\chapter{Numbers of countable models}


Our ultimate aim is to show that
\begin{equation}\label{eqn:ITN}
  I(T,\varN)\neq 2
\end{equation}
whenever $T$ is a countable, \emph{complete} theory.  The proof will
require several interesting general results.

Note that proving \eqref{eqn:ITN} requires $T$ to be complete:

\begin{example}
  Let $P$ be a singulary predicate, and in the signature $\{\lang\}$,
  let $T$ be axiomatized by
  \begin{equation*}
    \Forall x\Forall y(Px\land Py\to x=y).
  \end{equation*}
Then $T$ has non-isomorphic countably infinite models
$(\varN,\emptyset)$ and 
$(\varN,\{0\})$, and every countably infinite model is isomorphic to
one of these.
\end{example}

\section{Three models}

In the signature $\{<\}\cup\{c_n:n\in\varN\}$, let $T_3$ be the theory
axiomatized by
\begin{equation*}
  \TO^*\cup\{c_{n+1}<c_n:n\in\varN\}.
\end{equation*}
We shall see that $T_3$ is complete, and $I(T_3,\varN)=3$.
Let
  \begin{align*}
    A_0&=\{a\in\Q:0<a\}=\Q\cap(0,\infty),\\
A_1&=\Q\setminus\{0\},\\
A_2&=\Q.
  \end{align*}
Then each $A_k$ is the
universe of a model $\str A_k$ of $T_3$, where $<^{\str A_k}$ is the
usual ordering $<$, and
\begin{equation*}
  c_n{}^{\str A_k}=\frac 1{n+1}.
\end{equation*}
Then the set $\{c_n{}^{\str A_k}:n\in\varN\}$, in $\str A_k$,
\begin{mylist}
  \item
has no lower bound, if $k=0$;
\item
has a lower bound, but no infimum, if $k=1$;
\item
has an infimum, if $k=2$.
\end{mylist}
Hence the three structures are not isomorphic.  However, we shall be
able to show:
\begin{mylist}
\item
if $\str B\models T_3$ and is countable, then $\str B\cong\str
A_k$ for some $k$ in $3$;
  \item
$T_3$ is complete.
\end{mylist}
The proof of the first claim will be by the \defn{back-and-forth}
method.  The following gives the prototypical example:

\begin{theorem}\label{thm:Cantor}
  $\TO^*$ is $\varN$-categorical.
\end{theorem}

\begin{proof}
  Suppose $\str A,\str B\models\TO^*$ and $\size A=\varN=\size B$.  We
  shall show $\str A\cong\str B$.  

We can enumerate the universes:
\begin{equation*}
  A=\{a_n:n\in\varN\},\qquad B=\{b_n:n\in\varN\}.
\end{equation*}
We shall recursively define an order-preserving bijection $h$ from $A$
to $B$.  In particular, $h$ will be $\bigcup\{h_n:n\in\varN\}$, where,
notationally, we shall have
\begin{equation*}
  h_n=\{(a_k,b_k'):k<n\}\cup\{(a_k',b_k):k<n\}.
\end{equation*}
We let $h_0=\emptyset$.
Suppose we have $h_n$ so that the tuples
\begin{equation*}
  (a_0,a_0',\dots,a_{n-1},a_{n-1}'), 
\quad\text{ and }\quad (b_0',b_0,\dots,b_{n-1}',b_{n-1})
\end{equation*}
have the same \defn{order-type}.  This means that, if we write these tuples
as $(c_0,\dots,c_{2n-1})$ and $(c_0',\dots,c_{2n-1}')$ respectively,
then
\begin{equation*}
  c_i<c_j\Iff c_i'<c_j'
\end{equation*}
for all $i$ and $j$ in $2n$.  Since $\str B$ is a dense total order
without endpoints, we can chose $b_n'$ so that
\begin{equation*}
  (a_0,a_0',\dots,a_{n-1},a_{n-1}',a_n) \quad\text{ and }\quad
  (b_0',b_0,\dots,b_{n-1}',b_{n-1},b_n') 
\end{equation*}
have the same order-type.  Likewise, we can choose $a_n'$ so that
\begin{equation*}
  (a_0,a_0',\dots,a_n,a_n'), \quad\text{ and }\quad
  (b_0',b_0,\dots,b_n',b_n) 
\end{equation*}
have the same order-type.  Now let
$h_{n+1}=h_n\cup\{(a_n,b_n'),(a_n',b_n)\}$. 
\end{proof}

