%! TEX root = MT.tex
% vim: tw=50
% 27/02/2025 11AM

\begin{fcdefn}[Rado graph]
  \glssymboldefn{rg}%
  \label{defn:12.4}
  Let $\mathcal{L} = \{E\}$, with $E$ a binary
  relation. A \emph{Rado} or \emph{Random} graph
  is a graph $(V, E)$ such that $V \neq \emptyset$
  and for any finite disjoint $X, Y \subseteq V$
  there is some $v \in V$ such that:
  \begin{itemize}
    \item $E(v, x)$ for all $x \in X$
    \item $\neg(v, y)$  for all $y \in Y$
  \end{itemize}
  We denote the theory of Rado graphs as
  $\mathrm{RC}$. It consists of
  \begin{itemize}
    \item Graph axioms (irreflexive and
      symmetric).
    \item For any $k \ge 1$ and $l \ge 1$,
    \[
      \forall x_1, \ldots, x_k, \forall y_1,
      \ldots, y_l
      \left( \bigwedge_{i, j} x_i \neq y_j \to
      \exists v \left( \bigwedge_{i, j} E(v, x_i)
      \wedge \neg E(v, y_j) \right) \right)
    .\]
  \end{itemize}
\end{fcdefn}

\textbf{Facts:}
\begin{itemize}
  \item $\RG$ is $\aleph_0$-\gls{kcat}.
  \item If $M \models \RG$ then every finite graph
    is an induced subgraph.
  \item Suppose $\mathcal{M}, \mathcal{N} \models
    \RG$ are two coutnable models, and $f : X \to
    Y$ is a graph isomorphism between $X \subseteq
    M$ (finite) and $Y \subseteq M$ (finite). Then
    $f$ extends to an isomorphism from
    $\mathcal{M}$ to $\mathcal{N}$.
\end{itemize}

\begin{fcthm}[]
  \label{thm:12.5}
  $\RG$ \cloze{has \gls{qe}.}
\end{fcthm}

\begin{proof}
  Option 1: Show $\RG \cup \Diag(A)$, with $A$ a
  finite graph is \gls{compt}.

  Option 2: Use (ii) of \cref{thm:11.4}. Fix
  $\mathcal{M}, \mathcal{N} \models \RG$, $A
  \subseteq \mathcal{M} \cap \mathcal{N}$. Fix a
  quantifier-free formula $\varphi(x_1, \ldots,
  x_n, y)$, $\ol{a} \in A^n$. Assume that there
  exists $b \in M$, $\mathcal{M} \models
  \varphi(\ol{a}, b)$. Want to show $\exists c \in
  N$ such that $\mathcal{N} \models
  \varphi(\ol{a}, c)$.

  Write $\varphi(\ol{x}, y)$ in disjunction normal
  form:
  \[
    \bigvee_{s = 1}^k \bigwedge_{t = 1}^{l_s}
    \theta_{s, t}(\ol{x}, y)
  \]
  where $\theta_{s, t}(\ol{x}, y)$ is atomic or
  negated atomic. There is some $s \subseteq k$
  such that $M \models \bigwedge_{t = 1}^{l_s}
  \theta_{s, t} (\ol{a}, b)$.

  Each of $\theta_{s, t}(\ol{x}, y)$ is one of
  $x_i = x_j$, $x_i = y$, $E(x_i, x_j)$, $E(x_i,
  y)$ and negations.

  If $x_i = y$ appears then $b = a_i \in A
  \subseteq \mathcal{N}$ and so $\mathcal{N}
  \models \varphi(\ol{a}, b)$.

  We may assume $x_i = y$ does not appear in
  $\theta_{s, t}$. Let
  \begin{align*}
    X
    &= \{a_i : \mathcal{M} \models E(a_i, b)\}
    \subseteq \ol{a} \\
    Y
    &= \{a_i : \mathcal{M} \models \neg E(a_i, b)\}
    \subseteq \ol{a}
  \end{align*}
  Then $X$ and $Y$ are finite and disjoint. So we
  have $c \in N$ such that
  \begin{align*}
    N
    &\models E(a_i, c)
    \qquad \forall a_i \in X \\
    N
    &\models \neg E(a_i, c)
    \qquad \forall a_i \in Y \\
    c
    &\notin \{a_i, \ldots, a_n\}
  \end{align*}
  So $\mathcal{N} \models \bigwedge_{t = 1}^{l_s}
  \theta_{s, t}(\ol{a}, c)$ and thus $\mathcal{N}
  \models \varphi(\ol{a}, c)$.
\end{proof}

% Part 4: Types

\newpage
\section{Introduction to Types}

\begin{fcdefnstar}[$L$-formula with parameters
  from $A$]
  Given a language $\mathcal{L}$, an
  $\mathcal{L}$-structure $\mathcal{M}$ and a
  subset $A \subseteq M$, we call an
  $\mathcal{L}_A$-formula an
  \emph{$\mathcal{L}$-formula with parameters from
  $A$}.

