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\begin{flashcard}[chain-defn]
\begin{definition*}[Chain]
  \glsnoundefn{chain}{chain}{chains}
  \cloze{A subset $S$ of a \gls{poset} $X$ is a
  \emph{chain} if it is \glsref[linord]{linearly
  ordered} by the \gls{partord} on $X$.}
\end{definition*}
\end{flashcard}

\begin{example*}
  \phantom{}
  \begin{enumerate}[(1)]
    \item Every \glsref[linord]{linearly ordered}
      set is a \gls{chain} in itself.
    \item Any subset of a \gls{chain} in a
      \gls{poset}.
    \item In $\NN$ with $a \ple b \iff a \mid b$,
      $\{2^n \st n = 0, 1, 2, \ldots\}$ is a
      \gls{chain}.
    \item In $\PP(\{1, 2, 3\})$ with $\subset$,
      $\{\emptyset, \{1\}, \{1, 2\}, \{1, 2,
      3\}\}$ is a \gls{chain}.
    \item $\{a, c, d, e\}$ is a chain in
      \begin{center}
        \includegraphics[width=0.6\linewidth]{images/f1553fabe3384fa0.png}
      \end{center}
    \item In $\PP \QQ$, $\{(-\infty, x) \cap \QQ
      \st x \in \RR\}$ is an uncountable
      \gls{chain} in $\PP\QQ$.
  \end{enumerate}
\end{example*}

\begin{flashcard}[antichain-defn]
\begin{definition*}[Antichain]
  \glsnoundefn{achain}{antichain}{antichains}
  \cloze{A subset $S$ of a \gls{poset} $X$ is an
  \emph{antichain} if no two distinct members of
  $S$ are related, i.e. $\forall x, y \in S, x
  \ple y \implies x = y$.}
\end{definition*}
\end{flashcard}

\begin{example*}
  \phantom{}
  \begin{enumerate}[(1)]
    \item In a \glsref[linord]{linearly ordered}
      set there is no \gls{achain} of size $> 1$.
    \item In $\NN$ with $a \ple b \iff a \mid b$,
      the set of primes is an \gls{achain}.
    \item In $\PP(\{1, 2, \ldots, n\})$ with
      $\subset$, for any $k$, $0 \ple k \ple n$,
      \[ \mathcal{F}_k = \{A \subset \{1, \ldots,
      n\} \st |A| = k\} \]
      is an \gls{achain}.
    \item In 
      \begin{center}
        \includegraphics[width=0.6\linewidth]{images/f1553fabe3384fa0.png}
      \end{center}
      $\{b, d\}$ and $\{b, c\}$ are \glspl{achain}.
    \item In
      \begin{center}
        \includegraphics[width=0.6\linewidth]{images/d0a658d1b5c54a3b.png}
      \end{center}
      the whole set is an \gls{achain}.
  \end{enumerate}
\end{example*}

\begin{flashcard}[upper-bound-defn]
\begin{definition*}[Upper bound]
  \glsnoundefn{ub}{upper bound}{upper bounds}
  \cloze{Let $S$ be a subset of a \gls{poset} $X$. Say $x
  \in X$ is an \emph{upper bound} for $S$ if
  $\forall y \in S$, $y \ple x$.}
\end{definition*}
\end{flashcard}

\begin{flashcard}[lub-defn]
\begin{definition*}[Least upper bound]
  \glsnoundefn{lub}{least upper bound}{least upper
  bounds}
  \glsnoundefn{sup}{supremum}{suprema}
  \glssymboldefn{posup}{sup}{sup}
  \cloze{Say $x \in X$ is a \emph{least upper bound} or
  \emph{supremum} for $S$ if $x$ is an \gls{ub}
  for $S$ and $x \ple y$ for all \glspl{ub} $y$
  for $S$.