\begin{corollary}\label{cor:T3-3}
  $I(T_3,\varN)=3$.
\end{corollary}

\begin{proof}
Suppose $\str B$ is a countable model of $T_3$.  The
interpretation in $\str B$ of each formula
\begin{equation*}
  c_{n+1}<x\land x<c_n
\end{equation*}
is (when equipped with the ordering induced from $\str B$) a countable
model of $\TO^*$.  The same
is true for the formula $c_0<x$.  Finally, the set
\begin{equation*}
  \bigcap_{n\in\varN}\{b\in B:b<c_n\}
\end{equation*}
is one of the following:
\begin{mylist}
  \item
empty;
\item
a countable model of $\TO^*$;
\item
a countable dense total order with a greatest point, but no least
point. 
\end{mylist}
Then the previous theorem allows us to construct an isomorphism
between $\str B$ and $\str A_0$, $\str A_1$ or $\str A_2$
respectively.
\end{proof}

The following is really a corollary of Theorem~\ref{thm:TO-QE}:

\begin{theorem}\label{thm:T3-QE}
  $T_3$ admits elimination of quantifiers.
\end{theorem}

\begin{proof}
  Any formula $\phi(\tuple x)$ of $\{<,c_0,c_1,\dots\}$ can be
  considered as
  \begin{equation*}
    \theta(\tuple x,c_0,\dots,c_{n-1})
  \end{equation*}
for some formula $\theta$ of $\{<\}$.  By quantifier-elimination in
$\TO^*$, there is an open formula $\alpha$ of $\{<\}$ such that
\begin{equation*}
  \TO^*\models\Forall{\tuple x}\Forall {\tuple y}(\theta(\tuple
  x,\tuple y)\land\bigwedge_{i<n}y_{i+1}<y_i\iff \alpha(\tuple
  x,\tuple y)).
\end{equation*}
But $T_3\models c_{i+1}<c_i$, and $T_3\models\TO^*$; so 
\begin{equation*}
  T_3\models\Forall {\tuple x}(\theta(\tuple x,\tuple
  c)\iff\alpha(\tuple x,\tuple c)).
\end{equation*}
Thus $T_3$ admits quantifier-elimination.
\end{proof}

\begin{corollary}
  $T_3$ is complete.
\end{corollary}

\begin{proof}
  The three countable models $\str A_k$ form a chain:
  \begin{equation*}
    \str A_0\included\str A_1\included\str A_2.
  \end{equation*}
%By the comment preceeding Theorem~\ref{thm:T3-QE}, 
But here $\diag\str B\models\Th{\str B_B}$ for all models $\str B$ of
$T_3$, so by Theorem~\ref{thm:isom},
the chain is
elementary:
\begin{equation*}
  \str A_0\elsub\str A_1\elsub\str A_2.
\end{equation*}
In particular, the three structures are elementarily equivalent.  Now,
if $\str B$ is an arbitrary model of $T_3$, then it is infinite, so
$\str B\equiv\str C$ for some countably infinite structure $\str C$ by
Theorem~\ref{thm:LST}.  But $\str C\cong\str A_k$ for some $k$, by
Corollary~\ref{cor:T3-3}.   Hence $\str B\equiv\str A_0$ by
Theorem~\ref{thm:isom}.   Thus
\begin{equation*}
  T_3\models\Th{\str A_0};
\end{equation*}
so $T_3$ is complete.
\end{proof}

\section{Omitting types}

Since there is a sound, complete proof-system for first-order logic,
we may say that a set of sentences \defn{is consistent} to mean that
it has a model.   


An \defn{$n$-type} of a signature $\lang$ is a set of $n$-ary formulas
of $\lang$.  

An $n$-type $\Phi$ of $\lang$ is \defn{realized} by $\tuple a$ in an
$\lang$-structure $\str A$ if
\begin{equation*}
  \str A\models\phi(\tuple a)
\end{equation*}
for all $\phi$ in $\Phi$.  A type not realized in a structure is
\defn{omitted} by the structure.

