_{\mathbb{Q}}$ fixes all named rationals and preserves realization of the cut, so it cannot be isomorphic to $Q^{+}$. Thus $T$ is not countably categorical. Omission by itself would need this additional realization-and-countable-model argument to establish noncategoricity.

No infinite cardinal is categorical for this theory, as described by no categorical cardinal for the complete named rational diagram. For an arbitrary infinite $\kappa$, choose a dense linear order $I_{\kappa }$ without endpoints of size $\kappa$; for example the lexicographic order on $\kappa \times \mathbb{Q}$ works, since every rational block is dense and has no endpoints.

Construct one order by inserting $I_{\kappa }$ into a fixed irrational cut of the named rational order. Every element of this order is bounded above by some named rational. Construct a second order by putting an $I_{\kappa }$ tail above all named rationals, as the ordered sum $\mathbb{Q}+I_{\kappa }$. Both are dense linear orders without endpoints, both have cardinality $\kappa$, and both contain the named rational copy as an elementary substructure by quantifier elimination. Hence both satisfy its complete diagram $T$.

They are not isomorphic over the named constants: the second has an element greater than every named rational, and the first has none. This property need not be expressible by a single sentence in the infinite language to be an isomorphism invariant. Finite cardinalities have no models at all because the constants name infinitely many distinct Elements, so the existence clause excludes categoricity there too.

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.