\lecture{06}{2023-11-06}{}

\begin{theorem}[Zorn]
    \yalabel{Zorn's Lemma}{Zorn}{thm:zorn}
    Let $(a, \le )$ be a partial order with $a \neq \emptyset$.
    Assume that for all $b \subseteq a$ with $b \neq \emptyset$
    and $ b$ linearly ordered, $b$ has an upper bound.
    Then $a$ has a maximal element.
\end{theorem}
\begin{yarefproof}{thm:zorn}
    \gist{%
    Fix $(a, \le )$ as in the hypothesis.
    Let $A \coloneqq  \{ \{(b,x) : x \in b\}  : b \subseteq a, b \neq \emptyset\}$.
    Note that $A$ is a set (use separation on  $\cP(\cP(a) \times  \bigcup \cP(a))$).
    Note further that if $b_1 \neq  b_2$,
    then $\{(b_1, x) : x \in  b_1\} $
    and $\{(b_2, x) : x \in b_2\}$ are disjoint.
    Hence the \yaref{ax:c}
    gives us a choice function $f$ on $A$,
    i.e.~$\forall b \in  \cP(a) \setminus \{\emptyset\} .~(f(b) \in b)$.

    Now define a binary relation $\le^\ast$:
    We let $W$ denote the set of all well-orderings $\le'$
    of subsets $b \subseteq a$,
    such that for all $u,v \in b$
    if $u \le' v$ then $u \le v$
    and for all $u \in  b$
    and
    \[
    B_u^{\le'} \coloneqq  \{ w \in  a : w \text{ is an $\le$-upper bound of $\{v \in  b : v \le' u\}$}\}
    \]
    then $B_u^{\le'} \neq \emptyset$ and $f(B^{\le'}_u) = u$.

    \begin{claim}
        If $\le', \le'' \in W$,
        then $\le' \subseteq \le''$ or $\le'' \subseteq \le'$.
    \end{claim}
    \begin{subproof}
        Let $\le' \in W$ be a well-ordering of $b \subseteq a$
        and let $\le'' \in W$ be a well-ordering on $c \subseteq a$.
        We know that wlog.~$(b, \le') \cong (c, \le'')$
        or $\exists v \in c .~(b, \le') \cong (c, \le'')\defon{v}$.
        Let $g\colon b \to c$ or $g\colon b \to c\defon{v}$ be a witness.
        We want to show that $g = \id$.
        Suppose that $g \neq \id$.
        Let $u_0 \in  b$ be $\le'$-minimal such that $g(u_0) \neq u_0$.
        Writing $\overline{g} \coloneqq  g\defon{\{w \in b: w <' u_0\}}$,
        then $(b, \le ')\defon{u_0} \cong (c, \le'') \defon{g(u_0)}$
        and $\overline{g}$ is in fact the identity on $\{w \in b | w \le' u_0\}$
        but this means $\{w \in b | w <' u_0\} = \{w \in c | w <'' g(u_0)\}$
        and $B_{u_0}^{\le'} = B_{g(u_0)}^{\le''} \neq \emptyset$.
        Then $u_0 = f(B_{u_0}^{\le'}) = f(B_{g(u_0)}^{\le''}) = g(u_0)$.
        Thus $g$ is the identity.
    \end{subproof}
    Given the claim, we can now see that $\bigcup W$ is a well-order $\le^{\ast\ast}$
    of $a$.
    Let $B = \{w \in a | w \text{ is a $\le$-upper bound of $b$}\}$
    (this is not empty by the hypothesis).
    Suppose that $b$ does not have a maximum.
    Then $B \cap b = \emptyset$.
    Now $f(B) = u_0$
    and let
    \[
    \le^{\ast\ast} = \le^{\ast} \cup  \{(u,u_0) | u \in b\} \cup  \{(u_0,u_0)\}.
    \]
    Then $B = B_{u_0}^{\le^{\ast\ast}}$.
    So $\le^{\ast\ast} \in W$, but now $u_0 \in b$.
    So $b$ must have a maximum.
    \todo{Why does this prove the lemma?}
}{
    \begin{itemize}
        \item $A  \coloneqq  \{\{(b,x) : x \in b\}, b \subseteq a, b \neq \emptyset\}$.
        \item \AxC $\leadsto$ choice function on $A$,
            $f\colon  \cP(a) \setminus \{\emptyset\}  \to  a$, $f(b) \in b$.
        \item $\le^\ast$ on $a$:
            \begin{itemize}
            \item $W \coloneqq  \{\le' \text{wo on} b \subseteq a : \forall u,b \in b.~u \le' v \implies u \le v, B_u^{\le '} \neq \emptyset, u = f(B^{\le '}_u)\}$ where
                \item $B_u^{\le'} = \{w \in  a : w \text{ $\le $-upper bound of } \{v \in  b : v <' u\} \}$.
            \end{itemize}
        \item $\le', \le '' \in W \implies \le' \substack{\subseteq\\\supseteq} \le''$:
            \begin{itemize}
                \item $(b, \le') \overset{g}{\cong} (c, \le'') (\defon{v})$.
                \item $g = \id_b$:
                    \begin{itemize}
                        \item $u_0$ $\le'$ minimal with $g(u_0) \neq u_0$.
                        \item $\{w \in b : w <' u_0\} \overset{g \defon{\ldots}}{=} \{w \in c : w <'' g(u_0)\}$.
                        \item $B^{\le '}_{u_0}  = B_{g(u_0)}^{\le ''} \neq \emptyset$,
                        so $u_0 = f(B^{\le '}_{u_0}) = f(B^{\le ''}_{g(u_0)}) = g(u_0) \lightning$
                    \end{itemize}
            \end{itemize}
        \item  $\le^\ast \coloneqq  \bigcup W$ is wo on $b \subseteq a$.
        \item Suppose $b$ has no maximum. Then  $B \cap b = \emptyset$.
        \item $u_0 \coloneqq f(B)$, $\le^{\ast\ast} = \le^\ast \cup \{(u,u_0) | u \in b\}  \cup \{(u_0,u_0)\}$.
        \item $B = B_{u_0}^{\le^{\ast\ast}}$, so $\le^{\ast\ast} \in W$,
            but $u_0 \in b \lightning$. ?
    \end{itemize}
}

