\lecture{07}{2023-11-09}{}

\begin{notation}
    From now on, we will write $\alpha, \beta, \ldots$
    for ordinals.
\end{notation}
\begin{lemma}
    \label{lem:7:ordinalfacts}
    \begin{enumerate}[(a)]
        \item $0$ is an ordinal, and if $\alpha$ is
            an ordinal, so is $\alpha + 1$.
        \item If $\alpha$ is an ordinal and $x \in \alpha$,
            then $x$ is an ordinal.
        \item If $\alpha, \beta$ are ordinals
            and $\alpha \subseteq \beta$,
            then $\alpha = \beta$ or $\alpha \in \beta$.
        \item If $\alpha$ and $\beta$ are ordinals,
            then $\alpha \in \beta$, $\alpha = \beta$
            or $\alpha \ni \beta$.
    \end{enumerate}
\end{lemma}
\begin{yarefproof}{lem:7:ordinalfacts}
\gist{
    We have already proved (a) before.

    (b) Fix $x \in \alpha$. Then $x \subseteq \alpha$.
    So if $y, z \in x$, then $y \in  z \lor y = z \lor y \ni z$.
    Let $y \in x$. We need to see $y \subseteq x$.
    Let $z \in y$.
    \begin{claim}
        $z \in x$
    \end{claim}
    \begin{subproof}
    As $\alpha$ is transitive, we have that $z, y, x \in \alpha$.
    Thus $z \in x \lor z = x \lor z \ni x$.

    $z = x$ contradicts \AxFund:
    Consider $\{x,y\}$. Then $x \cap \{x,y\}$ is non empty,
    as it contains $y$.
    Furthermore $x \in y \cap \{x,y\} $

    $z \ni x$ also contradicts \AxFund:
    If $x \in z$, then $z \ni x \ni y \ni z \ni x \ni \ldots$.
    $\{x,y,z\}$ yields a contradiction,
    as $y \in x \cap \{x,y,z\}$, $z \in  y \cap \{x,y,z\}$,
    $x \in z \cap \{x,y,z\}$.

    So $z \in x$ as desired.
    \end{subproof}

    (c) Say $\alpha \subsetneq \beta$.
    Pick $\xi \in \beta \setminus \alpha$
    such that $\eta \in \alpha$ for every $\eta \in\xi \cap \beta$.
    (This exists by \AxFund).
    We want to see that $\xi = \alpha$.
    We have $\xi \subseteq \alpha$ by the choice of $\xi$.
    On the other hand $\alpha \subseteq \xi$:
    Let $\eta \in \alpha \subseteq \beta$.
    We have that $\eta \in \xi \lor \eta = \xi \lor \eta \ni \xi$.
    If $\xi \in \eta$, then since $\eta  \in \alpha$,
    we get $\xi \in \alpha$ contradicting the choice of $\xi$.
    If $\xi = \eta$, the $\xi = \eta \in \alpha$,
    which also is a contradiction.
    Thus $\eta \in \xi$.

    This yields $\alpha \in \beta$, hence $\alpha$ is an ordinal.


    (d) By (c) if $\alpha$ and $\beta$ are ordinals,
    then $\alpha \subseteq  \beta \iff (\alpha = \beta \lor \alpha \in \beta)$.
    We need tho see that if $ \alpha$, $\beta$ are ordinals,
    then $\alpha \subseteq \beta$ or $\beta \subseteq \alpha$.
    Suppose there are ordinals $\alpha$, $\beta$ such that
    this is not the case.

    Pick such an $\alpha$.

    Let $\alpha_0 \in  \alpha \cup \{\alpha\}$
    be such that there is some $\beta$
    with $\lnot( \beta \subseteq \alpha_0 \lor \alpha_0 \subseteq \beta)$
    for for all $\gamma \in \alpha_0$,
    $\forall \beta. ( \beta \subseteq \gamma \lor \gamma \subseteq \beta)$.
    Pick $\beta_0$ such that
    \[
    \lnot \left( \beta_0 \subseteq \alpha_0 \lor \alpha_0 \subseteq \beta_0 \right).
    \]

