Show that, if for every non-denumerable $\mathcal{M}$, $\mathcal{M} \not\models \Gamma$, then $\Gamma$ is unsatisfiable.

Question

Show that, if for every non-denumerable $\mathcal{M}$, $\mathcal{M} \not\models \Gamma$, then $\Gamma$ is unsatisfiable.

Or alternatively, if for every uncountable $\mathcal{M}$, $\mathcal{M} \not\models \Gamma$, then $\Gamma$ is unsatisfiable.

The following is my proposed proof, I would like to get some verification on whether it works. I would also be very interested in alternative proofs.

Let's prove this by contradiction. Assume $\Gamma$ has no non-denumerable models. Then supposing that $\Gamma$ has a model gets us a contradiction.

Taking non-denumerable to mean either finite or non-enumerable.

1. Suppose $\Gamma$ is satisfiable, and so has a model.
2. Then $\Gamma$ either has an enumerable or non-enumerable model.
3. If $\Gamma$ has a non-enumerable model, then we have a contradiction.
4. If $\Gamma$ has an enumerable model, then it has either a finite or denumerable model.
5. If $\Gamma$ has a denumerable model, then by Upward lowenheim skolem theorem, $\Gamma$ also has a non-denumerable model. This contradicts our assumption.
6. If $\Gamma$ has a finite model, then this contradicts our assumption too.
7. Therefore, $\Gamma$ is unsatisfiable.

I think this question reveals the inexpressibility of first-order theories. Particularly pertinent to this proof is how they can't express that they only have finite models. But they can express that they have models of cardinality, say, 5. All is fine so long as they are specifying integers. Once we go beyond specifying particular integers, they can no longer pin down the size of their models. "I only have finite models", "I only have denumerable models" or "I only have non-denumerable models" would be some inexpressible statements in first-order logic.

Answer 1

I'm going to use more standard language below:

• What you call "denumerable" is countably infinite.

• What you call "non-denumerable" is finite or uncountable.

So the result you're trying to prove can be rephrased via contraposition as:

Suppose $\Gamma$ is satisfiable. Then either $\Gamma$ has a finite model or $\Gamma$ has an uncountable model.

The reason we want to split "non-denumerable" up into the two components "finite" and "uncountable" is that they behave totally differently.

Specifically, we argue as follows. Suppose $\Gamma$ is satisfiable. Let $\mathcal{M}\models\Gamma$. If $\mathcal{M}$ is finite we're done; otherwise, upwards Lowenheim-Skolem applies and there is an uncountable $\mathcal{N}\equiv\mathcal{M}$ (and hence $\mathcal{N}\models\Gamma$).

The key point is that Lowenheim-Skolem can only be applied to an infinite structure, so we can't use it to argue that every satisfiable theory has an uncountable model (consider $\{\forall x,y(x=y)\}$).

---

Meanwhile, your specific argument is incorrect. It seems you're trying to pass from "$\Gamma$ has a finite model" to "$\Gamma$ has an infinite model" via the construction of $\Gamma^+$ and compactness. However, this doesn't work: $\Gamma^+$ need not be satisfiable. (Again, consider $\Gamma=\{\forall x,y(x=y)\}$.)

What is true by compactness is the following:

If $\Gamma$ has arbitrarily large finite models, then $\Gamma$ has an infinite model.

Finally, it's worth noting that we can have a theory which has a finite model, does not have arbitrarily large finite models, but does have infinite models: take e.g. the axiom $\lambda$ for a linear order with no greatest element and consider $$\{\lambda\vee\forall x,y(x=y)\}.$$

(Incidentally, things are simpler when we restrict attention to complete theories. If $\Gamma$ is complete, then either $\Gamma$ has exactly one finite model up to isomorphism or $\Gamma$ has only infinite models.)