  If it exists, we denote this by $\sup S$ or
  $\bigvee S$ (`join' of $S$).}
\end{definition*}
\end{flashcard}

\begin{example*}
  \phantom{}
  \begin{enumerate}[(1)]
    \item In $\RR$, $\posup [0, 1] = 1$,
      $\posup(0, 1) = 1$.
    \item $\QQ$ has no \gls{sup} in $\QQ$, as it
      doesn't even have any \gls{ub}.
    \item In
      \begin{center}
        \includegraphics[width=0.6\linewidth]{images/e38e4a67252b4e77.png}
      \end{center}
      $\{a, b\}$ has \glspl{ub}, for example $c,
      d$, but no \gls{lub}.
    \item If $X = \PP A$, $A$ any set, $S \subset
      X$, then $\posup S = \bigcup\{B \subset A \st
      B \in S\}$.
  \end{enumerate}
\end{example*}

\begin{flashcard}[complete-poset-defn]
\begin{definition*}[Complete Partial Order]
  \glsadjdefn{complete}{complete}{\gls{poset}}
  \cloze{A \gls{poset} $X$ is \emph{complete} if every $S
  \subset X$ has a \gls{sup}.}
\end{definition*}
\end{flashcard}

\begin{example*}
  \phantom{}
  \begin{enumerate}
    \item $\PP A$ for any $A$ is \gls{complete}.
    \item $[0, 1]$ is \gls{complete}.
    \item $\RR$ is not \gls{complete}.
    \item $\QQ \cap [0, 2]$ is not \gls{complete}.
  \end{enumerate}
\end{example*}

\begin{remark*}
  A \gls{complete} \gls{poset} $X$ has a greatest
  element $\posup X$ and a least element $\posup
  \emptyset$. In particular, $X \neq \emptyset$.
\end{remark*}

\begin{flashcard}[order-preserving-function]
\begin{definition*}[Order-preserving function]
  \glsadjdefn{op}{order-preserving}{function}
  \cloze{Let $f : X \to Y$ be a function between
  \glspl{poset} $X, Y$. Say $f$ is
  \emph{order-preserving} if $\forall x, y \in X$,
  $x \ple y \iff f(x) \ple f(y)$.}
\end{definition*}
\end{flashcard}

\begin{note*}
  $f$ need not be injective. But $f$ is \gls{op}
  injective if and only if $\forall x, y \in X$,
  $x \plt y \iff f(x) \plt f(y)$.
\end{note*}

\begin{example*}
  $f : \NN \to \NN$, $f(n) = n + 1$ (with the
  usual order).

  $g : \PP(A) \to \PP(A)$, $A \mapsto A \cup B$,
  $B$ fixed.
\end{example*}

\begin{flashcard}[fixed-point-defn]
\begin{definition*}[Fixed point]
  \glsnoundefn{fp}{fixed point}{fixed points}
  \cloze{Let $X$ be any set. Then a \emph{fixed point}
  for a function $f : X \to X$ is an element $x
  \in X$ such that $f(x) = x$.}
\end{definition*}
\end{flashcard}

\begin{flashcard}[kt-fp-thm]
\begin{theorem}[Knaster-Tarski Fixed Point Theorem]
  \label{kt_fp}
  \cloze{If $X$ is a \gls{complete} \gls{poset} and $f :
  X \to X$ is \gls{op}, then $f$ has a \gls{fp}.}
\end{theorem}

\begin{proof}
  \cloze{Let $S = \{x \in X \st x \ple f(x)\}$. Let $z =
  \posup S$. Let $x \in S$. Then $x \ple z$, so
  $f(x) \ple f(z)$. Since $x \in S$, $x \ple
  f(x)$, so by transitivity, $x \ple f(z)$. Thus
  $f(z)$ is an \gls{ub} for $S$, so $z \ple f(z)$.
  It follows that $f(z) \ple f(f(z))$. So $f(z)
  \in S$, and thus $f(z) \ple z$. So $z$ is a
  \gls{fp}.}
\end{proof}
\end{flashcard}

\begin{flashcard}[Schroder-Bernstein-thm]
\begin{corollary}[Schr\"oder-Bernstein Theorem]
  \label{schroeder_bernstein}
  \cloze{Let $A, B$ be sets and assume there exist
  injections $f : A \to B$ and $g : B \to A$. Then
  there exists a bijection $h : A \to B$.}
\end{corollary}