If a consistent theory $T$ of $\lang$ is specified, then
an \defn{$n$-type of $T$} is an $n$-type $\Phi$ that is
\defn{consistent with $T$}:  This means that $\Phi$ is realized in
some model of $T$.  Equivalently, it means that, if $\tuple c$ is an
$n$-tuple of new constants, then the set
\begin{equation*}
  T\cup\{\phi(\tuple c):\phi\in\Phi\}
\end{equation*}
is consistent.
  By Compactness, for $\Phi$ to be consistent with
$T$, it is sufficient that
\begin{equation*}
  T\cup\{\Exists{\tuple x}\bigwedge\Phi_0\}
\end{equation*}
be consistent for all finite subsets $\Phi_0$ of $\Phi$.

By Compactness also, for \emph{any} collection of
types consistent with $T$, there is a model of $T$ in which all of the
types are realized.

An $n$-type $\Phi$ of $T$ is \defn{isolated} in $T$ by an $n$-ary
formula $\psi$ if:
\begin{mylist}
  \item
$T\cup\{\Exists{\tuple x}\psi\}$ is consistent;
\item
$T\models\Forall{\tuple x}(\psi\to\phi)$
for all $\phi$ in $\Phi$.
\end{mylist}
Hence, if $\psi$ is satisfied by $\tuple a$ in a model of $T$, then
$\tuple a$ realizes $\Phi$.  Also, if $T$ is complete, then
$T\models\Exists{\tuple x}\psi$, so $\Phi$ is realized in \emph{every}
model of $T$.

We can call a theory \defn{countable} if its signature is countable.
(A more general definition is possible: $T$ is \tech{countable} if, in
its signature, only countably many formulas are inequivalent in $T$.)
It turns out that, in a \emph{countable} theory, being isolated is
the only barrier to being omitted by some model:

\begin{theorem}[Omitting Types]
  Suppose $T$ is a countable theory, and $\Phi$ is a non-isolated
  $1$-type of $T$.  Then $\Phi$ is omitted by some countable model of
  $T$.
\end{theorem}

\begin{proof}
  We adjust our proof of the Compactness Theorem.  As there, we
  introduce a set $C$ of new constants $c_n$ (where $n\in\varN$).  We
  enumerate $\Sn[\lang\cup C]$ as $\{\sigma_n:n\in\varN\}$.  We
  construct a chain
  \begin{equation*}
    T=\Sigma_0\included\Sigma_1\included\dotsb
  \end{equation*}
as follows.  Assume $\Sigma_{3n}$ is consistent.  Then let
\begin{equation*}
  \Sigma_{3n+1}=
  \begin{cases}
    \Sigma_{3n}\cup\{\sigma_n\},&\text{ if this is consistent;}\\
    \Sigma_{3n},                &\text{ otherwise.}
  \end{cases}
\end{equation*}
Now let
\begin{equation*}
  \Sigma_{3n+2}=
  \Sigma_{3n+1}\cup\{\phi(c_k)\},
\end{equation*}
where $k$ is minimal such that $c_k$ does not appear in
  $\Sigma_{3n+1}$, if $\sigma_n\in\Sigma_{3n+1}$ and $\sigma_n$ is
  $\Exists x\phi$; otherwise,
$\Sigma_{3n+2}=  \Sigma_{3n+1}$.  Finally, let
\begin{equation*}
  \Sigma_{3n+3}=\Sigma_{3n+2}\cup\{\lnot\psi(c_n)\},
\end{equation*}
where $\psi$ is an element of $\Phi$ such that
$\Sigma_{3n+2}\cup\{\lnot\psi(c_n)\}$ is consistent.   But we have to
check that there \emph{is} such a formula $\psi$ in $\Phi$.  \emph{If}
there is, then we can let
\begin{equation*}
  \Sigma^*=\bigcup_{n\in\varN}\Sigma_n.
\end{equation*}
Then $\Sigma^*$ has a countable model $\str A$ (as in the proof of
Compactness) such that every element of $A$ is $c^{\str A}$ for some
$c$ in $C$.  But by construction, no such element can realize $\Phi$;
so $\str A$ omits $\Phi$.

