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\textbf{Question:} Does there exist an uncountable
\gls{ordinal}, i.e. does there exist an
uncountable \glsref[wellord]{well-ordered} set?
Can we well order $\RR$?

\begin{flashcard}[uncountable-ordinal-existence-thm]
\begin{theorem}
  \label{uncountable_ordinal_existence}
  There exists an uncountable \gls{ordinal}.
\end{theorem}

\fcscrap{\textbf{Idea:} Assume $\alpha$ is an uncountable
\gls{ordinal}. Then there is a least such
$\alpha$:
\[ \{\beta \in \I_{\alpha} \cup\{\alpha\}
\st \text{$\beta$ uncountable}\} \neq \emptyset
,\]
so has a least element $\gamma$, say. So
$\I_\gamma$ is exactly the set of all countable
\glspl{ordinal}. If $X$ is a countable
\glsref[wellord]{well-ordered} set, then there
exists an injection $f : X \to \NN$. Then $Y =
f(X)$ is \glsref[wellord]{well-ordered} by $f(x)
\wlt f(y) \iff x \wlt y$ in $X$. Then $Y$ is
\gls{ordisic} to $X$.}

\begin{proof}
  \cloze{Let
  \[ A = \{(Y, \wlt) \in \PP\NN \times \PP(\NN
  \times \NN) \st \text{$Y$ is
  \glsref[wellord]{well-ordered} by $\wlt$}\} .\]
  Let $B = \{\OT(Y, \wlt) \st (Y, \wlt) \in A\}$.
  By above, $B$ is exactly the set of all
  countable \glspl{ordinal}. Let $\omega_1 = \sup
  B$. If $\omega_1 \in B$ then $\omega_1\osucc \notin
  B$, so $\omega_1\osucc$ is an uncountable
  \gls{ordinal}. In fact, $\omega_1$ is
  uncountable, since if $\omega_1$ is countable,
  then $\omega_1\osucc$ must be countable as well
  (countable set union with a single element is
  still countable).}
\end{proof}
\end{flashcard}

\begin{notation*}
  \glssymboldefn{omegaone}{$\omega_1$}{$\omega_1$}
  $\omega_1$ in the proof is the least uncountable
  \gls{ordinal}. In general, when we write
  $\omega_1$, we mean the least uncountable
  \gls{ordinal} (which may be constructed as in
  the previous proof).
\end{notation*}

\begin{remark*}
  Every proper \gls{is} of $\omegaone$ is
  countable. If $\alpha_1, \alpha_2, \alpha_3,
  \ldots \in \omegaone$, then
  \[ \sup\{\alpha_1, \alpha_2, \ldots\} = \OT
  \left( \bigcup_{i \in \NN} \I_{\alpha_i} \right) \]
  is countable, hence not equal to $\omegaone$.
\end{remark*}

\begin{flashcard}[hartogs-lemma-thm]
\begin{theorem}[Hartog's Lemma]
  \label{hartogs}
  \cloze{For any set $X$, there exists an \gls{ordinal}
  $\alpha$ such that $\alpha$ does not inject into
  $X$.}
\end{theorem}

\begin{proof}
  \cloze{Repeat the proof of
  \cref{uncountable_ordinal_existence} replacing
  $\NN$ with $X$.}
\end{proof}
\end{flashcard}

\begin{flashcard}[hartogs-notation]
\begin{notation*}
  \glssymboldefn{hartgamma}{$\gamma(X)$}{$\gamma(X)$}
  The least such $\alpha$ in \nameref{hartogs} is
  denoted by \cloze{$\gamma(X)$. For example
  $\gamma(\oomega) = \omegaone$.}
\end{notation*}
\end{flashcard}

\[ 0, 1, 2, \ldots, \oomega, \ldots, \eps_0 =
\omega^{\omega^{\omega^{\iddots}}}, \ldots,
\eps_1, \ldots, \eps_{\eps_{\eps_{\ddots}}}, \ldots,
\omegaone, \ldots, \omegaone \cdot 2, \ldots,
\omega_2 = \gamma(\omegaone), \ldots \]

\subsubsection*{Types of ordinals}

\begin{flashcard}[successor-limit-defn]
\begin{definition*}[Successor / limit ordinal]
  \glsnoundefn{succord}{successor
  ordinal}{successor ordinals}
  \glsnoundefn{limord}{limit ordinal}{limit
  ordinals}
  \cloze{Let $\alpha$ be an \gls{ordinal}, and consider
  whether $\alpha$ has a greatest element (i.e. if
  $X$ has \gls{ot} $\alpha$, does $X$ have a
  greatest element).
  \begin{enumerate}[If yes:]
    \item[If yes:] Let $\beta$ be the greatest
      element of $\I_\alpha$. Then $\I_\alpha =
      \I_\beta \cup \{\beta\}$. So $\alpha =
      \beta\osucc$, and $\alpha = (\sup
      \I_\alpha)\osucc$. We call such an $\alpha$
      a \emph{successor ordinal}.
    \item[If no:] Then $\I_\alpha = \sup
      \I_\alpha$, i.e. $\alpha = \sup\{\beta \st
      \beta < \alpha\}$. We say $\alpha$ is a
      \emph{limit ordinal}.
  \end{enumerate}}
\end{definition*}
\end{flashcard}

\begin{example*}
  $1 = 0\osucc$ is a \gls{succord}, $\oomega =
  \sup\{n \wlt \omega\}$ is a \gls{limord},
  $\omega\osucc$ is a \gls{succord}, $\omegaone$
  is a \gls{limord}.

