%! TEX root = LST.tex
% vim: tw=50
% 25/01/2024 09AM

\newpage
\section{Well-ordering and ordinals}
\label{sec2}

\begin{flashcard}[linear-order-defn]
\begin{definition*}[Linear order]
  \glsnoundefn{linord}{linear order}{N/A}%
  \glsadjdefn{lirref}{irreflexive}{order}
  \glsadjdefn{ltrans}{transitive}{order}
  \glsadjdefn{ltrich}{trichotomous}{order}
  \cloze{A \emph{linear order} of \emph{total order} on a
  set $X$ is a relation $<$ on $X$ that is:
  \begin{enumerate}[(i)]
    \item \emph{irreflexive}: $\forall x \in X$,
      $\neg (x < x)$.
    \item \emph{transitive}: $\forall x, y, z \in
      X$, $(x < y \wedge y < z) \implies (x < z)$.
    \item \emph{trichotomy}: $\forall x, y \in X$,
      $x < y$ or $x = y$ or $y < x$.
  \end{enumerate}
  }
\end{definition*}
\end{flashcard}

\begin{remark*}
  In (iii) exactly one holds: for example, if $x <
  y$ and $y < x$, then $x < $ by (ii) which
  contradicts (i).
\end{remark*}

\begin{notation*}
  \glssymboldefn{linord}{$<$}{$<$}
  We say $X$ is \glsref[linord]{linearly ordered}
  by $<$, or simply say $X$ is a
  \glsref[linord]{linearly ordered} set.
\end{notation*}

\begin{example*}
  $\NN$, $\ZZ$, $\QQ$, $\RR$ with their usual
  order ($\NN = \{1, 2, 3, \ldots\}$).
\end{example*}

\begin{note*}
  If $X$ is a set of size $\ge 2$, then on $\PP X
  = \{Y \st Y \subset X\}$ (power set of $X$),
  defining $a \llt b$ to mean $a \subset b$, $a \neq
  b$ is not \gls{ltrich}.
\end{note*}

\begin{notation*}
  \glssymboldefn{leetc}{$\le$}{$\le$}
  If $X$ is \glsref[linord]{linearly ordered} by $\llt$, then we write
  $x \lgt y$ for $y \llt k$, $x \lle y$ for $x <
  y$ or $x = y$, and $x \lge y$ for $x > y$ or $x
  = y$.
\end{notation*}

\begin{note*}
  \glsadjdefn{lerefl}{reflexive}{order}
  \glsadjdefn{leantisym}{antisymmetric}{order}
  \glsadjdefn{lerans}{transitive}{order}
  \glsadjdefn{letrich}{trichotomous}{order}
  Note that $\lle$ is:
  \begin{enumerate}
    \item \emph{reflexive}: $\forall x \in X$, $x
      \lle x$.
    \item \emph{antisymmetric}: $\forall x, y \in
      X$, $(x \lle y \wedge y \lle x) \implies (x
      = y)$.
    \item \emph{transitive}: $\forall x, y, z \in
      X$, $(x \lle y \wedge y \lle z) \implies (x
      \lle z)$.
    \item \emph{trichotomous}: $\forall x, y \in
      X$, $x \le y$ or $y \le x$.
  \end{enumerate}
\end{note*}

\begin{note*}
  If $X$ is \glsref[linord]{linearly ordered} by
  $\llt$, then any $Y \subset X$ is
  \glsref[linord]{linearly ordered} by $\llt$
  (more precisely, by the restriction of $\llt$ to
  $Y$).
\end{note*}

\begin{flashcard}[well-order-defn]
\begin{definition*}[Well-ordering]
  \glsnoundefn{wellord}{well-ordering}{well-orderings}
  \glsnoundefn{leastel}{least element}{least
  elements}
  \cloze{A \emph{well-ordering} on a set $X$ is a
  \gls{linord} $\llt$ on $X$ such that every
  non-empty subset $X$ has a least element:
  $\forall S \subset X$, $S \neq \emptyset$
  implies $\exists x \in S$ such that $\forall y
  \in S$, $x \le y$.}
\end{definition*}
\end{flashcard}

\begin{note*}
  This least element is always unique by
  \gls{leantisym}.
\end{note*}

\begin{notation*}
  \glssymboldefn{wellord}{$<$}{$<$}
  Say $X$ is \emph{well-ordered} by $\llt$, or
  simply say $X$ is a
  \glsref[wellord]{well-ordered} set.
\end{notation*}

\begin{example*}
  $\NN$ with the usual \gls{linord} is a
  \gls{wellord}.

  $\ZZ, \QQ, \RR$ are not (they have no least
  element). $\{x \in \RR \st x \ge 0\}$ is not
  \glsref[wellord]{well-ordered}, because for
  example, $\{x \in \RR \st x > 0\}$ has no least
  element.
\end{example*}

\begin{note*}
  Every subset of a \glsref[wellord]{well-ordered}
  set is \glsref[wellord]{well-ordered}. We'll see
  that $\QQ$ has a rich collection of well-ordered
  subsets.
\end{note*}

\begin{flashcard}[order-iso-defn]
\begin{definition*}[Order isomorphic]
  \glsadjdefn{ordisic}{order-isomorphic}{linearly
  ordered sets}
  \glsnoundefn{ordisism}{order-isomorphism}{order-isomorphisms}
  \cloze{Say \glsref[linord]{linearly ordered} sets $X, Y$
  are \emph{order-isomorphic} if there exists a
  bijection $f : X \to Y$ which is
  \emph{order-preserving}: $\forall x \llt y$ in
  $X$, $f(x) \llt f(y)$. Such an $f$ is called an
  \emph{order-isomorphism}. Then $f^{-1}$ is also
  an order-isomorphism.}
\end{definition*}
\end{flashcard}

\begin{note*}
  If \glsref[linord]{linearly ordered} sets $X, Y$
  are \gls{ordisic} and $X$ is
  \glsref[wellord]{well-ordered}, then
  so is $Y$.
\end{note*}

\begin{example*}
  $\NN$ and $\QQ$ are not \gls{ordisic}.