Now, in the definition of $\Sigma_{3n+3}$, the formula $\psi$ exists
as desired because the set $\Sigma_{3n+2}\setminus T$ can be assumed to be
\emph{finite}.  In particular, the formulas in this set use only
finitely many constants from $C$.  We may assume that these constants
form a tuple $(c_n,\tuple d)$.  Then we can write
$\bigwedge\Sigma_{3n+2}\setminus T$ as a sentence
\begin{equation*}
  \phi(c_n,\tuple d),
\end{equation*}
where $\phi$ is a certain formula of $\lang$.  Now, if
\begin{equation*}
  \Sigma_{3n+2}\models\psi(c_n)
\end{equation*}
for some formula $\psi$, then
\begin{equation*}
  T\models(\phi(c_n,\tuple d)\to\psi(c_n)),
\end{equation*}
hence
\begin{equation*}
  T\models\Forall x(\Exists {\tuple y}\phi(x,\tuple y)\to\psi(x)). 
\end{equation*}
Since $\Phi$ is not isolated in $T$, it is not isolated by
$\Exists{\tuple y}\phi$.  Therefore the set
$\Sigma_{3n+2}\cup\{\lnot\psi(c_n)\}$ must be consistent for some
$\psi$ in $\Phi$. 
\end{proof}

In the proof, it is essential that $\Sigma_n\setminus T$ is finite;
the proof can't be generalized to the case where $T$ is uncountable.
But the proof \emph{can} be generalized to yield the following:

\begin{porism}
  Suppose $T$ is a countable theory, and $\Phi_k$ is an $n$-type of
  $T$ for some $n$ (depending on $k$), for each $k$ in $\varN$.  Then
  $T$ has a countable model omitting each $\Phi_k$.
\end{porism}

An $n$-type $\Phi$ of a theory $T$ is called \defn{complete} if
\begin{equation*}
  \phi\notin\Phi\Iff\lnot\phi\in\Phi
\end{equation*}
for all $n$-ary formulas $\phi$ of $\lang$.  Any $n$-tuple $\tuple a$
of elements of a model $\str A$ of $T$ determines a complete $n$-type
of $T$, namely
\begin{equation*}
  \{\phi:\str A\models\phi(\tuple a)\};
\end{equation*}
this is the \defn{complete type of $\tuple a$ in $\str A$} and can be
denoted
\begin{equation*}
  \tp{\str A}{\tuple a}.
\end{equation*}
If $\Phi$ is an arbitrary $n$-type of $T$, then some $\tuple a$ from
some model $\str A$ of $T$ realizes $\Phi$, and therefore
\begin{equation*}
  \Phi\included  \tp{\str A}{\tuple a}.
\end{equation*}
In particular, every type of $T$ is included in a complete type of $T$.

The set of complete $n$-types of $T$ can
be denoted
\begin{equation*}
  \ts nT;
\end{equation*}
then we can let $\bigcup_{n\in\varN}\ts nT$ be denoted
\begin{equation*}
  \ts{}T.
\end{equation*}
So the Omitting-Types Theorem gives us that, if $T$ is countable and
$\size{\ts{}T}\leq\varN$, then $T$ has a countable model that omits
\emph{all} non-isolated types of $T$.

A structure $\str A$ that realizes only \emph{isolated} types of
$\Th{\str A}$ is called \defn{atomic}.  

\begin{examples}
  \mbox{}
  \begin{myenum}
\item
 $(\varN,',0)$ is atomic, since each element is named by a term.  For
 example, a $1$-type realized by $5$ is isolated by the formula
 $x=0'''''$. 
    \item
The theory of Example~\ref{example:binary} has \emph{no} atomic
models.
  \end{myenum}
\end{examples}


The following lemma hints at the characterization of countable atomic
models
that we shall see in the next section.

