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Does a complete type contain a complete theory i.e. are sentences considered formulas with a free variable in the definition of a complete type?

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Yes, in general the convention is that "the collection of formulas whose only free variables are $v_1,\ldots, v_n$" includes formulas where some or all of those variables don't occur. Even if it wasn't defined that way, it would effectively include them, since, e.g., for any sentence $\sigma,$ we'd have the formula $\sigma\land v_1=v_1.$

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  • $\begingroup$ Thank you. And so be extension complete $n$-types include a complete $k$-types for $k\leq n$? Or put more simply, complete $n$-types will include formulas with a subset of the free variables $v_1,\ldots v_n$? $\endgroup$ May 22 at 15:20
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    $\begingroup$ Yes. You can think about it semantically too. A complete n-type is realized by some tuple $(a_1,\ldots, a_n)$ in some model $M.$ Then you can consider the type of $(a_1,\ldots, a_k)$, etc. The type of the empty tuple would be the complete theory $\operatorname{Th}(M)$. $\endgroup$ May 22 at 22:56

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