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\begin{fcdefn}[]
  \label{defn:5.6}
  Let $(A_j)_{j \in J}$, $\mathcal{U}$
  (\gls{ultra} on $J$), $\sim$ as before, and let
  $n \in \Nbb$.

  For each $j \in J$, suppose $B_j \subseteq
  A_j^n$. Define
  \begin{align*}
    [(B_j)_{j \in J}]
    &= \left\{
      ([(a_j^1)_{j \in J}], \ldots, [(a_j^n)_{j
      \in J}]) \in \left( \prod_{j \in J} A_j /
      \mathcal{U} \right)^n
      :
      \{j \in J : (a_j^1, \ldots, a_j^n) \in B_j\}
      \in \mathcal{U}
    \right\} \\
    &\subseteq \left( \prod_{j \in J} A_j /
    \mathcal{U} \right)^n
  \end{align*}
\end{fcdefn}

\begin{note*}
  If $n = 0$, then we get that
  \begin{align*}
    B_j
    &= \{()\} \\
    [(B_j)_{j \in J}]
    &= \{()\}
  \end{align*}
\end{note*}

\begin{fcdefn}[]
  \label{defn:5.7}
  Let $n > 0$, $k \in \{1, \ldots, n\}$. Define:
  \begin{itemize}
    \item $\pi_k(x_1, \ldots, x_n) = (x_1, \ldots,
      x_{k - 1}, x_{k + 1}, \ldots, x_n)$.
    \item For $X$ a set of $n$-tuples,
      \[
        \pi_k(X) = \{(x_1, \ldots, x_{k - 1}, x_{k +
        1}, \ldots, x_n) : (x_1, \ldots, x_n) \in X\}
      .\]
  \end{itemize}
\end{fcdefn}

\begin{fcprop}[]
  \label{prop:5.8}
  Assuming:
  - $(A_j)_{j \in J}$, $\mathcal{U}$, $\sim$ as
  usual
  - $n \in \Nbb$
  - for each $j$ we have $B_j, C_j \subseteq A_j^n$
  Thens:[(1)]
  - $[(B_j)_{j \in J}] \cap [(C_j)_{j \in J}] =
  [(B_j \cap C_j)_{j \in J}]$
  - $[(B_j)_{j \in J}] \cup [(C_j)_{j \in J}] =
  [(B_j \cup C_j)_{j \in J}]$
  - $[(B_j)_{j \in J}] \setminus [(C_j)_{j \in J}] =
  [(B_j \setminus C_j)_{j \in J}]$
  - If $n > 0$, $k \in \{1, \ldots, n\}$ then
  \[
    \pi_k([(B_j)_{j \in J}])
    = [(\pi_k(B_j))_{j \in J}]
  .\]
\end{fcprop}

\begin{proof}
  (1) - (3): Straightforward. See \es{}.

  (4): Let $n > 0$, $k \in \{1, \ldots, n\}$.
  Let
  \[
    \alpha = ([(a_j^1)_{j \in J}], \ldots,
    [(a_j^{k - 1})_{j \in J}], [(j_j^{k + 1})_{j
    \in J}], \ldots, [(a_j^n)_{j \in J}])
    \in \pi_k([(B_j)_{j \in J}])
  .\]
  So we have some $[(a_j^k)_{j \in J}]$ such that
  \[
    ([(a_j^1)_{j \in J}], \ldots, [(a_j^n)_{j \in
    J}])
    \in [(B_j)_{j \in J}]
  .\]
  So $U = \{j \in J : (a_j^1, \ldots, a_j^n) \in
  B_j\} \in \mathcal{U}$.

  Consider
  \[
    V = \{j \in J : (a_j^1, \ldots, a_j^{k - 1},
    a_j^{k + 1}, \ldots, a_n) \in \pi(B_j)\}
    \supseteq \mathcal{U}
  .\]

  So $\alpha \in [(\pi_k(B_j))_{j \in J}]$, i.e.
  $\pi_k[(B_j)_{j \in J}] \subseteq
  [(\pi_k(B_j))_{j \in J}]$.

  Showing $[(\pi_k(B_j))_{j \in J}] \subseteq
  \pi([(B_j)_{j \in J}])$ is similar (do it as an
  exercise).
\end{proof}

\begin{fcdefn}[]
  \label{defn:5.9}
  Suppose $A_j = A$ for each $j \in J$. Then we
  call $\prod_{j \in J} A_j / \mathcal{U}$ an
  \emph{ultrapower} of $A$, and write
  $A^{\mathcal{U}}$.

