# If $\phi$ is satisfiable and $\mathscr{S}$ is countable, then the set of all models of $\phi$ has the cardinality of the continuum

I have just started reading Chang and Keisler and I'm already stuck in an exercise. Let $\mathscr{S}$ be a countable set of sentence letters (i.e. $\mathscr{S} = \{S_0, S_1, S_2, \dots\}$ or some such). Let $\phi$ be a wff built in the usual way (using negation and/or conjunction). Suppose $\phi$ is satisfiable, i.e. it has at least one model. On p. 17, exercise 1.2.7, they ask us to prove that the set of all models of $\phi$ has the cardinality of the continuum. Now, I'm a little bit confused by this. It seems obvious that, if $\phi$ is valid, then, indeed, the set of all models of $\phi$ is the cardinality of the continuum (there are two possible valuations for each sentence letter, thus the set of all models has cardinality $2^{|\mathscr{S}|}$). But if $\phi$ is only satisfiable, I don't see how the result follows. What am I missing?

Hint: Sentence symbols/letters which do not occur in $\phi$ play no role in determining the truth or falsity of $\phi$ in a model.
• So the idea is: $\phi$ is necessarily composed of a finite number of sentence letters. Let $S_i$ be a sentence letter which does not occur in $\phi$. We know that $\phi$ is satisfiable, i.e. true in at least one model. Call such a model $A$. Now, given each $S_i$, we can extend $A$ to $A'$ such that it either values $S_i$ as true or false. Since there are countably many $S_i$s, this means that $A$ can be extended to $2^{|\mathscr{S}|}$ models. – Nagase Mar 23 '14 at 1:33
• @Nagase: Yup! (Or, as I might put it, let $\mathscr{S}^\prime$ denote all of the sentence symbols from $\mathscr{S}$ which do not appear in $\phi$, and let $A_0$ be a subset of the sentence symbols from $\mathscr{S}$ which do appear in $\phi$ such that $A_0 \models \phi$. Then $A_0 \cup A \models \phi$ for all $A \subseteq \mathscr{S}^\prime$.) – user642796 Mar 23 '14 at 5:54