\end{yarefproof}

\begin{remark}
    Over $\ZF$ the \yaref{ax:c} and \yaref{thm:zorn}
    are equivalent.
\end{remark}

\begin{corollary}[Hausdorff's maximality principle]
    Let $a \neq  \emptyset$.
    Let $A \subseteq  \cP(a)$ be such that $\forall  B \subseteq  A$,
    if $x \subseteq y \lor y \subseteq x$
    for all $x,y \in B$,
    then there is some $z \in  A$
    such that $x \subseteq z$ for all $x \in B$.
    Then $A$ contains a $\subseteq$-maximal element.
\end{corollary}

\gist{%
\begin{remark}[Cultural enrichment]
    Other assertions which are equivalent
    to the \yaref{ax:c}:
    \begin{itemize}
        \item Every infinite family of non-empty sets
            $\langle  a_i : i \in I \rangle$
            has non-empty product,
            i.e.
            \[
            \prod_{i \in I} a_i \neq  \emptyset.%\footnote{This is clearly true.}
            \]
        \item Every set can be well-ordered.%\footnote{This is clearly false.}
    \end{itemize}
\end{remark}
}{}
% \begin{remark}
%     The axiom of choice is true.
% \end{remark}

\pagebreak
\subsection{The Ordinals}
\gist{
\begin{goal}
    We want to define nice representatives of the equivalence classes
    of well-orders.
    % TODO theorem
\end{goal}
Recall that \AxInf states the existence of an inductive set $x$.
We can hence form the smallest inductive set
\[
\omega \coloneqq  \bigcap \{ x : x \text{ is inductive}\}
\]
Note that $\omega$ exists, as it is a subset of the inductive
set given by \AxInf.
We call $\omega$ the set of \vocab{natural numbers}.
}{}
\begin{notation}
    We write $0$ for $\emptyset$,
    and $y + 1$ for $y \cup  \{y\}$.
\end{notation}
With this notation the \AxInf is equivalent to
\[
\exists  x_0.~(0 \in  x_0 \land \forall  n. ~(n \in x_0 \implies n+1 \in x_0)).
\]