  Weirdly, $0$ is a \gls{limord}. Some people
  prefer to add a special category for $0$,
  defining it as neither a \gls{succord} nor a
  \gls{limord}.
\end{example*}

\subsubsection*{Ordinal Arithmetic}

\begin{flashcard}[ordinal-addition-defn]
\begin{definition*}[Ordinal addition]
  \glssymboldefn{opl}{$+$}{$+$}
  \glsnoundefn{ordadd}{ordinal addition}{N/A}
  \glsnoundefn{indadd}{inductive addition}{N/A}
  \cloze{We define $\alpha + \beta$ for $\alpha, \beta$
  \glspl{ordinal} by recursion on $\beta$ with
  $\alpha$ fixed. We define:
  \begin{enumerate}[$\beta \neq 0$ limit:]
    \item[$\beta = 0$:] $\alpha + 0 = \alpha$,
    \item[$\beta = \gamma\osucc$:] $\alpha +
      \gamma\osucc = (\alpha + \gamma)\osucc$,
    \item[$\beta \neq 0$ {\glsref[limord]{limit}}:]
      $\alpha + \beta = \sup\{\alpha + \gamma \st
      \gamma < \beta\}$.
  \end{enumerate}}
\end{definition*}
\end{flashcard}

\begin{remark*}
  Technically, we fix $\alpha, \beta$ and define
  $\alpha + \gamma$ for all $\gamma \le \beta$ by
  \nameref{recursion} as above. We do this for all
  $\beta$. This gives a well-defined `$+$' by
  uniqueness in the \nameref{recursion}.

  Similarly, we can prove things by induction: Let
  $p(\alpha)$ be a statement for each
  \gls{ordinal} $\alpha$. Then
  \[ (\forall \alpha) ((\forall \beta)((\beta \wlt
  \alpha) \implies p(\beta)) \implies p(\alpha))
  \implies (\forall \alpha) p(\alpha) \]
  If not, then there exists $\alpha$ with
  $p(\alpha)$ false. Then there exists least such
  $\alpha$ ($\{\beta \wle \alpha \st p(\beta)
  \text{ false}\} \neq \emptyset$). Then
  $p(\beta)$ is true for all $\beta \wlt \alpha$.
  By assumption, $p(\alpha)$ is true,
  \contradiction.
\end{remark*}

\begin{example*}
  For any $\alpha$, $\alpha \opl 1 = \alpha \opl 0\osucc
  = (\alpha \opl 0)\osucc = \alpha\osucc$.

  If $m \wlt \oomega$, then we have $m \opl 0 = m$
  and for $n \wlt \omega$,
  \[ m \opl (n \opl 1) = m + \opl n\osucc = (m
  \opl n)\osucc = (m + n) + 1 = m + n + 1 \]
  So on $\oomega$, \gls{ordadd} is the usual
  addition.

  More examples:
  \begin{align*}
    \oomega \opl 2
    &= \oomega \opl 1\osucc = (\oomega \opl 1)\osucc =
    \oomega\osucc\osucc \\
    \oomega \opl \oomega
    &= \sup\{\oomega \opl n \st n \wlt \oomega\} =
    \sup\{\oomega, \oomega \opl 1, \oomega \opl 2,
    \ldots\} \\
    1 \opl \oomega
    &= \sup\{1 \opl n \st n \wlt \oomega\}
    = \sup\{1, 2, 3, \ldots\} = \oomega \neq
    \oomega \opl 1
  \end{align*}
  So `$\opl$' is not commutative.
\end{example*}

\begin{flashcard}[ordadd-ineq-prop]
\begin{proposition}
  \label{ordadd_ineq}
  $\forall \alpha, \beta, \gamma$ \glspl{ordinal},
  $\beta \wle \gamma \implies \cloze{\alpha \opl \beta
  \wle \alpha \opl \gamma}$.
\end{proposition}