  $\QQ$ and $\QQ \setminus \{0\}$ are
  \gls{ordisic} (see Numbers \& Sets Example
  Sheet).

  $A = \left\{ \frac{1}{2}, \frac{2}{3},
  \frac{3}{4}, \ldots \right\} = \left\{
  \frac{n}{n + 1} \st n \in \NN \right\}$ is
  \gls{ordisic} to $\NN$ ($n \mapsto \frac{n}{n +
  1}$).

  $B = A \cup \{1\}$ is
  \glsref[wellord]{well-ordered}, but not
  \gls{ordisic} to $\NN$ (it has a greatest
  element).

  $C = A \cup \{2\}$ is \gls{ordisic} to $B$.

  $D = A \cup (A + 1) = A \cup \left\{
  \frac{3}{2}, \frac{5}{3}, \frac{7}{4}, \ldots
  \right\}$ is \glsref[wellord]{well-ordered}, but not \gls{ordisic}
  to $A$ or $B$.
\end{example*}

\begin{flashcard}[initial-segment-defn]
\begin{definition*}[Initial segment]
  \glsnoundefn{is}{initial segment}{initial
  segments}
  \cloze{A subset $I$ of a \glsref[linord]{linearly
  ordered} set $X$ is an \emph{initial segment}
  (i.s.) of $X$ is $x \in I, y < x \implies y \in
  I$ for any $x, y \in X$.}
\end{definition*}
\end{flashcard}

\begin{example*}
  $\{1, 2, 3, 4\}$ is an \gls{is} of $\NN$.
  $\{1, 2, 3, 5\}$ is not.

  $[0, 1]$ is an \gls{is} of $\{x \in \RR \st x
  \ge 0\}$.
\end{example*}

\begin{flashcard}[initial-segment-notation]
\begin{notation*}
  \glssymboldefn{Ix}{$I_x$}{$I_x$}
  In general, for $x \in X$, $I_x = \cloze{\{y \in X \st
  y < x\}}$ is an \emph{is} of $X$ by \gls{ltrans}.
  \fcscrap{$I_x$ is a proper \gls{is} of $X$ (meaning $I_x
  \neq X$), because it does not contain $x$.}
\end{notation*}
\end{flashcard}

\begin{note*}
  In general, not every proper \gls{is} is of this
  form. For example, $(-\infty, 1]$ is a proper
  \gls{is} of $\RR$, but $(-\infty, 1] \neq \I_x$
  for any $x \in \RR$.
\end{note*}

\begin{remark*}
  If $X$ is \glsref[wellord]{well-ordered} and $I$ is a proper
  \gls{is} of $X$, then $I = \I_x$ where $x$ is
  the \gls{leastel} of $X \setminus I$.

  Indeed, if $y \in \I_x$ then $y < x$, so $y \in
  I$ by choice of $x$. If $y \in I$ and $y \ge x$,
  then $x \in I$ as $I$ is an \gls{is},
  contradiction. So $y < x$, i.e. $y \in \I_x$.
\end{remark*}

\begin{lemma}
  \label{ordisism_min_lemma}
  Let $X, Y$ be a \glsref[wellord]{well-ordered} set, $I$ an
  \gls{is} of $Y$ and $f : X \to Y$ an
  \gls{ordisism} between $X$ and $I$. Then for
  each $x \in X$, $f(x)$ is the \gls{leastel} of
  $Y \setminus \{f(y) \st y < x\}$.
\end{lemma}

\begin{proof}
  The set $A = Y \setminus \{f(y) \st y < x\}$ is
  $\neq\emptyset$ since $f(x) \in A$. Let $a$ be
  the \gls{leastel} of $A$. Then $a \le f(x)$ and
  $f(x) \in I$, and so $a \in I$. Thus $a = f(x)$
  for some $z \in X$. Note that $z > x$ implies $a
  = f(z) > f(x)$, contradiction. So $z \wle x$.

  If $z < x$, then $a = f(z) \in \{f(y) \st y <
  x\}$, \contradiction as $a \in A$. So $z = x$
  and $a = f(z) = f(x)$.
\end{proof}

\begin{flashcard}[proof-by-induction-prop]
\begin{proposition}[Proof by induction]
  \glsnoundefn{induction}{induction}{induction}
  \cloze{Let $X$ be a \glsref[wellord]{well-ordered} set and
  $S \subset X$ satisfying the following for every
  $x \in X$: $\forall y < x$, $y \in S$ implies $x
  \in S$. Then $S = X$.}
\end{proposition}

\fcscrap{
\begin{note*}
  Assume $S$ is given by a property $p$: $S = \{x
  \in X \st p(x)\}$. The above can be written as
  \[ (\forall x \in X)((\forall y < x, p(y))
  \implies p(x)) \implies (\forall x \in X, p(x)) \]
  (base case is included since the left hand side
  will be vacuously true for the \gls{leastel}).
\end{note*}
}

\begin{proof}
  \cloze{If $S \neq X$, then $X \setminus S$ has a
  \gls{leastel} $x$, say. If $y \wlt x$, then $y
  \in S$ by choice of $x$. By the assumption on
  $S$, $x \in S$, contradiction.}
\end{proof}
\end{flashcard}