The Löwenheim-Skolem-Tarski theorem, which says that

If a theory has a model of a given infinite cardinality, then it has models of any greater infinite cardinality

is proved by my textbook in the following way:

It is enough to add an amount of constants equating the desired cardinality and propositions asserting that every constant is different from each other. Because of compactness the extended theory is satisfiable, therefore has a model $\mathscr{M}$ of the desired cardinality. At this point it is sufficient to exctract from $\mathscr{M}$ a model of the initial theory, trascurating to give an interpretation to the added constants.

I do not see where the infiniteness of the initial model, which is a hypotheses that I guess we cannot omit, is used. Thank you very much for any explanation!


The assumption that the original theory has an infinite model is necessary in order to conclude by compactness that you can add new constants $c_i$ and axioms $c_i\ne c_j$ without making the theory inconsistent.

If all you had was a finite model with $n$ individuals, then once you add $n+1$ constants and distinctness axioms between them, you can't extend the model with an interpretation of all your new constants such that they're all different, so the argument that every finite selection of the new axioms is co-satisfiable with the original theory is not available.

For example if you original theory had the single axiom $$ (\exists x)(\exists y)(\forall z)\;z=x\lor z=y $$ then it only has models with exactly 1 or 2 elements, and Löwenheim-Skolem doesn't apply to that.

The assumption can be weakened to "the theory has an infinite model, or has models of arbitrarily large finite sizes", however.

  • 1
    $\begingroup$ Aww, this was exactly the example I had in mind for how the theorem fails for finite models. :-) $\endgroup$ – Asaf Karagila Feb 2 '15 at 19:41
  • $\begingroup$ Thank you very, very much! $\endgroup$ – Self-teaching worker Feb 2 '15 at 19:55

What you get is that the theory of the infinite model, and the theory of the larger model are the same. Namely, the same sentences are true in both models.

If one of the models is finite, then it satisfies the sentence "there are exactly $n$ objects in the universe" which cannot be satisfied by any larger model.

You can see this use when you apply compactness. In order to ensure that every finite subtheory of the augmented theory (the original one, along with the axioms stating the new constants are pairwise distinct), you need to be able and interpret any finitely many constants in a model. So you need models of arbitrarily large cardinality, or at least one of infinite cardinality.

  • $\begingroup$ Thank you so much! Two equally great answers: I chose the other one just because it contains an explicit example. $\endgroup$ – Self-teaching worker Feb 2 '15 at 19:55

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