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\newpage
\section{Type Spaces}

\begin{fcdefnstar}[]
  Let $\mathcal{M}$ be an $\mathcal{L}$-structure,
  $A \subseteq M$. Given an $\mathcal{L}_A$-formula
  $\varphi(x_1, \ldots, x_n)$, define
  \[
    [\varphi(x_1, \ldots, x_n)]
    = \mathcloze{\{p \in \Snm_n^{\mathcal{M}}(A) : \varphi(x_1,
    \ldots, x_n) \in p\}}
  .\]
\end{fcdefnstar}

We have the following basic properties:
\begin{enumerate}[(i)]
  \item $\Snm_n^{\mathcal{M}}(A) = [\bigwedge_{i =
    1}^n x_i = x_i]$.
  \item $[\varphi(\ol{x}) \wedge \varphi(\ol{x})]
    = [\varphi(\ol{x})] \cap [\psi(\ol{x})]$.
  \item $[\neg \varphi(\ol{x})] =
    \Snm_n^{\mathcal{M}}(A) \setminus
    [\varphi(\ol{x})]$.
\end{enumerate}
We define a topology on $\Snm_n^{\mathcal{M}}(A)$
(``the logic topology'') by taking
$[\varphi(\ol{x})]$ for all
$\mathcal{L}_A$-formulas $\varphi(\ol{x})$ as a
basis of open sets.

\begin{fcthm}[]
  \label{thm:14.1}
  $\Snm_n^{\mathcal{M}}(A)$ is a \cloze{totally
  disconnected compact Hausdorff space.}
\end{fcthm}

\begin{proof}
  Showing that it is a topology is left as an
  exercise.

  \textbf{Hausdorff:} Fix $p, q \in
  \Snm_n^{\mathcal{M}}(A)$ distinct. Then there is
  a $\varphi(\ol{x})$ $\mathcal{L}_A$-formula such
  that $\varphi(\ol{x}) \in p$ and $\neg
  \varphi(\ol{x}) \in q$. Then $p \in
  [\varphi(\ol{x})]$ and $q \in [\neg
  \varphi(\ol{x})]$ -- these are disjoint.

  \textbf{Compactness:} Sufficient to consider
  open covers consisting of basic open sets.
  SUppose we have $\mathcal{L}_A$-formulas
  $(\varphi_i(\ol{x}))_{i \in I}$ such that
  $\Snm_n^{\mathcal{M}}(A) = \bigcup_{i \in I}
  [\varphi_i(\ol{x})]$. Let $\Sigma = \{\neg
  \varphi_i(\ol{x}) : i \in I\}$.

  Claim: $\Sigma \cup \Th_A(\mathcal{M})$ is
  inconsistent.

  Proof of claim: Otherwise $\mathcal{N} \models
  \Th_A(\mathcal{M})$, $\ol{a} \in N^n$ such that
  $\ol{a} \models \Sigma$. Let $p =
  \tp^{\mathcal{N}}(\ol{a} / A) \in
  \Snm_n^{\mathcal{M}}(A)$. But $p \notin
  [\varphi_i(\ol{x})] ~\forall i \in I$,
  contradiction.

  So by compactness we have finite $I_0 \subseteq
  I$ with
  \[
    \{\neg \varphi_i(\ol{x}) : i \in I_0\} \cup
    \Th_A(\mathcal{M})
    \tag{$*$}
    \label{lec13eq1}
  \]
  inconsistent.

  Claim: $\Snm_n^{\mathcal{M}}(A) = \bigcup_{i \in
  I_0} [\varphi_i(\ol{x})]$.

  Proof: Fix $p \in \Snm_n^{\mathcal{M}}(A)$. We
  can realise $p$ in some $\mathcal{N} \models
  \Th_A(\mathcal{M})$, i.e. we have $\ol{a} \in
  N^n$ with $\ol{a} \models p$.

  By \eqref{lec13eq1}, we have $\mathcal{N}
  \models \varphi_i(\ol{a})$ for some $i \in I_0$.
  So $\varphi_i(\ol{x}) \in p$, so $p \in
  [\varphi_i(\ol{x})]$.

  \textbf{Totally disconnected:} In a compact
  Hausdorff space, we have totally disconnected
  if and only if two points are separated by a
  clopen set. All basic sets are clopen, so
  totally disconnected.
\end{proof}

\newpage
\section{Saturated Models}

\begin{fcdefn}[]
  \label{defn:15.1}
  \glsadjdefn{ksat}{saturated}{model}%
  Let $\mathcal{M}$ be an infinite
  $\mathcal{L}$-structure, and $\kappa >
  |\mathcal{L}| + \aleph_0$. We say $\mathcal{M}$
  is $\kappa$-saturated if for any $A \subseteq
  M$ with $|A| < \kappa$, every type in
  $\Snm_n^{\mathcal{M}}(A)$ is realised in
  $\mathcal{M}$ for all $n$.
\end{fcdefn}

\begin{remark}
  \label{remark:15.2}
  \phantom{}
  \begin{enumerate}[(i)]
    \item Restricting to complete types is not
      important, as every \gls{ntype} over $A$
      with respect to $M$ can be extended to a
      complete type.
    \item It suffices to assume $n = 1$ to prove
      $\kappa$-\glsref[ksat]{saturation}.
    \item If $\mathcal{M}$ is $\kappa$-\gls{ksat}
      then $|M| \ge \kappa$. $\{x \neq a : a \in
      M\}$ is a consistent
      \glsref[ntype]{$1$-type} over $M$ in
      $\mathcal{M}$.
  \end{enumerate}
\end{remark}