We have the following principle of induction:
\begin{lemma}
    \yalabel{Induction}{Induction}{lem:induction}
    Let $A \subseteq \omega$ such that $0 \in A$
    and for each $y \in A$, we have that $y + 1 \in A$.
    Then $A = \omega$.
\end{lemma}
\begin{proof}
    Clearly $A$ is an inductive set,
    hence $\omega \subseteq A$.
\end{proof}

\begin{definition}
    A set $x$ is \vocab{transitive},
    iff $\forall y \in x.~y \subseteq x$.
\end{definition}
\begin{definition}
    A set $x$ is called an \vocab{ordinal} (or \vocab{ordinal number})
    iff $x$ is transitive
    and for all $y, z \in x$,
    we have that $y = z$, $y \in z$ or $y \ni z$.
\end{definition}
\gist{
Clearly, the $\in$-relation is a well-order on an ordinal $x$.
\begin{remark}
    This definition is due to \textsc{John von Neumann}.
\end{remark}
}{}

\begin{lemma}
    Each natural number (i.e.~element of $\omega$)
    is an ordinal.
\end{lemma}
\begin{proof}
\gist{
    We use \yaref{lem:induction}.
    Clearly $\emptyset$ is an ordinal.
    Now let $\alpha$ be an ordinal.
    We need to show that $\alpha + 1$ is an ordinal.
    It is transitive, since $\alpha$ is transitive
    and $\alpha \subseteq (\alpha + 1)$.

    Let $x, y \in (\alpha+1)$.
    If $x, y \in \alpha$, we know that $x = y \lor x \in y \lor x \ni y$
    since $\alpha$ is an ordinal.
    Suppose $x = \alpha$.
    Then either $y = x$ or $y \in \alpha = x$.
}{Induction}
\end{proof}

\begin{lemma}
    $\omega$ is an ordinal.
\end{lemma}
\begin{proof}
    $\omega$ is transitive:

    Let $y \in \omega$. Let us show by \yaref{lem:induction},
    that $y \subseteq \omega$.
    For $y = \emptyset$ this is clear.

    Suppose that $y \in \omega$ with $y \subseteq \omega$.
    But now $\{y\} \subseteq \omega$,
    so $y + 1 = y \cup \{y\} \subseteq \omega$.


    $\omega$ is well-ordered by $\in$:

    We do a nested induction. First let
    \[
    \phi(y,z) \coloneqq  y \in z \lor y \ni z \lor y = z.
    \] 
    We want to show:
    \begin{enumerate}[(a)]
        \item $\phi(0,0)$
        \item $\forall z \in \omega. ~\phi(0, z) \implies \phi(0,z+1)$.
        \item $\forall  y \in \omega.~((\forall z' \in \omega.~\phi(y,z')) \implies (\forall z \in  \omega.~\phi(y+1, z)))$.
    \end{enumerate}
    (a) and (b) are trivial.
    Fix $y \in \omega$ and
    suppose that $\forall z' \in \omega .~\phi(y, z')$.
    We want to show that $\forall z \in  \omega .~\phi(y+1, z)$.

    We already know that $\forall  z \in \omega.~\phi(0,z)$ holds
    by (b).
    In particular, $\phi(0,y+1)$ holds,
    so $\phi(y+1, 0)$ is true, since $\phi$ is symmetric.
    Now if $\phi(y+1,z)$ is true,
    we want to show $\phi(y+1,z+1)$ is true as well.
    We have
    \[(y + 1 \in z) \lor (y + 1 = z) \lor (y + 1 \ni z)\]
    by assumption.
    \begin{itemize}
        \item If $y + 1 \in z \lor y+1 = z$, then clearly $y + 1 \in z + 1$.
        \item If $y +1 \ni z$, then either $z = y$ or $z \in y$.
        \begin{itemize}
            \item In the first case, $z+1 = y+1$.
            \item Suppose that $z \in y$.
                Then by the induction hypothesis $\phi(y, z+1)$ holds.
                If $y \in z+1$, then $\{y,z\}$ would violate \AxFund.
                If $y = z+1$, then $z + 1 \in y + 1$.
                If $z+1 \in y$, then $z+1 \in y+1$ as well.
        \end{itemize}
    \end{itemize}
\end{proof}