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\begin{remark*}
  The definition of $\alpha \opl \beta$ we gave
  last time is called the ``induction
  definition''.
\end{remark*}

\begin{flashcard}[synthetic-ordadd-defn]
\begin{definition*}[Synthetic ordinal addition]
  \glssymboldefn{sopl}{$+$}{$+$}
  \glssymboldefn{wocup}{$\sqcup$}{$\sqcup$}
  \glsnoundefn{synthadd}{synthetic addition}{N/A}
  \cloze{Given \glsref[wellord]{well-ordered} sets $X,
  Y$, the disjoint union $X \sqcup Y$ is the
  \glsref[wellord]{well-ordered} set
  $\stackrel{X}{\leftrightarrow}
  \stackrel{Y}{\leftrightarrow}$. Formally, it is
  the set $X \times \{0\} \cup Y \times \{1\}$
  with ordering:
  \[ (x, i) \wlt (y, j) \iff \begin{cases}
    \text{either $i = j = 0$ and $x \wlt y$ in
    $X$} \\
    \text{or $i = j = 1$ and $x \wlt y$ in $Y$} \\
    \text{or $i = 0, j = 1$ and $x \in X$, $y \in
    Y$}
  \end{cases} \]
  So this is a \glsref[wellord]{well-ordered} set
  $Z$ which has an \gls{is} $X'$ to $X$ and $Z
  \setminus X'$ is \gls{ordisic} to $Y$. This is
  unique up to \gls{ordisism}.

  For \glspl{ordinal} $\alpha, \beta \sopl \beta =
  \alpha \wocup \beta$ (more precisely, $\alpha
  \sopl \beta$ is the \gls{ot} of $X \wocup Y$
  where $\alpha = \OT(X)$, $\beta = \OT(Y)$).}
\end{definition*}
\end{flashcard}

\begin{note*}
  $\alpha\osucc = \alpha \wocup 1$. With this
  definition, it's easy to see that $\alpha \sopl
  (\beta \sopl \gamma) = (\alpha \sopl \beta)
  \sopl \gamma$ since $(\alpha \wocup \beta)
  \wocup \gamma$ is \gls{ordisic} to $\alpha
  \wocup (\beta \wocup \gamma)$.

  Also, we can easily prove $\beta \wle \gamma
  \implies \alpha \sopl \beta \wle \alpha \sopl
  \gamma$ as $\alpha \wocup \beta$ is an \gls{is}
  of $\alpha \wocup \gamma$.
\end{note*}

\begin{flashcard}[synthetic-inductive-ordadd-equiv-prop]
\begin{proposition}
  The \glsref[indadd]{inductive} and
  \glsref[synthadd]{synthetic} definitions of
  \gls{ordinal} addition coincide.
\end{proposition}

\begin{proof}
  \newcommand\sdopl{\stackrel{.}{\sopl}}
  \cloze{Temporarily, let $\alpha \sdopl \beta$ denote
  the \glsref[synthadd]{synthetic addition}, and
  $\alpha \opl \beta$ denote the
  \glsref[indadd]{inductive addition}. We prove
  $\forall \alpha, \beta$ $\alpha \opl \beta
  \sdopl \beta$ by induction on $\beta$ (with
  $\alpha$ fixed).
  \begin{enumerate}[$\beta \neq 0$ limit:]
    \item[$\beta = 0$:] $\alpha \opl 0 = \alpha =
      \alpha \wocup 0$.
    \item[$\beta = \delta\osucc$:] $\alpha \opl
      \beta = (\alpha \opl \delta)\osucc = (\alpha
      \sdopl \delta)\osucc = (\alpha \wocup
      \delta) \wocup 1 = \alpha \wocup (\delta
      \wocup 1) = \alpha \sdopl \delta\osucc =
      \alpha \sdopl \beta$.
    \item[$\beta \neq 0$ {\glsref[limord]{limit}}:]
      \phantom{} \\[-2.5\baselineskip]
      \begin{align*}
        \alpha \opl \beta
        &= \sup\{\alpha \opl
        \gamma \st \gamma \olt \beta\} \\
        &= \sup\{\alpha \sdopl \gamma \st \gamma
        \olt \beta\} \\
        \bigcup_{\gamma \olt \beta} \alpha \wocup
        \gamma \\
        &= \alpha \wocup \bigcup_{\gamma \olt
        \beta} \gamma \\
        &= \alpha \wocup \beta \\
        &= \alpha \sdopl \beta
      \end{align*}
      (as $\alpha \wocup \gamma$, $\gamma \olt
      \beta$ are nested). \qedhere
  \end{enumerate}}
\end{proof}
\end{flashcard}