\begin{fcdefn}[Partial elementary / homogeneous]
  \glsadjdefn{partelem}{partial
  elementary}{function}%
  \glsadjdefn{homog}{homogeneous}{model}%
  \label{defn:15.3}
  Let $\mathcal{M}, \mathcal{N}$ be
  $\mathcal{L}$-structures, $A \subseteq M$, $B
  \subseteq N$. A function $f : A \to B$ is
  \emph{partial elementary} if for every
  $\mathcal{L}$-formulas $\varphi(x_1, \ldots,
  x_n)$ and $a_1, \ldots, a_n \in A$ we have
  \[
    \mathcal{M} \models \varphi(\ol{a})
    \iff
    \mathcal{N} \models \varphi(f(\ol{a}))
  .\]
  Given $\kappa \ge |\mathcal{L}| + \aleph_0$,
  $\mathcal{M}$ is \emph{$\kappa$-homogeneous} if for any
  $A \subseteq M$ with $|A| < \kappa$, any partial
  elementary map $f  A \to M$ and any $c \in M$
  there is some $d \in M$ with $f \cup \{(c, a)\}$
  partial elementary. In other words, ``partial
  elementary maps can be extended''.
\end{fcdefn}

For the rest of this section, assume $T$ to be a
complete $\mathcal{L}$-theory with infinite
models.

\begin{fcdefn}[]
  \label{defn:15.4}
  Define $S_n(T) =
  \Snm_n^{\mathcal{M}}(\emptyset)$ for any / some
  $\mathcal{M} \models T$ (because if $M, N
  \models T$, then
  $\Snm_n^{\mathcal{N}}(\emptyset) =
  \Snm_n^{\mathcal{M}}(\emptyset)$ as
  $\Th(\mathcal{M}) = \Th(\mathcal{N}) = T$).
\end{fcdefn}

\begin{fcprop}[]
  \label{prop:15.5}
  Assuming:
  - $T$ a complete $\mathcal{L}$-theory with infinite models
  - $\mathcal{M} \models T$
  Then:
  \begin{iffc}
    \lhs
    $\mathcal{M}$ is
    $\aleph_0$-\gls{ksat}
    \rhs
    $\mathcal{M}$ is $\aleph_0$-\gls{homog} and
    $\mathcal{M}$ realises all types in $S_n(T)$, $n
    \ge 1$.
  \end{iffc}
\end{fcprop}

\begin{iffproof}
  \rightimpl
  Assume $\mathcal{M}$ is $\aleph_0$-\gls{ksat}.
  In particular, $\mathcal{M}$ realises all types
  in $S_n(T)$ ($\emptyset$ is finite).

  Fix some finite $A \subseteq M$, and a
  \gls{partelem} map $f : A \to M$. (Aim: find $d
  \in M$ such that $f \cup \{(c, d)\}$ is
  \gls{partelem}). Given $c \in M$, define $p \in
  \Snm_1^{\mathcal{M}}(f(A))$ to be such that
  \[
    \varphi(x, f(\ol{a}))
    \iff
    \mathcal{M} \models \varphi(c, \ol{a})
  \]
  (for all $\varphi(y, \ol{x})$
  $\mathcal{L}$-formulas and $\ol{a} \in A$).
  Write $p = f(\tp^{\mathcal{M}}(c / \ol{a}))$.

  To show $p \in \Snm_1^{\mathcal{M}}(f(A))$,
  consider $\varphi(x, \ol{a}) \in p$. Then
  $\mathcal{M} \models \varphi(c, \ol{a})$, so
  $\mathcal{M} \models \exists x, \varphi(x,
  \ol{a})$, os $\mathcal{M} \models \exists
  \varphi(x, f(\ol{a}))$ as $f$ is \gls{partelem}.

  So $p$ is finitely satisfiable, and completeness
  follows from $\tp^{\mathcal{M}}(c / A)$ being
  complete. So as $\mathcal{M}$ is
  $\kappa$-\gls{ksat} we have $d \models p$ for
  some $d \in M$. Then $f \cup \{(c, d)\}$ is a
  \gls{partelem} map.

  \leftimpl
  Fix $a_1, \ldots, a_n \in M$, $p \in
  \Snm_1^{\mathcal{M}}(\{a_1, \ldots, a_n\})$. We
  want to show $p$ is realised in $\mathcal{M}$.

  Set
  \[
    q = \{\varphi(y, x_1, \ldots, x_n) :
    \varphi(y, \ol{a}) \in p\} \in \Snm_{n +
    1}^{\mathcal{M}}(T)
  .\]
  So by assumption, we have $d, b_1, \ldots, b_n$
  with $(d, \ol{b}) \models q$.

  Consider $\tp(\ol{a} / \emptyset) = \tp(\ol{b} /
  \emptyset)$. So $f : b_i \to a_i$ is
  \gls{partelem}. Let $c \in M$ such that $f \cup
  \{(d, c)\}$ is \gls{partelem}. Then
  $\tp^{\mathcal{M}}(c, \ol{a}) =
  \tp^{\mathcal{M}}(d, \ol{b})$. So $(c, \ol{a})
  \models q$ and $\tp(c / \ol{a}) = p$.
\end{iffproof}
