# Why can any type be realized?

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?

-

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.

-
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 $L'=L\cup\{c_a: a\in A\}\cup\{c_{v_1},...c_{v_n}\}$.

-

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.

-
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