\subsubsection*{Ordinal Multiplication}

We give two definitions: inductive and syntetic.

\begin{flashcard}[inductive-mult-defn]
\begin{definition*}[Inductive multiplication]
  \glssymboldefn{ocdot}{$\cdot$}{$\cdot$}
  \cloze{Define $\alpha \cdot \beta$ by recursion on
  $\beta$ ($\alpha$ fixed):
  \begin{itemize}
    \item $\alpha \cdot 0 = 0$
    \item $\alpha \cdot \beta\osucc = \alpha \cdot
      \beta \opl \alpha$
    \item $\alpha \cdot \beta = \sup\{\alpha \cdot
      \gamma \st \alpha \olt \beta\}$ (for $\beta
      \neq 0$ \gls{limord})
  \end{itemize}}
\end{definition*}
\end{flashcard}

\begin{example*}
  For $m, n \olt \omega$, we have $m \ocdot 0 = 0$,
  $m \ocdot (n \opl 1) = m \ocdot n\osucc = m
  \ocdot n \opl m$. This gives the usual
  multiplication.
  \[ \oomega \ocdot 2 = \oomega \ocdot 1\osucc =
  \oomega \ocdot 1 \opl \oomega = \oomega \cdot
  0\osucc \opl \oomega = (\oomega \ocdot 0 \opl
  \oomega) \opl \oomega = \oomega \opl \oomega \]
  \[ 2 \ocdot \oomega = \sup\{2 \ocdot n \st n
  \olt \oomega\} = \oomega \neq \oomega \ocdot 2 \]
  So multiplication is not commutative.
\end{example*}

\begin{flashcard}[synth-mult-defn]
\begin{definition*}[Synthetic multiplication]
  \cloze{Given \glsref[wellord]{well-ordered} sets $X,
  Y$, we \glsref[wellord]{well-order} $X \times Y$
  by
  \[ (x, y) \wlt (w, z) \iff \begin{cases}
    \text{either $y = z$ and $x \wlt w$ in $X$} \\
    \text{or $y \wlt z$ in $Y$}
  \end{cases} \]
  For \glspl{ordinal} $\alpha, \beta$ define
  $\alpha \socdot \beta = \alpha \times \beta$
  (the \gls{ot} of $X \times Y$ where $X$ has
  \gls{ot} $\alpha$, $Y$ has \gls{ot} $\beta$).}
\end{definition*}
\end{flashcard}

\begin{note*}
  As before, the two definitions coincide (proof
  by inductionon $\beta$).
\end{note*}

\textbf{Properties:}
\[ \alpha \ocdot (\beta \ocdot \gamma) = (\alpha
\ocdot \beta) \ocdot \gamma \]
\[ \beta \ole \gamma \implies \alpha \ocdot \beta
\ole \alpha \ocdot \gamma \]
On \es{2}, you will check whether the following
are true:
\[ (\alpha \opl \beta) \ocdot \gamma = \alpha
\ocdot \gamma \opl \beta \ocdot \gamma \]
\[ \alpha \ocdot (\beta \opl \gamma) = \alpha
\ocdot \beta \opl \alpha \ocdot \gamma \]

\subsubsection*{Ordinal Exponentiation}

Define $\alpha^\beta$ by recursion on $\beta$
($\alpha$ fixed):
\begin{itemize}
  \item $\alpha^0 = 1$
  \item $\alpha^{\beta\osucc} = \alpha^\beta
    \ocdot \alpha$
  \item $\alpha^\beta = \sup\{\alpha^\gamma \st
    \gamma \olt \beta\}$ (for $\beta \neq 0$
    \gls{limord})
\end{itemize}

\begin{example*}
  For $m, n \olt \oomega$, $m^n$ has usual
  meaning.
  \[ \oomega^2 = \oomega^{1\osucc}  = \oomega^1
  \ocdot \oomega = \oomega^{0\osucc} \ocdot
  \oomega = (\oomega^0 \ocdot \oomega) \ocdot
  \oomega = \oomega \ocdot \oomega \]
  \[ 2^{\oomega} = \sup\{2^n \st n \wlt \oomega\}
  = \oomega \]
  which is countable!
\end{example*}

\subsubsection*{** Non-examinable **}

Let $X$ be a separable Banach space, then $X
\hookrightarrow C[0, 1]$ (universal property for
separable Banach spaces).