  Write these as $\varphi(\ol{x}, \ol{a})$ for
  $\varphi(\ol{x}, \ol{y})$ an
  $\mathcal{L}$-formula, and $\ol{a} \in A$
  (identify with $\ul{a}^M$).
\end{fcdefnstar}

Suppose $\mathcal{N} \esup \mathcal{M}$. What does
$\mathcal{N}$ look like from the point of
$\mathcal{M}$?

SIngle formulas don't give you much insight:
suppose $a \in \mathcal{N}$, $\mathcal{N} \models
\phi(a)$. Then there is some $a' \in \mathcal{M}$
with $\mathcal{M} \models \phi(a')$.

This changes if you consider sets of infinitely
many formulas.

\begin{notation}
  \phantom{}
  \begin{itemize}
    \item Let $p$ be a set of formulas in free variables
      $x_1, \ldots, x_n$. We often write
      $\ul{p(x_1, \ldots, x_n)}$ and $\ul{p}$
      interchangeably.
    \item Given $\mathcal{M}$ and $a_1, \ldots,
      a_n \in M$, we write $\mathcal{M} \models
      p(a_1, \ldots, a_n)$ if $\mathcal{M} \models
      \phi(a_1, \ldots, a_n)$ for every $\phi \in
      p$.
    \item We say $p$ is consistent if it is
      realised in some $\mathcal{L}$-structure.
  \end{itemize}
\end{notation}

Exercise: Show $p$ is consistent if and only if
every finite subest of $p$ is consistent (\es{2},
Q8).

\begin{fcdefn}[$n$-type]
  \glsnoundefn{ntype}{$n$-type}{$n$-types}%
  \glsadjdefn{comptype}{complete}{$n$-type}%
  \glssymboldefn{Snm}%
  \label{defn:13.2}
  Let $\mathcal{M}$ be an $\mathcal{L}$-structure
  and $A \subseteq M$. An \emph{$n$-type} over $A$
  with respect to $\mathcal{M}$ is a set of
  $\mathcal{L}$-formulas with parameters from $A$,
  in free variables $x_1, \ldots, x_n$ such that
  $p \cup \Th_A(\mathcal{M})$ is consistent.

  An $n$-type is \emph{complete} if for every
  $\mathcal{L}_A$-formula with $n$ variables
  $\phi$, either $\phi \in p$ or $\neg \phi \in
  p$.

  Let $S_n^{\mathcal{M}}(A)$ denote the set of all
  complete $n$-types over $A$ with respect to
  $\mathcal{M}$.
\end{fcdefn}

\begin{fcdefn}[$tp^M$]
  \glssymboldefn{tp}%
  \label{defn:13.3}
  Given $a_1, \ldots, a_n \in \mathcal{M}$, let
  $\mathrm{tp}^{\mathcal{M}}(a_1, \ldots, a_n / A)$ be the
  set of all $\mathcal{L}_A$-formulas $\phi(x_1,
  \ldots, x_n)$ such that $\mathcal{M} \models
  \phi(a_1, \ldots, a_n)$ (usually $a_i \notin
  A$).

  $\mathrm{tp}^{\mathcal{M}}(\ol{a} / A) \in
  \Snm_n^{\mathcal{M}}(A)$ and $\ol{a} \models
  \mathrm{tp}^{\mathcal{M}}(\ol{a} / A)$.
\end{fcdefn}

\begin{fcprop}[]
  \label{prop:13.4}
  Assuming:
  - $p \in \Snm_n^{\mathcal{M}}(A)$
  Then:
  there is $\mathcal{N} \esup \mathcal{M}$ with
  $\ol{a} \in N^n$ such that $p =
  \tp^{\mathcal{N}}(\ol{a} / A)$.
\end{fcprop}