  If $B  \subseteq A$, we write $[B]$ for $[(B)_{j
  \in J}]$.
\end{fcdefn}

\begin{fcthm}[]
  \label{thm:5.10}
  Assuming:
  - $A^{\mathcal{U}}$ be as above
  - $J = \Nbb$
  - $\{C \subseteq \Nbb : \Nbb \setminus C \text{
  is finite}\} \subseteq \mathcal{U}$
  - $n \in \Nbb$
  - for each $m \in \Nbb$, let $B_m \subseteq A^n$
  satisfying:
  \begin{enumerate}[(1)]
    \item $[B_m] \neq \emptyset ~\forall m \in
      \Nbb$
    \item $[B_k] \subseteq [B_m]$ for all $m, k
      \in \Nbb$ with $m \le k$
  \end{enumerate}
  Then:
  $\bigcap_{m \in \Nbb} [B_m] \neq \emptyset$.
\end{fcthm}

\noproof

\begin{proof}
  Omitted. For $n = 1$, this is a potential
  presentation topic.
\end{proof}

\newpage
\section{Ultraproduct Structures}

\begin{fcdefn}[]
  \label{defn:6.1}
  Let $\mathcal{M}_j = (M_j, I_j)$ be
  $\mathcal{L}$-structures for each $j \in J$. Let
  $\mathcal{U}$ be an ultraproduct on $J$. Define
  an interpretation $I_{\mathcal{U}}$ of
  $\mathcal{L}$ on $\prod_{j \in J} M_j /
  \mathcal{U}$. Let $S \in \mathcal{L}$.
  \begin{citems}
    \item If $S$ is an $n$-ary relation:
      \[
        I_{\mathcal{U}}(S)
        = [(I_j(S))_{j \in J}]
        \subseteq \left( \prod_{j \in J} M_j / \mathcal{U} \right)^n
      .\]
    \item If $S$ is a constant:
      \[
        I_{\mathcal{U}}(S)
        = [(I_j(S))_{j \in J}]
        \in \prod_{j \in J} M / \mathcal{U}
      .\]
    \item Functions are a bit less clear. However,
      we can always turn a function into a
      relation by looking at its graph (i.e. $f :
      M^n \to M$ has graph $R_f = \{(\ol{x}, y)
      \in M^{n + 1} : f(\ol{x}) = y\}$).

      So if $S$ is a function: for each $j \in J$,
      define the graph of $I_j(S)$ as
      \[
        G_j(S)
        = \{(a_1, \ldots, a_n, b) \in M_j^{n + 1}
        : I_j(S)(a_1, \ldots, a_n) = b\}
      .\]
      Then $[(G_j(S))_{j \in J}]$ is the graph of
      a function
      \[
        \left( \prod_{j \in J} M_j / \mathcal{U}
        \right)^n
        \to
        \prod_{j \in J} M_j / \mathcal{U}
      .\]
      (Checking this is left as an exercise). Now
      define $I_{\mathcal{U}}(S)$ to be the
      function corresponding to $[(G_j(S))_{j \in
      J}]$.
  \end{citems}
\end{fcdefn}

\begin{example}
  \label{eg:6.2}
  $\mathcal{L} = \{+, R\}$ (where $+$ is a
  function and $R$ is a unary relation). Let $\mathcal{C}_n
  = (C_n, I_n)$ with $C_n = \Zbb / n\Zbb$, with
  addition modulo $n$, and let $I_n(R) = \{x \in
  C_n : \exists y \in C_n, 2y = x\}$.

  Consider $\mathcal{C} = \left( \prod_{n \in
  \Nbb_{>0}} C_n / \mathcal{U}, I_{\mathcal{U}}
  \right)$.

  What does the set $I_{\mathcal{U}}(R)$ look
  like?

  If $\gcd(n, 2) = 1$, then $I_n(R) = C_n$.

  If $\gcd(n, 2) = 2$, then $I_n(R) \neq C_n$ (for
  example, $1 \notin I_n(R)$).

  \begin{itemize}
    \item If $\mathcal{U} = \{A \subseteq \Nbb : 3 \in
      A\}$, then $\mathcal{C} \cong \mathcal{C}_3$, so
      $I_{\mathcal{U}}(R) = \mathcal{C}$.
    \item Homework: Can you think of two
      non-principal \glspl{ultra}, with one of
      them having $I_{\mathcal{U}}(R) =
      \mathcal{C}$, and the other having
      $I_{\mathcal{U}}(R) \neq \mathcal{C}$.
  \end{itemize}