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I couldn't find this question asked previously, which means it's probably an especially daft question.

Given an $\mathcal{L}$-structure $\mathcal{M}$, my textbook defines an $n$-type over $A\subseteq M$ to be a set $p$ of sentences all in the same $n$ free variables such that $p\cup Th_A(\mathcal{M})$ is satisfiable. ($Th_A(\mathcal{M})$ is here the complete theory of $\mathcal{M}$ considered as a structure in the language $\mathcal{L}\cup\{c_a: a\in A\}$.)

The proof in the book proceeds by showing that since the union of $p$ and the elementary diagram of $\mathcal{M}$ is satisfiable, there's a model $\mathcal{N}$ into which $\mathcal{M}$ is elementarily embedded and which obviously satisfies $p\cup Th_A(\mathcal{M})$. This much I understand perfectly.

The (imo, important) step of showing that there's an $\overline{a}\in N^n$ such that it satisfies every formula in $p$ is sort of brushed over. "Now let $c_i\in N$ be the interpretations of $v_i$. Then $(c_1,\ldots,c_n)$ is a realization of $p$." (This is David Marker; Chang & Keisler are even less helpful.)

I have clearly misunderstood something important; I know what the interpretation of $v_i$ with respect to a sequence $\overline{a}\in N^m$ for $m > i$ is, but I don't see a warrant in the proof or in the definitions surrouding interpretation for such a thing as "the" intepretation of a free variable. Without such a thing, though, I'm not sure what actually guarantees realization of $p$.

So, what's the step I'm missing here?

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3 Answers 3

up vote 2 down vote accepted

It all just hinges on the definition of satisfiable formula. Note that you misquoted Marker (you wrote sentence instead).

There is of course only one sensible possible definition (which I didn't manage to locate in Marker, unfortunately):

A set of formulae $\Phi$ in the free variables $(x_n)_n$ is satisfiable if there exist a model $\mathcal M$ and elements $(m_n)_n \in M$ such that for each $\phi \in \Phi$: $$\mathcal M \models \phi(m_1,\ldots,m_n)$$ (where $n$ is the highest index of a variable occurring in $\phi$).

Now the desired existence of $\bar a$ is an immediate consequence of the definition.

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Hm, so this makes more sense, but now I fail to see how $p\cup Th_A(M)$ is satisfiable in this sense (given what I have in Marker). Is @azarel's answer the appropriate one in this case? –  Malice Vidrine Nov 28 '13 at 18:05
@MaliceVidrine Both answers are essentially the same. Since we can (by definition of $n$-type) realise $p\cup Th_A(\mathcal M)$ in $\mathcal M$ (say by $(m_n)_n$), it must also be realised in $N$, by $(\psi m_n)_n$ (if $\psi:\mathcal M\to\mathcal N$ is the elementary embedding). Effectively this works by the passage to the elementary diagram of $\mathcal M$: we have $\mathcal M \models \phi(c_{m_1},\ldots, c_{m_n})$ for each $\phi \in \Phi$ (with $c_m$ the constant for $m$ in the elementary diagram). –  Lord_Farin Nov 28 '13 at 18:12
Okay... So if $p\cup Th_A(\mathcal{M})$ implies that $\mathcal{M}$ realizes $p$, what exactly is even the import of this theorem? (This exchange has essentially unlearned me all my model theory :/) –  Malice Vidrine Nov 28 '13 at 18:27
Lord_Farin, I think @MaliceVidrine is confused because you said $p \cup Th_{A}(\mathcal{M})$ can be realized in $\mathcal{M}$, which in general isn't true. Right? –  Quinn Culver Nov 28 '13 at 22:27
@QuinnCulver: Following the gist of the responses here, I'm gathering that a type is most commonly expected to be finitely satisfied in $\mathcal{M}$ (in the sense of having its finite subsets realized), and generally realizable in some extension of $\mathcal{M}$. Does this sound like the right picture? –  Malice Vidrine Nov 28 '13 at 22:35

The definition as stated is a bit sloppy. In order to make it more precise, take new constants $c_{v_1},...,c_{v_n}$ representing the variables $v_1,..v_n$. So $p\cup Th_A(M)$ is consistent really means $\{\phi(c_{v_1},...,c_{v_n}): \phi(v_1,...,v_n)\in p\}\cup Th_A(M)$ is consistent in the language $L'=L\cup\{c_a: a\in A\}\cup\{c_{v_1},...c_{v_n}\}$.

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Since you understand that $\mathcal N$ satisfies $p\cup Th_A(\mathcal M)$, there must be at least one value assignment for the free variables in $p$ such that every formula in $p$ is true under this value assignment. That's what satisfiability means.

The sentence you quote then supposed that you choose one such value assignment and define the $c_i$s as the values the variables take in that particular value assignment. Since you're proving an existential statement, you don't have to come up with a unique $\bar a$ that satisfies every formula in $p$ -- it is enough to show that at least some $\bar a$ exists here.

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So there being a model $\mathcal{M}$ and a sequence $\overline{a}$ such that $\mathcal{M}\vDash\phi(\overline{a})$ is (hopefully) what is meant by a formula $\phi$ being satisfiable. But for a set $T$ of formula I had been lead to think that it was "$\exists\mathcal{M}\forall\phi\in T\exists\overline{a}(\mathcal{M}\vDash\phi(\overline{a}))$" that was "$T$ is satisfiable". It appears not? –  Malice Vidrine Nov 28 '13 at 18:17
@MaliceVidrine: Yes, I'd say your quantifiers on $\phi$ and $\bar a$ need to be swapped there, if it is to make sense that $T$ itself is satisfiable. It doesn't seem to be very useful to me to split the quantifiers on $\mathcal M$ and $\bar A$ away from each others. –  Henning Makholm Nov 28 '13 at 22:42
Gotcha, thank you. –  Malice Vidrine Nov 28 '13 at 22:47

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