Is there a relationship between the Compactness Theorem and the upward Lowenheim-Skolem Theorem in FOL?

I was thinking of another post of mine "Why accept the axiom of infinity?" when I though, "Well, if someone accepts arbitrary large finite numbers, what stops them from the jumping into the infinite?"

After all, doesn't compactness imply that if you have a theory $T$ with arbitrarily large finite models, then $T$ has arbitrarily large models—infinite or finite? This sounds a lot like the upward Lowenheim-Skolem theorem to me.

Is there a connection?

  • $\begingroup$ Haven't you answered your own question? The compactness theorem implies that if a theory $T$ has arbitrarily large finite models, then it has an infinite model. So it gives a special case of upward Löwenheim-Skolem, but as far as I can tell the argument doesn't extend to larger cardinalities. $\endgroup$ – Qiaochu Yuan Jun 15 '12 at 14:09

Well, one might say everything is connected, so it depends on which kind of connections you're looking for.

There certainly is the connection that one can derive upward Löwenheim-Skolem from downward Löwenheim-Skolem and compactness:

Let $T$ be a theory with equality that admits some infinite normal model $M$. For any $\aleph_\beta \ge |T|$ we can construct a model of cardinality $\aleph_\beta$ in two steps. (1) Construct a new theory by adjoining $\aleph_\beta$ new constant letters with axioms that say they are all different. By compactness, the extended theory is consistent and so has a model $N$. Clearly this model must have cardinality at least $\aleph_\beta$, and in particular it is a model of $T$. (2) If $|N|>\aleph_\beta$, use downward Löwenheim-Skolem to cut it down to size as a model of $T$.

  • $\begingroup$ If you formulate the upward Skolem with inequality (there exists an elementary superstructure (/ a model of the theory) of cardinality at least $\kappa$...), you can derive it straight from compactness, without downward Skolem in-between. $\endgroup$ – tomasz Jun 15 '12 at 14:15
  • 2
    $\begingroup$ @tomasz: But it is strictly weaker, in the sense that if all the ultrapowers of a countable model would be at least of size continuum - how can you prove the existence of a model of size $\aleph_1$? $\endgroup$ – Asaf Karagila Jun 15 '12 at 14:19

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