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\begin{remark*}
  \cref{wole_leq_or_geq} and
  \cref{wole_leq_geq_impl_eq} together show that
  $\wole$ is a \gls{linord} (\gls{lerefl},
  \gls{leantisym}, \gls{ltrans} and
  \gls{letrich}), provided we identify
  \glsref[wellord]{well-ordered} sets that are
  \gls{ordisic} to each other.
\end{remark*}

\begin{notation*}
  \glssymboldefn{wolt}{$<$}{$<$}
  We introduce `$X \wolt Y$` to mean $X \wole Y$
  and $X$ is not \gls{ordisic} to $Y$. So $X \wolt
  Y$ if and only if $X$ \gls{ordisic} to a proper
  \gls{is} of $Y$.
\end{notation*}

\vspace{-1em}
\textbf{Question:} Do the
\glsref[wellord]{well-ordered} sets form a set? If
so, is it a \glsref[wellord]{well-ordered} set?

First we construct new
\glsref[wellord]{well-ordered} sets from old ones.

`there's always another one':

\begin{flashcard}[successor-ordinal-defn]
\begin{definition*}[Successor ordinal]
  \glssymboldefn{succ}{$X^+$}{$X^+$}
  \cloze{Let $X$ be a
  \glsref[wellord]{well-ordered} set, fix $x_0
  \notin X$, and set $X^+ = X \cup \{x_0\}$, which
  we \glsref[wellord]{well-order} by extending
  $\wlt$ on $X$ to $X^+$ by letting $x \wlt x_0$
  for all $x \in X$. This is unique up to
  \gls{ordisism} and $X \wlt X^+$.}
\end{definition*}
\end{flashcard}

\vspace{-1em}
\textbf{Upper bounds:} Given a set $\{X_i \st i
\in I\}$ of \glsref[wellord]{well-ordered} sets,
we seek a \glsref[wellord]{well-ordered} set $X$
such that $X_i \wole X$ for all $i \in I$.

\begin{flashcard}[extends-defn]
\begin{definition*}[Extends]
  \glsverbdefn{extends}{extends}{extends}
  \cloze{Given \glsref[wellord]{well-ordered} sets $(X, \wlt_X)$
  and $(Y, \wlt_Y)$, say $Y$ \emph{extends} $X$ if
  $X \subset Y$, $\wlt_X$ is the restriction to $X$
  of $\wlt_Y$ and $X$ is an \gls{is} of $Y$.}
\end{definition*}
\end{flashcard}

\begin{flashcard}[nested-defn]
\begin{definition*}[Nested]
  \glsadjdefn{nested}{nested}{well-ordered sets}
  \cloze{We say $\{X_i \st i \in I\}$ is \emph{nested} if
  $\forall i, j \in I$ either $X_j$ \gls{extends}
  $X_i$ or $X_i$ \gls{extends} $X_j$.}
\end{definition*}
\end{flashcard}

\begin{flashcard}[wellord-union-prop]
\begin{proposition}
  \label{wellord_union}
  Let $\{X_i \st i \in I\}$ be a \gls{nested} set
  of \glsref[wellord]{well-ordered} sets. Then
  there exists a \glsref[wellord]{well-ordered} set
  $X$ such that $X_i \wole X$ for all $i \in I$.
\end{proposition}

\begin{proof}
  \cloze{Let $X = \bigcup_{i \in I} X_i$ and define
  $\wlt$ on $X$ as follows: $x \wlt y$ if and only
  if $\exists i \in I$ such that $x, y \in X_i$
  and $x \wlt_i y$ where $\wlt_i$ is the
  \gls{wellord} of $X_i$. Since the $X_i$ are
  \gls{nested}, this is well-defined, is a linear order
  and each $X_i$ is an \gls{is} of $X$.

  Given $S \subset X$, $S \neq \emptyset$, since
  $S = \bigcup_{i \in I} (S \cap X_i)$, there
  exists $i \in I$ such that $S \cap X_i \neq
  \emptyset$. Let $x$ be a \gls{leastel} of $S
  \cap X_i$ (since $X_i$ is
  \glsref[wellord]{well-ordered}). Then $x$ is a
  \gls{leastel} of $S$ since $X_i$ is an \gls{is}
  of $X$.}
\end{proof}
\end{flashcard}

\begin{remark*}
  \cref{wellord_union} holds even if the $X_i$ are
  not \gls{nested} (see \cref{sec5}).
\end{remark*}

