Omitting types theorem for types with parameters

Question

Does the omitting types theorem as exposed e.g. in Hodges consider types with parameters or is it just about types over the empty set?

Answer 1

As it stands your question is too vague to answer. You should post the version in Hodges along with a modified statement of the theorem which reflects what you mean by "with parameters". You will have to pay attention to the cardinalities of the things involved though. For example:

1) The argument with the Henkin construction for the easiest form of the omitting types theorem works only when the language is countable.

2) The size of the model matters. e.g. Consider the $L$ structure $(\omega, <)$ with $<$ being interpreted canonically. Now there are countable models that realize and omit the type $\{x>n: n \in{\omega} \}$ (Even though it looks like I'm using parameters I'm not doing so. The $n$'s are definable by formulas: $0$ is the least element, $1$ is the least element if you disregard $0$, etc) for example $(\omega^{\smallfrown}\mathbb{Z},<)$ realizes the above type but $(\omega, <)$ omits it. However any uncountable model of $Th(\omega,<)$ will always realize this type.