\begin{proof}
  \cloze{We prove this by \gls{induction} on
  $\gamma$ (with $\alpha, \beta$ fixed).
  \begin{enumerate}[$\gamma\neq 0$ limit:]
    \item[$\gamma = 0$:] If $\beta \wle \gamma$,
      then $\beta = 0$, so result is true.
    \item[$\gamma = \delta\osucc$] If $\beta \wle
      \gamma$, then either $\beta = \gamma$ and
      we're done or $\beta \wle \delta$ and so
      $\alpha \opl \beta \wle \alpha + \delta <
      (\alpha + \delta)\osucc = \alpha +
      \delta\osucc = \alpha + \gamma$.
    \item[$\gamma \neq 0$
      {\glsref[limord]{limit}}] If $\beta \wle
      \gamma$, then without loss of generality
      $\beta \wlt \gamma$, so $\alpha \opl \beta
      \wle \sup\{\alpha \opl \delta \st \delta
      \wlt \gamma\} = \alpha \opl \gamma$.
      \qedhere
  \end{enumerate}
  }
\end{proof}
\end{flashcard}

\begin{remark*}
  From \cref{ordadd_ineq}, we get $\beta \wlt
  \gamma \implies \alpha \opl \beta \wlt \alpha
  \opl \gamma$. Indeed,
  \[ \alpha \opl \beta \wlt (\alpha \opl
  \beta)\osucc = \alpha \opl \beta\osucc \wle
  \alpha \opl \gamma .\]
  Note that $1 \wlt 2$ but $1 \opl \oomega = 2 +
  \oomega = \oomega$, the proposition is not true
  when the order is swapped.
\end{remark*}

\begin{flashcard}[ordadd-sup-lemma]
\begin{lemma}
  Let $\alpha$ be an \gls{ordinal} and $S$ a
  nonempty set of \glspl{ordinal}. Then
  \[ \alpha + \sup S = \sup\{\alpha \opl \beta \st
  \beta \in S\} .\]
\end{lemma}

\begin{proof}
  \cloze{If $\beta \in S$, then $\alpha \opl \beta \wle
  \alpha \opl \sup S$ (by \cref{ordadd_ineq}).
  Hence
  \[ \sup\{\alpha \opl \beta \st \beta \in S\} \wle
  \alpha \opl \sup S .\]
  For the reverse inequality, consider two cases.
  If $S$ has a greatest element, $\beta$ say, then
  \[ \alpha \opl \sup S = \alpha \opl \beta .\]
  For all $\gamma \in S$, $\gamma \wle \beta$, so
  by \cref{ordadd_ineq}, $\alpha \opl \gamma
  \wle \alpha \opl \beta$. It follows that
  \[ \sup\{\alpha \opl \gamma \st \gamma \in S\} =
  \alpha \opl \beta .\]
  If $S$ has no greatest element, then $\lambda =
  \sup S$ is a $\neq 0$ \gls{limord} (if $\lambda
  = \gamma\osucc$ then $\gamma \wlt \lambda$, so
  there exists $\delta \in S$ with $\gamma \wlt
  \delta$, then $\lambda = \gamma\osucc \wle
  \delta$, so $\lambda \in S$, contradiction). So
  \[ \alpha \opl \sup S = \sup\{\alpha \opl \beta
  \st \beta \wlt \lambda\} \]
  by definition. If $\beta \wlt \gamma$, then
  there exists $\delta \in S$, $\beta \wlt
  \delta$. By \cref{ordadd_ineq}, $\alpha \opl
  \beta \le \alpha \opl \delta$. It follows that
  \[ \sup\{\alpha \opl \beta \st \beta \wlt
  \lambda\}\ wle \sup\{\alpha \opl \delta \st
  \delta \in S\} \qedhere \]}
\end{proof}
\end{flashcard}

\begin{flashcard}[ordadd-assoc-prop]
\begin{proposition}
  $\forall \alpha, \beta, \gamma$, $(\alpha \opl
  \beta) \opl \gamma = \alpha \opl (\beta \opl
  \gamma)$.
\end{proposition}

\begin{proof}
  \cloze{By induction on $\gamma$.
  \begin{enumerate}[$\gamma \neq 0$ limit:]
    \item[$\gamma = 0$:] $(\alpha \opl \beta) \opl
      0 = \alpha \opl \beta = \alpha \opl (\beta
      \opl 0)$.
  \item[$\gamma = \delta\osucc$:] $(\alpha \opl
    \beta) \opl \delta\osucc = ((\alpha \opl
    \beta) \opl \delta)\osucc = (\alpha \opl
    (\beta \opl \delta))\osucc = \alpha \opl
    (\beta \opl \delta)\osucc = \alpha \opl (\beta
    \opl \gamma)$.
  \item[$\gamma \neq 0$ {\glsref[limord]{limit}}:]
    \phantom{} \\[-2.5\baselineskip]
    \begin{align*}
      (\alpha \opl \beta) \opl \gamma
      &= \sup\{(\alpha \opl \beta) \opl \delta \st
      \delta \wlt \gamma\} \\
      &= \sup\{\alpha \opl (\beta \opl \delta) \st
      \delta \wlt \gamma\} \\
      &= \alpha \opl
      \sup\{\beta \opl \delta \st \delta \wlt
      \gamma\} \\
      &= \alpha \opl (\beta \opl \delta) \qedhere
    \end{align*}
  \end{enumerate}
  }
\end{proof}
\end{flashcard}