Past exam of the model theory and Gödel's incompleteness theorems course of the University of Oxford / 2010 / B / 6

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.

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.

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.