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% 04/02/2025 11AM

% Part 2: Ultraproducts

\newpage
\section{Filters}

\begin{fcdefn}[Filter]
  \label{defn:4.1}
  \glsnoundefn{filt}{filter}{filters}%
  Let $J$ be a set. A \emph{filter} $\mathcal{F}$ on $J$ is
  a non-empty subset of $\mathcal{P}(J)$ such that:
  \begin{itemize}
    \item $\emptyset \notin \mathcal{F}$.
    \item $\forall A, B \in \mathcal{F}, A \cap
      \in \mathcal{F}$
      (``closed under finite intersections'').
    \item $\forall A \in \mathcal{F}$, if $A \subseteq B
      \subseteq B \subseteq J$, then $B \in
      \mathcal{F}$ (``closed under super set'').
  \end{itemize}
\end{fcdefn}

\begin{example}
  \label{eg:4.2}
  \phantom{}
  \begin{itemize}
    \item For $J$ infinite,
      \[
        \mathcal{F} \defeq \{A \subseteq J : J
        \setminus A \text{ is finite}\}
      \]
      is a \gls{filt}.
    \item For $J$ non-empty, and any $i \in J$,
      \[
        \mathcal{F} = \{A \subseteq J : i \in J\}
      \]
      is a \gls{filt}.
  \end{itemize}
\end{example}

\begin{fcdefn}[Ultrafilter]
  \label{defn:4.3}
  \glsnoundefn{ultra}{ultrafilter}{ultrafilters}%
  Let $J$ be an infinite set and $\mathcal{F}$ a
  \gls{filt} on $J$. We say $\mathcal{F}$ is an
  \emph{ultrafilter} if every \gls{filt} $\mathcal{G}$
  on $J$ satisfying $\mathcal{F} \subseteq
  \mathcal{G}$ also satisfies $\mathcal{G} =
  \mathcal{F}$.
\end{fcdefn}

\begin{fcprop}[]
  \label{prop:4.4}
  Assuming:
  - $J$ a set
  - $\mathcal{F}$ a \gls{filt} on $J$
  Then:
  \begin{iffc}
    \lhs
    $\mathcal{F}$ is an \gls{ultra}
    \rhs
    for every $A \subseteq J$ either $A \in
    \mathcal{F}$ or $J \setminus A \in
    \mathcal{F}$.
  \end{iffc}
\end{fcprop}

\noproof

\begin{proof}
  \es{1}.
\end{proof}

\begin{fcprop}[]
  \label{prop:4.5}
  Assuming:
  - $J$ is a set
  - $\mathcal{F}$ a \gls{filt} on $J$
  Then:
  there is an \gls{ultra} $\mathcal{U}$ such that
  $\mathcal{F} \subseteq \mathcal{U}$.
\end{fcprop}

\begin{proof}[Proof (sketch)]
  Let $\mathcal{X}$ be the set of \glspl{filt}
  extending $\mathcal{F}$, and partially order it
  by inclusion. Note every chain has an upper
  bound (take the union), so by Zorn's lemma we
  have a maximal element, which is a \gls{ultra}
  (by definition).
\end{proof}

\newpage
\section{Ultraproducts}

\begin{definition}[]
  \label{defn:5.1}
  Let $(A_j)_{j \in J}$ be a family of sets ($J
  \neq \emptyset$),  with $A_j \neq \emptyset$ for
  all $j \in J$. Take $\mathcal{J}$ to be an
  \gls{ultra} on $J$, and define the following
  equivalence relation $\sim$ on $\prod_{j \in J}
  A_j$:
  \[
    (a_j)_{j \in J} \sim (b_j)_{j \in J}
    \qquad \text{iff} \qquad
    \{j \in J : a_j = b_j \in \mathcal{U}\}
  .\]
\end{definition}

\begin{proposition}[]
  \label{prop:5.2}
  The relation $\sim$ defined above is an
  equivalence relation on $\prod_{j \in J} A_j$.
\end{proposition}

\begin{proof}
  Symmetric / reflexive is obvious.

  Transitivity: let $(a_j)_{j \in J}, (b_j)_{j \in
  J}, (c_j)_{j \in J} \in \prod_{j \in J} A_j$,
  and suppose $(a_j)_{j \in J} \sim (b_j)_{j \in
  J}$ and $(b_j)_{j \in J} \sim (c_j)_{j \in J}$.

