Vaught's theorem on number of countable nonisomorphic models

Question

D.Marker Model Theory:
Theorem 4.4.6: $I(T,\aleph_0)\neq2$
Proof. Suppose that $I(T,\aleph_0)=2$. By Corollary 4.3.8 ii), there is $\mathcal{N}$ a prime model of $T$ and $\mathcal{M}$ a countable saturated model of $T$. Because $T$ is not $\aleph_0$-categorical, by Theorem 4.4.1, there is a nonisolated type $p \in S_n(T)$ for some $n$. The type $p$ is realized in $\mathcal{M}$ and omitted in $\mathcal{N}$. Let $a \in M$ realize $p$. Let $T^*$ be the $L_ \overline{a}$-theory of $\mathcal{M}_\overline{a}$ (in the notation of the previous lemma).
By Theorem 4.4.1, there are infinitely many $T$-inequivalent formulas in the free variables $v_1,...,v_n$. As they are still $T^*$-inequivalent, $T^*$ is not $\aleph_0$-categorical. By Lemma 4.4.5, $M_\overline{a}$ is a saturated $L_\overline{a}$-structure. Thus, by Corollary 4.3.8 i), $T^*$ has a countable atomic model $\mathcal{A}$. Let $\mathcal{B}$ denote the $L$-reduct of $\mathcal{B}$. Because $\mathcal{A} \models T^*$, $\mathcal{B}$ contains a realization of $p$, thus $\mathcal{B} \not\cong \mathcal{N}$. Because $T^∗$ is not $\aleph_0$-categorical, there is a nonisolated $L_\overline{a}-$type. This type is not realized in $\mathcal{A}$. Thus $\mathcal{A}$ is not saturated. If $\mathcal{B}$ were saturated, then, by Lemma 4.4.5, $\mathcal{A}$ would be saturated. Thus, $\mathcal{B} \not\cong \mathcal{M}$ and $I(T,\aleph_0) \geq 3$.

I have a questions on this proof, which I would be happy for your hints or answers.

Why is this true that "Because $\mathcal{A} \models T^*$, $\mathcal{B}$ contains a realization of $p$"?

Answer 1

Remember that $T^*$ is the $L_{\overline{a}}$-theory of $\mathcal{M}_{\overline{a}}$. So the language of $T^*$ contains a new sequence $\overline{c}$ of constant symbols (intended to refer to the elements of the tuple $\overline{a}$), and - among other things - contains the sentence "$\varphi(\overline{c})$" whenever $\varphi\in p$ (since whenever $\varphi\in p$, we have $\mathcal{M}_{\overline{a}}\models\varphi(\overline{a})$).

So suppose $\mathcal{A}\models T^*$. Then the tuple in $\mathcal{A}$ corresponding to the constant symbols $\overline{c}$ - call this tuple "$\overline{u}$" - satisfies each $\varphi\in p$. But these $\varphi$s are $L$-formulas! So in the reduct $\mathcal{B}$, the same tuple $\overline{u}$ still satisfies each $\varphi$. So $\overline{u}$ realizes $p$ in $\mathcal{B}$.

Remember that $\mathcal{B}$ is the same structure as $\mathcal{A}$, with the same elements - only we "forget" about the structure beyond $L$. But "realizing $p$" is purely $L$-ish; so if $\overline{u}$ realizes $p$ in $\mathcal{A}$, it still realizes $p$ in $\mathcal{B}$.

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Based on the comments below, I think it would be useful to give an example of this process.

Suppose my structure $\mathcal{M}$ is $(\mathbb{N}; +, \times, 0, 1)$. So the language has two binary function symbols, and two constant symbols, say $L=\{\oplus, \otimes, d_0, d_1\}$, interpreted in $\mathcal{M}$ in the obvious way.

And suppose my $T$ is the whole theory of $\mathcal{M}$ (that is, true arithmetic) and my $a$ (a single element for simplicity) is $17$. Then:

• $L_a$ is the same language as $L$, but with a new constant symbol $c$; so $L_a=\{\oplus, \otimes, d_0, d_1, c\}$.

• $\mathcal{M}_a$ has the same underlying set as $\mathcal{M}$, and $+$, $\times$, $0$, $1$ are interpreted in the same way; and the constant $c$ is interpreted as $17$.

• $T^*$ contains $T$, but also contains new sentences involving the symbol "$c$". For example, it contains each of the following sentences:

  • $\forall x\forall y(x\otimes y=c\implies(x=d_1\vee y=d_1))$ $\quad$("$c$ is prime")

  • $\exists x[((d_1\oplus d_1)\otimes x)\oplus d_1=c]$ $\quad$("$c$ is odd")

  • $(d_1\oplus d_1)\otimes (d_1\oplus d_1)\otimes (d_1\oplus d_1)\otimes (d_1\oplus d_1)\oplus d_1=c$ $\quad$("$c$ is seventeen"),

  because they are each true statements about $17$ in the structure $\mathcal{M}$.

Now, some other model $\mathcal{A}$ of $T^*$ will be a model of $T$, together with a "named element" $u$ (the thing named by the symbol "$c$"), and $\mathcal{A}$ will believe that $u$ satisfies every formula that $17$ did in $\mathcal{M}$ - because every such formula, with the variable replaced by the symbol "$c$", is in $T^*$, which $\mathcal{A}$ is a model of. For instance, the formula "$x$ is odd" will be true, in $\mathcal{A}$, of $u$, because the sentence "$c$ is odd" is in $T^*$, and in $c^\mathcal{A}=u$.

So in particular, in the reduct $\mathcal{B}$ of $\mathcal{A}$, the element $u$ (which is still in $\mathcal{B}$ - the underlying set of the structure hasn't changed at all, only the language is different) satisfies every formula that $17$ satisfies in $\mathcal{M}$. So - if $p$ is the type of $17$ in $\mathcal{M}$ - $u$ realizes $p$ in $\mathcal{B}$.