\subsubsection*{Ordinals}

\begin{flashcard}[ordinal-defn]
\begin{definition*}[Ordinal]
  \glsnoundefn{ordinal}{ordinal}{ordinals}
  \cloze{An \emph{ordinal} is a
  \glsref[wellord]{well-ordered} set but we
  consider two ordinals the same if they're
  \gls{ordisic}.}
\end{definition*}
\end{flashcard}

\begin{remark*}
  A formal definition will be given in
  \cref{sec5}. You could think of the term
  `ordinal' as a shorthand (for now).
\end{remark*}

\begin{flashcard}[order-type-defn]
\begin{definition*}[Order type]
  \glspropdefn{ot}{order type}{well-ordered
  set}
  \cloze{The \emph{order type} of a
  \glsref[wellord]{well-ordered} set $X$ is the
  unique \gls{ordinal} $\alpha$ \gls{ordisic} to
  $X$. Write `$\alpha$ is the order type (O.T.) of
  $X$`.}
\end{definition*}
\end{flashcard}

\begin{example*}
  For $k \in \NN \cup \{0\}$, we let $k$ be the
  \gls{ot} of a \glsref[wellord]{well-ordered} set
  of size $k$ (this is unique). Let $\omega$ be
  the \gls{ot} of $\NN$ (also the \gls{ot} of $\NN
  \cup \{0\}$). The set $A = \left\{ \frac{1}{2},
  \frac{2}{3}, \frac{3}{4}, \ldots\right\}$ in
  $\QQ$ also has \gls{ot} $\omega$.
\end{example*}

\begin{flashcard}[omega-notation]
\begin{notation*}
  \glssymboldefn{oomega}{$\omega$}{$\omega$}
  We write $\omega$ for \cloze{the \gls{ot} of any
  set which is \gls{ordisic} to $\NN$}.
\end{notation*}
\end{flashcard}

\begin{notation*}
  \glssymboldefn{ole}{$\le$}{$\le$}
  \glssymboldefn{olt}{$<$}{$<$}
  \glssymboldefn{osucc}{$\alpha^+$}{$\alpha^+$}
  For \glspl{ordinal} $\alpha, \beta$ we write
  $\alpha \ole \beta$ is $X \wole Y$ where $X$ is
  a \glsref[wellord]{well-ordered} set of \gls{ot}
  $\alpha$, $Y$ is a
  \glsref[wellord]{well-ordered} set of \gls{ot}
  $\beta$. This is well-defined. We also write
  $\alpha \olt \beta$ is $X \wolt Y$. We let
  $\alpha\osucc$ be the \gls{ot} of $X\wsucc$.
\end{notation*}

\begin{remark*}
  $\le$ is a \gls{linord}; if $\alpha \ole \beta$
  and $\beta \ole \alpha$ then $\alpha = \beta$.
\end{remark*}

\begin{flashcard}[I-alpha-ot-thm]
\begin{theorem}
  \label{I_alpha_ot}
  Let $\alpha$ be an \gls{ordinal}. The
  \glspl{ordinal} $\olt \alpha$ form a
  \glsref[wellord]{well-ordered} set of \gls{ot}
  $\alpha$.
\end{theorem}

\begin{proof}
  \cloze{Fix a \glsref[wellord]{well-ordered} set $X$
  with \gls{ot} $\alpha$. Let
  \[ \tilde{X} = \{Y \subset X \st \text{$Y$ is a
  proper \gls{is} of $X$}\} .\]
  Then $\wolt$ (defined for
  \glsref[wellord]{well-ordered} sets) is a
  \gls{linord} on $\tilde{X}$. Note that $x
  \mapsto \I_x : X \to \tilde{X}$ is an
  \gls{ordisism}. So $\tilde{X}$ is a
  \glsref[wellord]{well-ordered} set of \gls{ot}
  $\alpha$. So
  \[ \{\OT(Y) \st Y \in \tilde{X}\} \]
  is the set of \glspl{ordinal} $\olt \alpha$ and
  $Y \mapsto \OT(Y)$ is an \gls{ordisism} from
  $\tilde{X}$ to this set.}
\end{proof}
\end{flashcard}

\begin{flashcard}[I-alpha-notation]
\begin{notation*}
  \glssymboldefn{oI}{$I_\alpha$}{$I_\alpha$}
  $\I_\alpha = \cloze{\{\beta \st \beta <
  \alpha\}}$ `A
  nice example of a \glsref[wellord]{well-ordered}
  set of \gls{ot} $\alpha$'.
\end{notation*}
\end{flashcard}

\begin{flashcard}[ordinals-wellord-defn]
\begin{proposition}
  \label{ordinals_wellord}
  A non empty set $S$ of \glspl{ordinal} has a
  \gls{leastel}.
\end{proposition}