  Let
  \begin{align*}
    F_{ab}
    &= \{j \in J : a_j = b_j\} \\
    F_{bc}
    &= \{j \in J : b_j = c_j\} \\
    F_{ac}
    &= \{j \in J : a_j = c_j\}
  \end{align*}
  Note $F_{ab}, F_{bc} \in \mathcal{U}$. Also,
  \[
    \ub{F_{ab} \cap F_{bc}}_{\in \mathcal{U}}
    = \{j \in J : a_j = b_j = c_j\}
    \subseteq F_{ac}
  \]
  hence $F_{ac} \in \mathcal{U}$, i.e. $(a_j)_{j
  \in J} \sim (c_j)_{j \in J}$.
\end{proof}

\begin{fcdefn}[]
  \label{defn:5.3}
  Let $(A_j)_{j \in J}$ be a non-empty family of
  non-empty sets and $\mathcal{U}$ an \gls{ultra}
  on $J$.
  \begin{citems}
    \item Write $\prod_{j \in J} A_j /
      \mathcal{U}$ to be $\prod_{j \in J} A_j /
      \sim$ (where $\sim$ is defined as in
      \cref{defn:5.1}).
    \item $[(a_j)_{j \in J}]_{\mathcal{U}}$ is the
      equivalence class of $(a_j)_{j \in J}$ with
      respect to $\sim$.
    \item Let $B_j \subseteq A_j$ for every $j \in
      J$. Then
      \[
        [(B_j)_{j \in J}]_{\mathcal{U}}
        = \left\{[(a_j)_{j \in J}] \in \prod_{j \in J}
        A_j / \mathcal{U} : \{j \in J : a_j \in
        B_j\} \in \mathcal{U}\right\}
      .\]
  \end{citems}
\end{fcdefn}

Is the third item well-defined?

\begin{fcprop}[]
  \label{prop:5.4}
  Assuming:
  - $(A_j)_{j \in J}$, $(B_j)_{j \in J}$ satisfy
  $B_j \subseteq A_j$ for all $j$
  - $(a_j)_{j \in J}, (b_j)_{j \in J} \in \prod_{j
  \in J} A_j$ satisfying $(a_j)_{j \in J} \sim
  (b_j)_{j \in J}$
  Then:
  \begin{iffc}
    \lhs
    $\{j \in J : a_j \in B_j\} \in \mathcal{U}$
    \rhs
    $\{j \in J : b_j \in B_j\} \in \mathcal{U}$.
  \end{iffc}
\end{fcprop}

\begin{proof}
  Know $(a_j)_{j \in J} \sim (b_j)_{j \in J}$.
  Define
  \begin{align*}
    U
    &= \{j \in J : a_j = b_j\} \in \mathcal{U} \\
    V
    &= \{j \in J : a_j \in B_j\} \\
    W
    &= \{j \in J : b_j \in B_j\}
  \end{align*}
  Note $U \cap V \subseteq W$ and $U \cap W
  \subseteq V$.

  If $V \in \mathcal{U}$ then $U \cap V \in
  \mathcal{U}$ so $W \in \mathcal{U}$. Similarly,
  if $W \in \mathcal{U}$ then $V \in \mathcal{U}$.
\end{proof}

So $[(B_j)_{j \in J}]$ is well-defined.

\begin{fcprop}[]
  \label{prop:5.5}
  Assuming:
  - $(A_j)_{j \in J}$, $\mathcal{U}$, $\sim$ as
  usual.
  - $B_j, C_j \subseteq A_j$
  Thens:[(1)]
  - $[(B_j)_{j \in J}] \cap [(C_j)_{j \in
  J}] = [(B_j \cap C_j)_{j \in J}]$.
  - $[(B_j)_{j \in J}] \cup [(C_j)_{j \in
  J}] = [(B_j \cup C_j)_{j \in J}]$.
  - $[(B_j)_{j \in J}] \setminus [(C_j)_{j \in
  J}] = [(B_j \setminus C_j)_{j \in J}]$.
\end{fcprop}

\noproof

\begin{proof}
  \es{1}.
\end{proof}
