The question follows: "Prove the following, by using a compactness argument: If T has an infinite model, then for every cardinal k, T has a model of cardinality at least k."
I don't get what the compactness argument part wants, and I feel like using upward Löwenheim-Skolem theorem is enough, since it gives us elementary extensions of this model (M) for every cardinality larger than the max of the cardinalities of the model M and the vocabulary.
Is my reasoning incorrect? Can I not use Löwenheim-Skolem here?
Best Answer
Let $M$ be an infinite model of $T$. Let $\kappa$ be any cardinal and let $\{c_i : i < \kappa\}$ be a set of new constant symbols. Define $$ \Sigma = \{c_i \neq c_j : i < j < \kappa\}. $$ We will prove that $T \cup \Sigma$ is satisfiable. Let $\Sigma_0 \subseteq \Sigma$ be finite. Then only a finite number of the new constant symbols is mentioned in $\Sigma_0$. By interpreting these as distinct elements in $M$, which we can do because $M$ is infinite, we get that $M \models T \cup \Sigma_0$. So every finite subset of $T \cup \Sigma$ is satisfiable, and so by compactness $T \cup \Sigma$ itself is satisfiable. Let $N$ be a model of $T \cup \Sigma$. Then $N$ is a model of $T$ and because $N \models \Sigma$ it has at least $\kappa$ distinct elements.
Bonus exercise: can you adjust the proof so that $N$ is actually an elementary extension of $M$?