How to prove that in a first-order logic, the models of a theory cannot be only the interpretations with finite domains?
1
-
$\begingroup$ While a nice question, it’s not really research-level, so this sort of question would be more appropriate at math.stackexchange.com. $\endgroup$– Peter LeFanu LumsdaineCommented Dec 2, 2010 at 16:43
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
3
If the theory $T$ proves (for some natural number $n \in {\mathbb Z}_+$) the sentence
$$(\forall y_0) \cdots (\forall y_n) \bigvee_{0 \leq i < j \leq n} y_i = y_j$$
then every mcodel of $T$ will be an interpretation with a finite domain. Otherwise, the compactness theorem in the language ${\mathcal L}(\{ c_i : i \in {\mathbb N} \})$ where ${\mathcal L}$ is the language of $T$ and the $c_i$'s are new constant symbols applied to the set of sentences
$$T \cup \{ c_i \neq c_j : 0 \leq i < j \in {\mathbb N} \}$$ shows that there must be infinite models of $T$.
-
1$\begingroup$ The collection of disjunctions on the first two lines is often proved in the equivalent form "if it has any finite models, it has models of unboundedly large finite size". $\endgroup$– AdamCommented Dec 2, 2010 at 3:20
-
$\begingroup$ I'm not sure if this is correct: I think you mean "if it has a model of arbitrary large size then it has models of unboundedly large size!?" – Anna Fred 0 secs ago $\endgroup$ Commented Dec 2, 2010 at 20:07
-
$\begingroup$ Any way, how this could help in the proof? $\endgroup$ Commented Dec 2, 2010 at 20:08