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  Options to think about:
  \begin{enumerate}[(a)]
    \item Every $U \in \mathcal{U}$ contains an
      even number.
    \item Every $U \in \mathcal{U}$ contains an
      odd number.
    \item There is a $U \in \mathcal{U}$ all even.
    \item There is a $U \in \mathcal{U}$ all odd.
  \end{enumerate}
  Consider $b = (1, 1, 1, \ldots) \in \prod_{n \in
  \Nbb} C_n$.

  Suppose (a), so for every $K \in \mathcal{U}$ we
  have $i \in K$ even, i.e. $b_i \notin
  I_R(C_1)$, so $I_R(\mathcal{C}) \neq
  \mathcal{C}$. Note (a) is equivalent to (c).

  By similar reasoning, (b) implies
  $I_R(\mathcal{C}) = \mathcal{C}$ (and also (b)
  is equivalent to (c)).
\end{example}

\newpage
\section{\L o\'s's Theorem and Consequences}

Question, how does $\varphi \left( \prod_{j \in J}
\mathcal{M}_j / \mathcal{U}\right)$ relate to
$[\varphi(\mathcal{M}_j)_{j \in J}]$?

\begin{fcthm}[Los Lemma]
  \label{thm:7.1}
  Assuming:
  - $\mathcal{L}$ a language
  - $\varphi$ an $\mathcal{L}$-formula
  - $(\mathcal{M}_j)_{j \in J} = (M_j, I_j)_{j \in
  J}$ a non-empty family of
  $\mathcal{L}$-structures
  - $\mathcal{U}$ an \gls{ultra} on $J$
  - $\mathcal{M} = \left( \prod_{j \in J}
  M_j / \mathcal{U}, I_{\mathcal{U}} \right)$
  Then:
  \[
    \varphi(\mathcal{M}) =
    [\varphi(\mathcal{M}_j)_{j \in J}]
  .\]
\end{fcthm}

\begin{proof}[Proof (sketch)]
  Induction on
  \begin{itemize}
    \item Complexity of terms
    \item Formulas
  \end{itemize}
  (essentially \cref{prop:5.8}).
\end{proof}

\begin{fccoro}[]
  \label{coro:7.2}
  Assuming:
  - $\sigma$ an $\mathcal{L}$-sentence
  - $(\mathcal{M}_j)_{j \in J}$ a family of
  non-empty $\mathcal{L}$-structures
  - $\mathcal{U}$ an \gls{ultra} on $J$
  - $\mathcal{M} = \prod_{j \in J} \mathcal{M}_j /
  \mathcal{U}$
  Then:
  \begin{iffc}
    \lhs
    $\mathcal{M} \models \sigma$
    \rhs
    $\{j \in J : \mathcal{M}_j \models \sigma\}
    \in \mathcal{U}$.
  \end{iffc}
\end{fccoro}

\begin{iffproof}
  \rightimpl
  Suppose $\mathcal{M} \models \sigma$. Then
  $\sigma(\mathcal{M})$ is a non-empty subset of
  $\mathcal{M}^0$, i.e. its the empty tuple (). So
  $\{[(\sigma(\mathcal{M}_j))_{j \in J}]\} =
  \{()\}$. So
  \[
    \{j \in J : () \in \sigma(\mathcal{M}_j)\}
    =
    \{j \in J : \mathcal{M}_j \models \sigma\}
  .\]
  Since the LHS is in $\mathcal{U}$, we get that
  the RHS is too.

  \leftimpl
  Similar
\end{iffproof}

\begin{fcthm}[Compactness -- ultraproduct proof]
  Assuming:
  - $\mathcal{L}$ a language
  - $\Sigma$ a set of $\mathcal{L}$-sentences
  Then:
  \begin{iffc}
    \lhs
    $\Sigma$ is consistent
    \rhs
    every finite subset of $\Sigma$ is consistent.
  \end{iffc}
\end{fcthm}

\begin{iffproof}
  \rightimpl
  Clear.

  \leftimpl
  Assume every finite subset of $\Sigma$ is
  consistent. Let $J$ be the set of all finite
  subsets of $\Sigma$. For each $j \in J$, let
  \[
    \hat{j}
    = \{k \in J : j \subseteq k\}
  .\]
  Let $\mathcal{B} = \{\hat{j} : j \in J\}$ and
  let
  \[
    \mathcal{F} = \{A \subseteq J : \exists B \in
    \mathcal{B}, B \subseteq A\}
  .\]
  Exercise: $\mathcal{F}$ is a \gls{filt}.

  Let $\mathcal{U}$ be an \gls{ultra} extending
  $\mathcal{F}$. For each $j \in J$, let
  $\mathcal{M}_j \models j$. Let $\mathcal{M} =
  \left( \prod_{j \in J} \mathcal{M}_j /
  \mathcal{U} \right)$.

  Claim: $\mathcal{M} \models \Sigma$.

  Let $\sigma \in \Sigma$. Then $\{\sigma\} \in
  J$ and $\widehat{\{\sigma\}} \subseteq \{j \in J
  : \mathcal{M}_j \models \sigma\}$.

  So $\{j \in J : \mathcal{M}_j \models \sigma\}
  \in \mathcal{U}$, so $\mathcal{M} \models
  \sigma$.

  So: $\forall \sigma \in \Sigma, \mathcal{M}
  \models \sigma$.
\end{iffproof}

\newpage
\section{More Constructions}

Let $\mathcal{L}$ be a language, $\mathcal{M}$ an
$\mathcal{L}$-structure. Fix a collection
$(\mathcal{M}_i)_{i \in I}$ of \glspl{substruc} of
$\mathcal{M}$. Let $N = \bigcap_{i \in I} M_i$,
and assume $N$ is non-empty.

Then we have a canonical $\mathcal{L}$-structure,
with universe $N$ and interpretiation:
\begin{itemize}
  \item For $f$ a function, $f^N = f^{\mu} |_N$
    (which equals $f^{\mathcal{M}_i}|_N$ for each
    $i \in I$)
  \item For $R$ an $n$-ary relation, $R^N =
    R^{\mathcal{M}} \cap N^n$ (which equals
    $R^{\mathcal{M}_i}$ for each $i \in I$)
  \item For $c$ a constant, $c^N =
    c^{\mathcal{M}}$ (which equals
    $c^{\mathcal{M}_i}$ for each $i \in I$)
\end{itemize}
Note $N$ is also a \gls{substruc}.

\begin{fcdefn}[Generated by]
  \label{defn:8.1}
  Given an $\mathcal{L}$-structure $\mathcal{M}$,
  a non-empty $A \subseteq M$, the \emph{\gls{substruc}
  generated by $A$} is the intersection of all
  \glspl{substruc} containing $A$.
\end{fcdefn}

\begin{fcdefn}[Chain, Elementary chain]
  \label{defn:8.2}
  Let $\alpha$ be a limit ordinal.

  A collection $(\mathcal{M}_i)_{i < \alpha}$ of
  $\mathcal{L}$-structures is a \emph{chain} if
  $\mathcal{M}_i \subseteq \mathcal{M}_j$
  (\gls{substruc}) for all
  $i < j$, and is an \emph{elementary chain} if
  $\mathcal{M}_i \esub \mathcal{M}_j$ for all $i <
  j$.

  If $(\mathcal{M}_i)_{i < \alpha}$ is a chain
  then $\bigcup_{i < \alpha} \mathcal{M}_j$ is a
  well-defined $\mathcal{L}$-structure.
\end{fcdefn}

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