    Consider $\alpha_0 \cup \beta_0$.
    \begin{claim}
        $\alpha_0 \cup \beta_0$
        is an ordinal.
    \end{claim}
    \begin{subproof}
        $\alpha_0 \cup \beta_0$ is clearly transitive.
        Let $\gamma,\delta \in \alpha_0 \cup \beta_0$.
        We claim that $\gamma \in \delta \lor \gamma = \delta \lor \gamma \in \delta$.
        This can only fail if $\gamma \in \alpha_0$ and $\delta \in \beta_0$
        (or the other way around).
        But then $\gamma \in \delta \lor \gamma = \delta \lor \delta \in \gamma$
        by the choice of $\alpha_0$.
    \end{subproof}

    \begin{claim}
        $\alpha_0 = \alpha_0 \cup \beta_0$
        or $\beta_0 = \alpha_0 \cup \beta_0$.
    \end{claim}
    \begin{subproof}
        If that is not the case,
        then $\alpha_0 \in \alpha_0 \cup \beta_0$
        and $\beta_0 \in  \alpha_0 \cup \beta_0$.
        $\alpha_0 \in  \alpha_0$ violates \AxFund.
        Hence $\alpha_0 \in \beta_0$.
        By the same argument, $\beta_0 \in \alpha_0$.
        But this violates \AxFund,
        as $\alpha_0 \in  \beta_0 \in \alpha_0$.
    \end{subproof}
}{Long and tedious, but not many ideas.}
\end{yarefproof}

\begin{lemma}
    Let $X$ be a set of ordinals,
    $X \neq \emptyset$.
    Then $\bigcap X$ and $\bigcup X$ are ordinals.
\end{lemma}
\begin{proof}
    Easy.
\end{proof}
It is actually the case that $\bigcap X \in X$:
Pick $\alpha \in X$ such that $\alpha \subseteq \beta$
for all $\beta \in X$. This exists
by \AxFund and since all ordinals are comparable.
Then $\alpha = \bigcap X$.

\begin{notation}
    We write $\min(X)$ for  $\bigcap X$
    and $\sup(X)$ for $\bigcup X$.
\end{notation}

It need not be the case that $\bigcup X \in X$,
for example $\bigcup \omega = \omega$.
% but  $\bigcup 2 = 1$.

\begin{definition}
    An ordinal $\alpha$ is called
    a \vocab[Ordinal!successor]{successor ordinal},
    iff $\alpha = \beta \cup \{\beta\}$
    for some $\beta \in \alpha$.
    Otherwise $\alpha$ is called
    a \vocab[Ordinal!limit]{limit ordinal}.
\end{definition}
\gist{
\begin{observe}
    Note that $\alpha$ is a limit ordinal iff for all
    $\beta \in \alpha$, $\beta + 1 \in \alpha$:
    If there is $\beta \in \alpha$ such that $\beta+1 \not\in \alpha$,
    then either $\alpha = \beta+1$ (i.e.~$\alpha$ is a successor)
    or $\alpha \in \beta+1$,
    in which case $\beta \in \alpha \in \beta \cup \{\beta\} \lightning$.

    Also if $\alpha$ is a successor,
    then by definition there is some $\beta \in \alpha$,
    with $\beta + 1 = \alpha$, so $\beta + 1 \not\in \alpha$.
\end{observe}
}{}
\gist{
\begin{notation}
    If $\alpha, \beta$ are ordinals,
    we write $\alpha < \beta$ for $\alpha \in \beta$
    (equivalently $\alpha \subsetneq \beta$).
    We also write $\alpha \le \beta$
    for $\alpha \in \beta \lor \alpha = \beta$
    (i.e.~$\alpha \subseteq \beta$).
\end{notation}

\begin{example}
    Limit ordinals:
    \begin{itemize}
        \item $0$,
        \item $\omega$,
        \item $\omega + \omega = \sup(\omega \cup \{\omega, \omega + 1, \ldots\})$,%
            \footnote{To show that this exists, we need the recursion theorem
                and replacement.}
            $\omega + \omega + \omega, \omega + \omega + \omega + \omega, \ldots$
    \end{itemize}

    Successor ordinals:
    \begin{itemize}
        \item $1 = \{0\}, 2 = \{0,1\}, 3, \ldots$
        \item $\omega +1 = \omega \cup \{\omega\} , \omega + 2, \ldots$,
    \end{itemize}
\end{example}
}{}