Upward Löwenheim Skolem

Question

I'm trying to understand the proof of (a version) of the upward Löwenheim Skolem Theorem, which states that given a language $\mathscr{L}$ and a set of $\mathscr{L}$-sentences $\Sigma$ with a countably infinite model $\mathscr{A}$, there also exists an uncountable model of $\Sigma$.

I understand that first one has to enrich the language with uncountably many new constant-symbols $\{c_i|i\in \mathbb{R}\}$ to get a language $\mathscr{L^*}$, and then considers $\Sigma'=\Sigma \cup\{c_i\neq c_j|i\neq j\}$ and shows that it has an uncountable $\mathscr{L}^*$-model by adding new elements $\{a_i|i\in I\}$ to $\mathscr{A}$, proving that this is a $\mathscr{L}^*$model of every finite subset of $\Sigma'$, and hence due to the compactness theorem also a $\mathscr{L}^*$model of $\Sigma'$. Then the restriction of $\Sigma'$ is an $\mathscr{L}$-model of $\Sigma$ with the same cardinality as $\mathscr{A}\cup\{a_i|i\in I\}$, i.e. uncountable.

But I am missing the point where I need that there has to exist a countably infinite model in the first place. Why can't it just be finite?

Answer 1

Your reproduction of the proof is a bit off, which is why you're confused.

You're not supposed to add new elements to the model in the step where you consider the finite subsets of $\Sigma'$. The point here is that the original $\mathscr A$ can be a model of every finite subset of $\Sigma'$, without adding new individuals, because it has enough elements already to assign meanings to the $c_i$ constants such that sufficiently many of them are different. But this step only works because $\mathscr A$ is infinite; if it were finite there might not be space to make sufficiently many of the $c_i$s map to different different elements.

After you have done this, you know that $\Sigma'$ is consistent (by compactness), and the completeness theorem now gives you a new model for the entire $\Sigma'$. In this new model all the $c_i$s are different, so it must have at least the same cardinality as $I$.

(And then you may need to use the downwards Löwenheim-Skolem theorem to cut your new model down to size exactly $|I|$).

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And now you should also be able to see why the hypothesis can be weakened from "$\Sigma$ has an infinite model" to "for every finite $n$, $\Sigma$ has a model of cardinality at least $n$".

Answer 2

Consider the sentence $\forall x \forall y.x = y$. The only models of this sentence have exactly one element, so the upward Löwenheim-Skolem theorem must fail for the set comprising just this sentence. The proof breaks down because $c_1 \neq c_2$ is not satisfiable in any model of this sentence.