\begin{lemma}\label{lem:types-embeddings}
  If $\str A$ embeds elementarily in $\str B$, then $\str B$ realizes
  all types that $\str A$ realizes.
\end{lemma}

\begin{proof}
  Suppose $h$ is an elementary embedding of $\str A$ in $\str B$, and
  $\tuple a$ realizes the type $\Phi$ in $\str A$.  Then
  \begin{equation*}
    \{\phi(\tuple a):\phi\in\Phi\}\included\Th{\str A_A},
  \end{equation*}
so $h(\tuple a)$ realizes $\Phi$ in $\str B$ by Theorem~\ref{thm:elsub}.
\end{proof}

\section{Prime structures}

A structure is \defn{prime} if it embeds elementarily in
every model of its theory; if that theory is $T$, then the structure
is a \defn{prime model of $T$}.  (Note then that only complete
theories can have prime models, simply because the prime model is
elementarily equivalent to all other models.)

\begin{examples}
\mbox{}
\begin{myenum}
  \item
  If $T$ admits quantifier-elimination, then by
  Corollary~\ref{cor:QE-MC}, all embeddings of models of $T$ are
  elementary 
  embeddings.  Hence, for example, a countably infinite set is a prime
  model of the theory of infinite sets.  Also, $(\Q,<)$ embeds in
  every model of $\TO^*$, so it is a prime model.
\item
It is possible to show that, if $\size{\lang}\leq\kappa\leq\size B$,
then $\str B$ is an elementary extension of some structure $\str A$
such that $\size A=\kappa$.  Hence, a model of a countable theory $T$ is
prime, provided it embeds elementarily in all \emph{countable} models
of $T$.  In particular then, if $T$ is $\varN$-categorical, then its
countable model is prime.
\end{myenum}
\end{examples}

\begin{theorem}\label{thm:prime-atomic}
  Suppose $T$ is a countable complete theory.  Then the prime models
  of $T$ are precisely the countable atomic models of $T$.
\end{theorem}

\begin{proof}
Suppose $\str A\models T$.

  ($\Rightarrow$)  If $\str A$ is not countable,
  then $\str A$ cannot embed in countable models of $T$ (which must
  exist, by Theorem~\ref{thm:LST}), so $\str A$ cannot be prime.

If $\str A$ is not atomic, then $\str A$ realizes some non-isolated
type $\Phi$ of $T$.  But by the Omitting-Types Theorem, $T$ has a
countable model $\str B$ that omits $\Phi$.  Then $\str A$ cannot
embed elementarily in $\str B$, by Lemma~\ref{lem:types-embeddings}.

($\Leftarrow$)   Suppose $\str A$ is countable and atomic, and $\str
B\models T$.  We construct an elementary embedding of $\str A$ in
$\str B$ by the back-and-forth method, except that the construction is
in only one
direction. Write $A$ as
$\{a_n:n\in\varN\}$.  Then each $\tp{\str A}{a_0,\dots,a_{n-1}}$ is
isolated in $T$ by some formula $\phi_n$.  Then we have
\begin{mylist}
  \item
$T\models\Exists{\tuple x}\phi_n$;
\item
$T\models\Forall{\tuple x}(\phi_n\to\Exists {x_n}\phi_{n+1})$. 
\end{mylist}
Hence we can recursively find $b_k$ in $B$ so that
\begin{equation*}
  \str B\models\phi_n(b_0,\dots,b_{n-1})
\end{equation*}
for all $n$ in $\varN$.

Now, every sentence in $\Th{\str A_A}$ is $\theta(a_0,\dots,a_{n-1})$
for some formula $\theta$ of $\lang$.  Then
\begin{equation*}
  T\models\Forall {\tuple x}(\phi_n\to\theta),
\end{equation*}
so $\str B\models\theta(\tuple b)$.  Therefore the map $a_k\mapsto
b_k:A\to B$ is an elementary embedding of $\str A$ in $\str B$.
\end{proof}

\begin{porism}\label{por:prime-isom}
  All prime models of a countable complete theory are isomorphic.
\end{porism}

\begin{proof}
 In the proof that $\str A$ embeds elementarily in $\str B$, if we
 assume also that $\str B$ is countable and atomic, then the full
 back-and-forth method gives an isomorphism between the structures.  
\end{proof}

\begin{lemma}\label{lem:I-S}
  If $I(T,\varN)\leq\varN$, then $\size{\ts{}T}\leq\varN$.
\end{lemma}

