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

Question

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?

Answer 1

Hint: Sentence symbols/letters which do not occur in $\phi$ play no role in determining the truth or falsity of $\phi$ in a model.

Answer 2

Another possibility is: If $\phi$ is valid, the result is true. If $\phi$ is only satisfiable, we can prove, using induction in the complexity of $\phi$, that the set of all models of $\phi$ is in one-to-one correspondence with the sets of all models of $\lnot \phi$. Since their union has cardinality $2^{\aleph_0}$ and each has the same cardinality, they both have the cardinality of the continuum.