\begin{proof}
  \cloze{Pick $\alpha \in S$. If $\alpha$ is not a
  \gls{leastel} of $S$, then $S \cap \oI_\alpha
  \neq \emptyset$, and hence (by
  \cref{I_alpha_ot}) it has a \gls{leastel}
  $\beta$. Then $\beta$ is a \gls{leastel} of $S$:
  if $\gamma \in S$, $\gamma \olt \alpha$, then
  $\gamma \in \oI_\alpha \cap S$, and so $\beta
  \ole \gamma$.}
\end{proof}
\end{flashcard}

\begin{flashcard}[burali-forti-paradox-thm]
\begin{theorem}[Burati-Forti paradox]
  \label{burati_forti}
  \cloze{The \glspl{ordinal} do not form a set.}
\end{theorem}

\begin{proof}
  \cloze{Assume otherwise and let $X$ be the set of
  \glspl{ordinal}. Then $X$ is a
  \glsref[wellord]{well-ordered} set by
  \cref{ordinals_wellord} (and earlier results).
  Let $\alpha$ be the \gls{ot} of $X$. Then $X$ is
  \gls{ordisic} to $\oI_\alpha$, which is a proper
  \gls{is} of $X$, \contradiction.}
\end{proof}
\end{flashcard}

\begin{remark*}
  Let $S = \{\alpha_i \st i \in I\}$ be a set of
  \glspl{ordinal}. Then by \cref{wellord_union}
  the \gls{nested} set $\{\oI_{\alpha_i} \st i \in
  I\}$ has an upper bound. So there exists an
  \gls{ordinal} $\alpha$ such that $\alpha_i \wle
  \alpha$ for all $i \in I$. By \cref{I_alpha_ot} we can
  take the least such $\alpha$. We take the
  \gls{leastel} of
  \[ \{\beta \in \oI_\alpha \cup \{\alpha\} \st
  \forall i \in I, \alpha \ole \beta\} .\]
  We denote by $\sup S$ the least upper bound on
  $S$. Note if $\alpha = \sup S$ then $\oI_\alpha
  = \bigcup_{i \in I} \oI_\alpha$.
\end{remark*}

\subsubsection*{A list of some ordinals}

\[ 0, 1, 2, 3, \ldots, \oomega, \oomega,
\oomega\osucc = \oomega + 1, \oomega + 2, \oomega + 3,
\ldots, \]
\[ \oomega + \oomega = \oomega \cdot 2 = \sup
\{\oomega + n \st n \olt \oomega\}, \oomega \cdot 2
+ 1, \oomega \cdot 2 + 2, \ldots, \oomega \cdot 3,
\ldots, \oomega \cdot 4, \ldots\]
\[ \oomega \cdot \oomega = \oomega^2 = \{\oomega \cdot n \st n
\olt \oomega\}, \oomega^2 + 1, \oomega^2 + 2, \ldots
\oomega^2 + \oomega, \ldots, \oomega^2 + \oomega
\cdot 2, \ldots, \oomega^2 + \oomega \cdot 3, \ldots \]
\[ \oomega^2 \cdot 2, \ldots, \oomega^2 \cdot 3,
\ldots, \oomega^3, \ldots, \oomega^4, \ldots,
\oomega^\oomega = \sup\{\oomega^n \st n \olt
\oomega\}, \oomega^\oomega + 1, \ldots \]
\[ \oomega^\oomega + \oomega, \ldots,
\oomega^\oomega + \oomega^2, \ldots,
\oomega^\oomega \cdot 2, \ldots, \oomega^\oomega
\cdot \oomega = \oomega^{\oomega + 1}, \ldots,
\oomega^{\oomega + 2}, \ldots, \oomega^{\oomega
\cdot 2}, \ldots, \oomega^{\oomega \cdot 3}, \ldots \]
\[ \oomega^{\oomega^2}, \ldots,
\oomega^{\oomega^3}, \ldots,
\oomega^{\oomega^\oomega}, \ldots,
\oomega^{\oomega^{\oomega^\oomega}}, \ldots,
\eps_0 =
\sup\{\ub{\oomega^{\oomega^{\iddots^\oomega}}}_{n}
\st n \olt \omega\}, \ldots \]
\[ \eps_1, \ldots, \eps_2, \ldots, \eps_\oomega,
\ldots, \eps_{\eps_0}, \ldots, \eps_{\eps_{\eps_0}},
\ldots \]
Remarkably, all of these are countable! This can
be seen by checking that each of them is a
countable supremum of countable \glspl{ordinal},
hence must be countable.