\begin{proof}
  \textbf{Exercise.}
\end{proof}

\begin{theorem}\label{thm:prime-existence}
  Suppose $T$ is a countable complete theory.  Then $T$ has a prime
  model if $\ts{}T$ is countable.
\end{theorem}

\begin{proof}
  \textbf{Exercise.}
\end{proof}

\section{Saturated structures}

A \defn{saturated} structure is the opposite of an atomic structure.
Atomic structures realize as \emph{few} types as possible.  Saturated
structures realize as \emph{many} types as possible; moreover, these
types are allowed to have parameters from the structure. 

To be precise, let $\str M$ be an infinite $\lang$-structure, and let
$A\included M$.  In this context, the set $\ts n{\Th{\str M_A}}$ can be
denoted
\begin{equation*}
  \ts nA.
\end{equation*}
Consider the special case where $A$ is $M$ itself.  The set $\ts 1M$,
for example, contains types that include the type
\begin{equation*}
  \{x\neq a:a\in M\}.
\end{equation*}
These types cannot be realized in $\str M$.  So we say that $\str M$
is \defn{saturated}, provided that, whenever $A\included M$ and $\size
A\leq\size M$, each type in $\ts{}A$ is realized in $\str M$.  (In
particular, if
$\str M$ is countable here, then the sets $A$ should be finite.)

\begin{theorem}\label{thm:saturated}
  Suppose $T$ is countable and complete, and
  $\size{\ts{}T}\leq\varN$.  Then
  $T$ has a 
  countable saturated model.
\end{theorem}

\begin{proof}
  Suppose $\str M$ is a countable model of $T$.  If $A$ is a finite
  subset $\{a_k:k<n\}$ of $M$, then each element of $\ts mA$
  is
  \begin{equation*}
    \{\phi(x_0,\dots,x_{m-1},a_0,\dots,a_{n-1}):\phi\in p\}
  \end{equation*}
for some $p$ in $\ts{m+n}{T}$.  Hence $\size{\ts{}A}$ is countable.
Therefore the set
\begin{equation*}
  \bigcup\{\ts{}A:A\text{ is a finite subset of }M\}
\end{equation*}
is countable.  So all of the types in this set are realized in a
countable elementary extension $\str M'$ of $\str M$.

Thus, if $\str M_0$ is a countable model of $T$, then we can form an
\defn{elementary chain}
\begin{equation*}
  \str M_0\elsub \str M_1\elsub\str M_2\elsub\dotsb.
\end{equation*}
It is straightforward then to define the \defn{union} of this chain:
this is a structure $\str N$ whose universe $N$ is $\cup_{n\in\varN}M_n$,
and that is an elementary extension of each $\str M_n$.  Every finite
subset of $N$ is a subset of some $\str M_n$, and so the types of
$\ts{}A$ are realized in $\str M_{n+1}$, hence in $\str N$.  So $\str
N$ is saturated.
\end{proof}

If $A$ is a finite subset $\{a_k:k<n\}$ of $M$, and $\tuple a$ is
$(a_0,\dots,a_{n-1})$, we can denote $\str M_A$ by
\begin{equation*}
  (\str A,\tuple a).
\end{equation*}
If $\str M$ is countable, then $\str M$ is called \defn{homogeneous}
if
\begin{equation*}
  \tp{\str M}{\tuple a}=\tp{\str M}{\tuple b}\implies(\str M,\tuple
  a)\cong(\str M,\tuple b)
\end{equation*}
for all $n$-tuples $\tuple a$ and $\tuple b$ from $M$, for all $n$ in
$\varN$. 

\begin{theorem}\label{thm:homog}
  Countable saturated structures are homogeneous.
\end{theorem}

\begin{proof}
  The back-and-forth method.
\end{proof}

\section{One model}

For the sake of stating and proving the following theorem more easily,
we can use the following notation.  Suppose $T$ is a theory of $\lang$.  Then
equivalence in $T$ is an equivalence-relation on the set of $n$-ary
formulas of $\lang$.  Let the set of corresponding equivalence-classes
be denoted
\begin{equation*}
  \LT nT.
\end{equation*}


\begin{theorem}
  Suppose $T$ is a countable complete theory.  The following
  statements are equivalent:
  \begin{myenum}\setcounter{myenum}{-1}
    \item
$I(T,\varN)=1$.
\item
All types of $T$ are isolated.
\item
Each set $\LT nT$ is finite.
\item
Each set $\ts nT$ is finite.
  \end{myenum}
\end{theorem}

\begin{proof}
(0)$\Rightarrow$(1):  If $\ts{}T$ contains a non-isolated type, then it
  is realized in
  some, but not all, countable models of $T$, so $I(T,\varN)>1$.

