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\newpage
\section{Categoricity}

\begin{fcdefn}[$kappa$-categorical]
  \label{eg:3.1}
  % FIXME
  \glsnoundefn{kcat}{categorical}{categorical}%
  An $\mathcal{L}$-theory is
  \emph{$\kappa$-categorical} if it has a unique
  model of size $\kappa$ up to isomorphism.
\end{fcdefn}

For now, assume our theories have infinite models
and that $\kappa \ge \aleph_0 + |\mathcal{L}|$.

\begin{example}
  \label{eg:3.2}
  \phantom{}
  \begin{itemize}
    \item $\Th(\Nbb)$ when $\mathcal{L} =
      \emptyset$ is $\kappa$-\gls{kcat} for all
      $\kappa$.
    \item $\Th(\Qbb, +, 0)$ in
      $\mathcal{L}_{\text{group}}$ is
      $\kappa$-\gls{kcat} if and only if $\kappa =
      \aleph_0$.
    \item $\Th(\Qbb, <)$ in
      $\mathcal{L}_{\text{o}}$ is
      $\kappa$-\gls{kcat} if and only if $\kappa =
      \aleph_0$.
    \item $\Th(\Zbb, +, 0)$ in
      $\mathcal{L}_{\text{groups}}$ is
      $\kappa$-\gls{kcat} for no $\kappa$.
  \end{itemize}
\end{example}

So we can find four different cases...
surprisingly this is all.

\begin{fcthmstar}[Morley's Categoricity Theorem 1965]
  Assuming:
  - $T$ is a \gls{compt} theory in a countable
  language
  - $T$ is $\kappa$-\gls{kcat} for some
  uncountable $\kappa$
  Then:
  it is $\kappa$-\gls{kcat} for all uncountable
  $\kappa$.
\end{fcthmstar}

\noproof

We do not prove this theorem in this course. The
statement is examinable, but the proof is not.

\subsubsection*{Dense linear orders (with no
endpoints)}

\begin{fcdefn}[Theory of dense linear orders]
  \glsnoundefn{dlo}{DLO}{NA}%
  \label{defn:3.3}
  Let $\mathcal{L} = \{<\}$. We define the theory
  in axioms:
  \begin{cenum}[(i)]
    \item \emph{Irreflexive:} $\forall x, \neg (x < x)$.
    \item \emph{Transitive:} $\forall x, \forall
      y, \forall z, ((x < y
      \wedge y < z) \to x < z)$.
    \item \emph{Antisymmetric:} $\forall x,
      \forall y, (x \neq y \to (x <
      y \vee y < x))$.
    \item \emph{Dense:} $\forall x, \forall y, (x < y \to (\exists
      z (x < z < y))$.
    \item \emph{No endpoints:} $\forall x, \exists
      y, \exists z, (z < x < y)$.
  \end{cenum}
\end{fcdefn}

\begin{note*}
  \gls{dlo} is consistent, because $(\Qbb, <)
  \models \text{\gls{dlo}}$.
\end{note*}

\begin{fcthm}[Cantor 1895]
  \label{thm:3.4}
  \gls{dlo} is $\aleph_0$-\gls{kcat}.
\end{fcthm}

\begin{proof}
  Let $\mathcal{M}, \mathcal{N} \models
  \text{\gls{dlo}}$ with $M, N$ countable. We need
  to construct an $\mathcal{L}$-\gls{iso} $h : M
  \to N$, i.e. an order preserving bijection.

  We will use the \emph{back and forth method}.

  Let $M = \{a_1, a_2, \ldots\}$ and $N = \{b_1,
  b_2, \ldots\}$.

  We construct a series of functions $(h_n)_{n =
  0}^\infty$ such that:
  \begin{enumerate}[(i)]
    \item $h_n : X_n \to Y_n$ is an
      order-preserving bijection.
    \item $X_n \subseteq X_{n + 1}, Y_n \subseteq
      Y_{n + 1}, h_n \subseteq h_{n + 1}$ for each
      $n$.
    \item $a_n \in X_n$, $b_n \in Y_n$.
  \end{enumerate}
  Once we have done this, $h = \bigcup_{n =
  0}^\infty h_n$ is an order-preserving bijection
  $h : M \to N$ (i.e. an $\mathcal{L}$-\gls{iso}).

  Use induction.

  Base case: $X_0 = \{x_0\}$, $Y_0 = \{b_0\}$,
  $h_0(a_0) = b_0$.

  Inductive step: Suppose $h_n : X_n \to Y_n$ as
  required.

  ``Forth'': Construct an order preserving
  bijection $h_* : X_* \to Y_*$ extending $h_n$
  with $a_{n + 1} \in X_*$. Enumerate $X_n = \{x_1,
  \ldots, x_k\}$ with $x_1 <^M x_2 <^M x_3 <^M
  \cdots <^M x_k$. Let $y_i = h(x_i)$ so that $y_1
  <^M y_2 <^M \cdots <^M y_k$.
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/17cf53da753c4b3b.png}
  \end{center}
  Define $h_* = h_n \cup \{(a_{n + 1}, b)\}$ where
  $b \in N$ is chosen according to the following
  cases:
  \begin{itemize}
    \item If $a_{n + 1} = x_i$ for some $i \in \{1,
      \ldots, k\}$, then put $b = x_i$.
    \item If $a_{n + 1} < x_i$ for all $i \in \{1,
      \ldots, k\}$ then choose $b$ such that $b <
      y_i$ for all $i \in \{1, \ldots, k\}$
      (possible since no endpoints).
    \item If $x_i < a_{n + 1}$ for all $i \in \{1,
      \ldots, k\}$, then choose $b$ such that $y_i
      < b$ for all $i \in \{1, \ldots, k\}$
      (possible since no endpoints).
    \item If there is some $i \in \{1, \ldots,
      k - 1\}$ such that $x_i < a_{n + 1} < x_{i +
      1}$, then choose $b$ such that $y_i < b <
      y_{i + 1}$ (possible since $M$ is dense).
  \end{itemize}
  Then $h_*$ is an order-preserving bijection and
  $a_{n + 1} \in X_{n + 1}$ as desired.

  ``Back'': We need to construct an
  order-preserving map $h_{n + 1} \to Y_{n + 1}$
  extending $h_*$ with $b_{n + 1} \in Y_{n + 1}$.

  Exercise.

  Then $h_{n + 1}$ satisfies the conditions.
\end{proof}

\begin{note*}
  We used that $N, M$ were countable.
\end{note*}

The theory \gls{dlo} is not uncountably
\gls{kcat}:

Consider $(\Rbb, <)$, and
consider $\Rbb \times \Qbb$ with the lexicographic
order ($(a, b) < (c, d)$ if and only if $a < c$ or
$a = c$ and $b < d$). These are both models of
\gls{dlo} (and have the same cardinality), but are
not isomorphic (e.g. because the first does not
have any countable intervals, or because the
second does not have all bounded suprema).

\begin{fccoro}[]
  \label{coro:3.5}
  \gls{dlo} is \gls{compt}.
\end{fccoro}

\begin{proof}
  No finite models (because of the no end points
  axiom).

  If $\mathcal{M}, \mathcal{N} \models
  \text{\gls{dlo}}$, with both countable, then $M
  \mcong N$ and hence $M \ee N$.

  So by \nameref{thm:1.8}, \gls{dlo} is
  \gls{compt}.
\end{proof}