Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 23 / 5

Expand the language by constants for $A$ and write $T_A=\mathrm{Th}(\mathcal{M},a{)}_{a\in A}$. An A-type over a structure is a set of formulas in a fixed finite tuple $\bar{x}$, with parameters from $A$, consistent with this complete theory.

Let $Q=\{{\phi }_1(\bar{x}),\dots ,{\phi }_m(\bar{x})\}\subseteq P$. If no tuple of $M$ realized $Q$, thenThat sentence belongs to $T_A$ and contradicts the consistency of $T_A\cup P$. ConsequentlyFor $Q=\varnothing$, any tuple works because domains are nonempty. This proves finite satisfiability; realization of the entire partial type may require an elementary extension.

The Stone topology on a type space has basic open setswhere $\phi$ ranges over formulas with parameters in $A$ and $n$ free variables. Here points are complete types consistent with $T_A=\mathrm{Th}(\mathcal{M},a{)}_{a\in A}$. Arbitrary open sets are unions of these basic sets.

The identitiesshow that the basic sets are clopen sets and form a basis. Distinct complete types contain opposite choices for some formula, so this topology is Hausdorff. The compactness theorem gives compactness: a family of basic closed sets with the finite intersection property determines a consistent set of formulas, which extends to a complete type. Thus logical formulas provide a basis of clopen tests for distinguishing points of the type space.

A precise countable-language form of the omitting types theorem is the following. Let $T$ be a consistent first-order theory in a countable first-order language, and let $(p_j({\bar{x}}_j){)}_{j\in \mathbb{N}}$ be a countable family of nonprincipal partial types, each in finitely many variables. Then there is a finite or countably infinite model of $T$ omitting all the $p_j$.

A partial type is a set of formulas jointly consistent with $T$. It is principal if there is a formula $\theta (\bar{x})$ such that $T\cup \{\exists \bar{x}\theta \}$ is consistent andFor a complete type, this is exactly being an isolated type. Nonprincipal means that no such formula exists. A model omits $p$ if no tuple in that model satisfies every formula in $p$; different tuples may fail different members of the type. The model in the theorem need not omit each individual formula everywhere.

For parameter types, use the expanded language naming the parameters. Countability then applies to that expanded language, and hence requires a countable parameter set as well. There is no countability restriction in the finite-satisfiability argument of part (a).

Let $p(\bar{x})$ be an isolated type over $A$, with isolating formula $\theta (\bar{x})$ over $A$. Its consistency with $T_A$ and the complete theory property implyotherwise $\neg \exists \bar{x}\theta$ would belong to $T_A$, contradicting consistency. Choose a tuple $\bar{b}\in M$ satisfying $\theta$. By isolation,so $\bar{b}$ satisfies every member of $p$. Therefore every isolated A-type is realized in the original structure. The same reasoning works for a principal partial type whose isolating formula need not itself be a member of the type.

Use the first-order language with one constant $c$, take $\mathcal{M}=(\mathbb{N},c=0)$ with equality only, and let $A=\varnothing$. SetThis complete type contains $x=c$ and is isolated by that formula. In any first-order structure, at most one element can satisfy $x=c$. In every $\mathcal{N}\equiv \mathcal{M}$, its interpretation of $c$ realizes $P$, by elementary equivalence. Hence no elementarily equivalent model has infinitely many realizations of $P$. An algebraic type can have a fixed finite number of realizations in every model.

A counterexample can even use a nonalgebraic isolated type. Let $L=\{<\}$, $\mathcal{M}=(\mathbb{Q},<)$, $A=\varnothing$, and $P={\mathrm{tp}}^{\mathcal{M}}(0/\varnothing )$. The quantifier elimination for dense linear orders without endpoints shows that all elements have the same complete type over the empty set: a one-variable quantifier-free formula has no order information except $x=x$ and $x<x$.

Thus $x=x$ isolates $P$. Every $\mathcal{N}\equiv \mathcal{M}$ is a dense linear order without endpoints, and every element of $N$ realizes $P$. Such a model cannot omit it. Even a type with infinitely many realizations need not be omittable; isolation is the relevant obstruction.

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.