(1)$\Rightarrow$(0):  If all types of $T$ are isolated, then all
   models of $T$ are atomic, so all \emph{countable} models of $T$ are
   prime and therefore isomorphic.

(2)$\Rightarrow$(3):  Immediate.

(3)$\Rightarrow$(1)\&(2):  Suppose $\ts nT=\{p_0,\dots,p_{m-1}\}$.  For
   each $i$ and $j$ in $m$, if $i\neq j$, then there is a formula
   $\phi_{ij}$ in $p_i\setminus p_j$.  Let $\psi_i$ be the formula
   \begin{equation*}
     \bigwedge_{j\in m\setminus\{i\}}\phi_{ij}.
   \end{equation*}
Then $\psi_i$ is in $p_j$ if and only if $j=i$.  If $\str A\models T$,
and $\tuple a$ is an $n$-tuple from $A$, then $\str A$ realizes some
unique $p_i$, and then
$\str A\models \psi_i(\tuple a)$.
Conversely, if $\str A\models\psi_i(\tuple a)$, then $\tuple a$ must
realize $p_i$.  Therefore $\psi_i$ isolates $p_i$.

If $\chi$ is an
arbitrary $n$-ary formula, let $I=\{i\in m:\chi\in p_i\}$.  Then
\begin{equation*}
  T\models\Forall{\tuple x}(\chi\iff \bigvee_{i\in I}\psi_i).
\end{equation*}
There are only finitely many possibilities for $I$, so $\LT nT$ is
finite. 

(1)$\Rightarrow$(3):  Suppose infinitely many complete $n$-types are isolated
in $T$.  Since $T$ is countable, there must be countably many such
types.  Say they compose the set
   $\{p_k:k\in\varN\}$, and each $p_k$
   is isolated by $\phi_k$.  Then the type
   \begin{equation*}
     \{\lnot \phi_k:k\in\varN\}
   \end{equation*}
is consistent with $T$.  It is not included in any of the $p_k$, so it
must be included in a non-isolated type.
\end{proof}

\section{Not two models}

\begin{theorem}
  Suppose $T$ is a countable complete theory.  Then $I(T,\varN)\neq2$.
\end{theorem}

\begin{proof}
  Suppose if possible that $T_2$ has just two non-isomorphic countable
  models.  One of them, $\str A$, is prime, by
Lemma~\ref{lem:I-S} and Theorem~\ref{thm:prime-existence}.  The other
  one, $\str B$, is
  saturated, by Theorem~\ref{thm:saturated}.  Since $\str A$ embeds
  elementarily in $\str B$, we may assume $\str A\elsub\str B$.

Since $\str A\not\cong\str B$, there is a non-isolated type $\Phi$
realized by some $\tuple b$ in $\str B$, by
Theorem~\ref{thm:prime-atomic} and Porism~\ref{por:prime-isom}. 
Let $T^*=\Th{\str B,\tuple b}$.  Suppose $(\str C,\tuple c)$ is a
countable model of
$T^*$.  Then $\str C\models T_2$, so $\str C$ is
isomorphic to $\str A$ or $\str B$.  In any
case, $\str A$ embeds elementarily in $\str C$.  But $\Phi$ is
realized by $\tuple c$ in $\str C$.  Hence $\str C\cong\str B$  by
Lemma~\ref{lem:types-embeddings}.  Let the isomorphism take $\tuple c$
to $\tuple a$.  Then it is enough to show $(\str B,\tuple a)\cong(\str
B,\tuple b)$.  But this follows from Theorem~\ref{thm:homog}.
\end{proof}