\fcscrap{
\begin{center}
  \includegraphics[width=0.6\linewidth]{images/dce563ec742745bd.png}
\end{center}
}

\begin{proof}
  \cloze{We seek partitions $A = P \cup \QQ$, $B = R \cup
  S$ such that ($P \cap Q = \emptyset$, $R \cap S
  = \emptyset$), $f(P) = R$, $g(S) = Q$. Then we
  will have that
  \[ h : A \to B, \qquad h = \begin{cases}
    f & \text{on $P$} \\
    g^{-1} & \text{on $Q$}
  \end{cases} \]
  Such partitions exist if and only if there
  exists $P \subset A$ such that
  \[ A \setminus g(B \setminus f(P)) = P .\]
  Let $X = \PP A$ with ordering by $\subset$.
  Define $H : X \to X$,
  \[ H(P) = A \setminus g(B \setminus f(P)) .\]
  $H$ is \gls{op} and $X$ is \gls{complete}, so by
  \nameref{kt_fp}, we can find such $P$.}
\end{proof}
\end{flashcard}

\subsubsection*{Zorn's Lemma}

\begin{flashcard}[maximal-element-defn]
\begin{definition*}[Maximal element]
  \glsnoundefn{pomaxel}{maximal element}{maximal
  elements}
  \glsadjdefn{pomaxal}{maximal}{element}
  \cloze{Say an element $x$ in a \gls{poset} $X$ is
  \emph{maximal} if $\forall y \in X$, $x \ple y
  \implies x = y$. In other words, there is no $y
  \in X$ with $y \pgt x$.}
\end{definition*}
\end{flashcard}

\begin{example*}
  In $\PP A$, $A$ is \gls{pomaxal}, $A$ is even a
  greatest element. In general, ``greatest''
  $\implies$ \gls{pomaxal}, but the other way
  round does not hold.
\end{example*}

\begin{example*}
  In:
  \begin{center}
    \includegraphics[width=0.2\linewidth]{images/c5160a04d2c54623.png}
  \end{center}
  $c, d$ are both \gls{pomaxal}, but there does
  not exist a greatest element.
\end{example*}

\begin{flashcard}[Zorns-lemma]
\begin{theorem}[Zorn's Lemma]
  \label{ZL}
  \cloze{Let $X$ be a (non-empty) \gls{poset} such that
  every \gls{chain} in $X$ has an \gls{ub} in $X$.
  Then $X$ has a \gls{pomaxel}.}
\end{theorem}

\begin{remark*}
  \cloze{$\emptyset$ is a \gls{chain} in $X$, so it has
  an \gls{ub}, so $X \neq \emptyset$. Often we
  check the \gls{chain} condition by checking it
  for $\emptyset$ (i.e. that $X \neq \emptyset$)
  and then for $\neq \emptyset$ \glspl{chain}.}
\end{remark*}

\begin{proof}
  \cloze{Assume $X$ has no \gls{pomaxel}. For each $x \in
  X$, fix $x' \pgt x$. We also fix an \gls{ub} $u(C)$
  for every \gls{chain} $C \subset X$. Let $\gamma
  = \hartgamma(X)$ (from \nameref{hartogs}). Define
  $f : \gamma \to X$ by \nameref{recursion}:
  \begin{itemize}
    \item $f(0) = u(\emptyset)$.
    \item $f(\alpha + 1) = f(\alpha)'$.
    \item $f(\lambda) = u(\{f(\alpha) \st \alpha \plt
      \lambda\})'$ ($\lambda \neq 0$
      \gls{limord}).
  \end{itemize}
  An easy \gls{induction} shows that $\forall \alpha
  \plt \beta$ (in $\gamma$), $f(\alpha) \plt
  f(\beta)$ (on $\beta, \alpha$, $\alpha$ fixed).
  This also shows $\{f(\alpha) \st \alpha \plt
  \beta\}$ is a \gls{chain} for all $\beta \plt
  \alpha$. Hence $f$ is an injection. This
  contradicts the definition of $\hartgamma(X)$.}
\end{proof}
\end{flashcard}

\begin{remark*}
  Technically, for $\lambda \neq 0$ a
  \gls{limord}, $f(\lambda)$ should be defined as
  above if $\{f(\alpha) \st \alpha \plt \gamma\}$
  is a \gls{chain} and $f(\lambda) = u(\emptyset)$
  otherwise. Then by \gls{induction}, $\alpha \plt
  \beta \implies f(\alpha) \plt f(\beta)$, so the
  `otherwise' clause never happens.
\end{remark*}