\textbf{Question:} Does there exist a universal
space for separable reflexive spaces?

\newcommand{\Sz}{\mathrm{Sz}}
\textbf{Answer:} No (Szlenk).

To each Banach space $X$ you associate an
\gls{ordinal} $\Sz(X)$ (Szlenk index of $X$). For
all separable $X$, $\Sz(X) \ole \omegaone$.
\[ \Sz(X) \olt \omegaone \iff X^* \text{ separable} \]
\[ X \hookrightarrow Y \implies \Sz(X) \ole \Sz(Y) \]
$\forall \alpha \olt \omegaone$, there exists
separable reflexive $X_\alpha$ such that
$\Sz(X_\alpha) \ogt \alpha$. If $Z$ is separable
reflexive and for all separable reflexive $X$, $X
\hookrightarrow Z$ then $X_\alpha \hookrightarrow
Z$ for all $\alpha \olt \omegaone$, so $\Sz(Z)
\oge \Sz(X_\alpha) \ogt \alpha$. So $\Sz(Z) =
\omegaone$, contradiction.

\textbf{This is the end of the non-examinable
part.}

\newpage
\section{Posets and Zorn's Lemma}
\label{sec3}

\begin{flashcard}[partial-order-defn]
\begin{definition*}[Partial order]
  \glsnoundefn{partord}{partial order}{partial
  orders}
  \glssymboldefn{ple}{$\le$}{$\le$}
  \cloze{A \emph{partial order} on a set $X$ is a
  relation $\le$ that is:
  \begin{enumerate}[\bfseries antisymmetric:]
    \item[\bfseries reflexive:] $\forall x \in X$, $x
      \le x$
    \item[\bfseries antisymmetric:] $\forall x, y \in
      X$, $(x \le y \wedge y \le x) \implies x =
      y$
    \item[\bfseries transitive:] $\forall x, y, z \in
      X$, $(x \le y \wedge y \le z) \implies x \le
      z$
  \end{enumerate}
  We will write $x < y$ for ``$x \le y$ and $x
  \neq y$''. This is:
  \begin{enumerate}[\bfseries irreflexive]
    \item[\bfseries irreflexive:] $\forall x$,
      $\neg(x < x)$.
  \item[\bfseries transitive:] $\forall x, y, z$,
    $((x < y) \wedge (y < z)) \implies x <
    z$.
  \end{enumerate}
  }
\end{definition*}
\end{flashcard}

\begin{flashcard}[poset-defn]
\begin{definition*}[Partially ordered set]
  \glsnoundefn{poset}{partially ordered
  set}{partially ordered sets}
  \cloze{A \emph{partially ordered set} or \emph{poset}
  is a set $X$ with a \gls{partord}.}
\end{definition*}
\end{flashcard}

\textbf{Examples:}
\begin{enumerate}[(1)]
  \item Every \glsref[linord]{linearly ordered}
    set.
  \item $\NN$ with $a \ple b \iff a \mid b$.
  \item For a set $X$ $\PP X$ with $a \ple b
    \iff a \subset b$.
  \item Every subset of a \gls{poset}: for
    example, if $G$ is a group, then
    \[ \{H \in \PP G \st \text{$H$ is a subgroup
    of $G$}\} \]
  \item Posets given by Hasse diagrams. For
    example
    \begin{center}
      \includegraphics[width=0.3\linewidth]{images/fc4f4be901ff4087.png}
    \end{center}
    $X = \{a, b, c, d, e, f\}$. $b, c \pgt a$,
    $d \pgt b, c$, $e \pgt c$, $f \pgt d, e$ and
    all relations that follow by transitivity.
    ($e \not \pgt b$, $f \pgt a$).

    In general, a Hasse diagram for a
    \gls{poset} $X$ is a grawing of elements of
    $X$ where we join $x$ to $y$ with an upward
    line if $y > x$ and $\nexists z$ with $y > z
    > x$. For example:
    \begin{center}
      \includegraphics[width=0.3\linewidth]{images/18df7313851645a3.png}
    \end{center}
  \item \phantom{}
    \begin{center}
      \includegraphics[width=0.3\linewidth]{images/51f5d36910b049e8.png}
    \end{center}
  \item \phantom{}
    \begin{center}
      \includegraphics[width=0.3\linewidth]{images/6f6f07bba80e43b5.png}
    \end{center}
  \item \phantom{}
    \begin{center}
      \includegraphics[width=0.3\linewidth]{images/27b2db064c934841.png}
    \end{center}
\end{enumerate}