\lecture{12}{2023-11-27}{}

\subsection{Ordinal arithmetic}

We define $+$, $\cdot $ and exponentiation
for ordinals as follows:

Fix an ordinal $\beta$.
We recursively define
\begin{IEEEeqnarray*}{rClClr}
    \beta &+& 0 &\coloneqq & \beta\\
    \beta &+& (\alpha + 1)&\coloneqq &(\beta + \alpha) + 1,\\
    \beta &+& \lambda &\coloneqq & \sup_{\alpha < \lambda} \beta + \alpha &~ ~\text{for limit ordinals $\lambda$}
\end{IEEEeqnarray*}
(Recall that $\alpha + 1 = \alpha \cup \{\alpha\}$
was already defined.)
\begin{IEEEeqnarray*}{rClClr}
    \beta &\cdot&  0 &\coloneqq&  0,\\
    \beta &\cdot&  (\alpha+1) &\coloneqq & \beta\cdot \alpha + \beta,\\
    \beta &\cdot& \lambda &\coloneqq &\sup_{\alpha < \lambda} \beta\cdot \alpha  &~ ~\text{for limit ordinals $\lambda$}
\end{IEEEeqnarray*}
and
\begin{IEEEeqnarray*}{rClr}
    \beta^0 &\coloneqq & 1,\\
    \beta^{\alpha+1} &\coloneqq & \beta^{\alpha} \cdot \beta,\\
    \beta^{\lambda} &\coloneqq & \sup_{\alpha < \lambda} \beta^{\alpha} &~ ~\text{for limit ordinals $\lambda$}.
\end{IEEEeqnarray*}

\gist{%
\begin{example}
    \leavevmode
    \begin{itemize}
        \item $2+ 2 =4$,
        \item $196883 + 1 = 196884$,
        \item $1 + \omega = \sup_{n < \omega} 1 + n = \omega \neq \omega +1$,
        \item $2 \cdot \omega = \sup_{n < \omega} 2\cdot n = \omega$,
        \item $\omega \cdot 2 =\omega \cdot  1 + \omega = \omega + \omega$.
    \end{itemize}
\end{example}
}{}

\gist{%
\begin{warning}
    Cardinal arithmetic and ordinal arithmetic are very different!
    The symbols are the same, but usually we will distinguish
    between the two by the symbols used for variables
    (i.e.~$\alpha,\beta, \omega, \omega_1$ are viewed primarily as ordinals
    and $\kappa,\lambda, \aleph_\alpha$ as cardinals).
\end{warning}

We will very rarely use ordinal arithmetic.
}{}

\subsection{Cofinality}

\begin{definition}
    Let $\alpha$, $\beta$ be ordinals.
    We say that $f\colon  \alpha \to \beta$ is \vocab{cofinal}
    iff
    \gist{%
        for all $\xi < \beta$, there is some $\eta < \alpha$
        such that $f(\eta) \ge \xi$.%
    }{%
        \[
        \forall \xi < \beta.~\exists \eta < \alpha.~f(\eta) \ge \xi.
        \]
    }
\end{definition}
\gist{%
\begin{remark}
    If $\beta$ is a limit ordinal,
    this is equivalent to
    \[
    \forall  \xi < \beta .~\exists \eta  < \alpha.~f(\eta) > \xi.
    \]
\end{remark}
\begin{example}
    \begin{enumerate}[(a)]
        \item Look at $\omega + \omega$.
            \begin{IEEEeqnarray*}{rCl}
                f\colon \omega &\longrightarrow & \omega + \omega \\
                n &\longmapsto & \omega + n
            \end{IEEEeqnarray*}
            is cofinal.
        \item Look at $\aleph_\omega$.
            Then
            \begin{IEEEeqnarray*}{rCl}
                f\colon \omega &\longrightarrow & \aleph_{\omega} \\
                n &\longmapsto & \aleph_n
            \end{IEEEeqnarray*}
            is cofinal.
    \end{enumerate}
\end{example}
}{}
\begin{definition}
    Let $\beta$ be an ordinal.
    The \vocab{cofinality} of $\beta$,
    denoted $\cf(\beta)$,
    is the least ordinal $\alpha$
    such that there exists a cofinal
    $f\colon \alpha \to \beta$.
\end{definition}
\begin{example}
    \begin{itemize}
        \item $\cf(\aleph_\omega) = \omega$.
            In fact $\cf(\aleph_{\lambda}) \le \lambda$
            for limit ordinals $\lambda \neq 0$
            (consider $\alpha \mapsto \aleph_\alpha$).
        \item $\cf(\aleph_{\omega + \omega}) = \omega$.
    \end{itemize}
\end{example}
\begin{lemma}
    For any ordinal $\beta$,
    $\cf(\beta)$ is a cardinal.
\end{lemma}
\begin{proof}
\gist{%
    Let $f\colon \alpha \to \beta$ be cofinal.
    Then $\tilde{f}\colon |\alpha| \to \beta$,
    the composition with $\alpha \leftrightarrow|\alpha|$
    is cofinal as well and $|\alpha| \le \alpha$.
}{Trivial.}
\end{proof}

\gist{%
\begin{question}
    How does one imagine ordinals with
    cofinality $> \omega$?
\end{question}
No idea.
}{}

\begin{definition}
    An ordinal $\beta$ is \vocab{regular}
    iff $\cf(\beta) = \beta$.
    Otherwise $\beta$ is called \vocab{singular}.
\end{definition}
In particular, a regular ordinal is always a cardinal.
\begin{lemma}
    Let $\beta$ be an ordinal
    Then $\cf(\beta)$ is a regular cardinal,
    i.e.
    \[\cf(\cf(\beta)) = \cf(\beta).\]
\end{lemma}
\begin{proof}
    Suppose not.
    Let $f\colon \cf(\beta) \to \beta$ be cofinal
    and $g\colon \cf(\cf(\beta)) \to \cf(\beta)$.

    Consider
    \begin{IEEEeqnarray*}{rCl}
        h\colon \cf(\cf(\beta)) &\longrightarrow & \beta \\
        \eta &\longmapsto & \sup \{f(\xi): \xi \le g(\eta)\}  < \beta.
    \end{IEEEeqnarray*}
    Clearly this is cofinal.
\end{proof}
\begin{warning}
    Note that in general, a composition of cofinal maps
    is not necessarily cofinal.
\end{warning}

\begin{theorem}
    Let $\kappa > \aleph_0$.
    Then $\kappa^+$ is regular.
\end{theorem}
\begin{proof}
    Suppose that $\cf(\kappa^+) < \kappa^+$.
    Then $\cf(\kappa^+) \le \kappa$,
    i.e.~there is a cofinal function $f\colon \kappa \to  \kappa^+$.
    By the axiom of choice,
    there is a function $g$ with domain $\kappa$,
    such that $g(\eta)\colon \kappa \twoheadrightarrow f(\eta)$ is onto.
    Now define
    \begin{IEEEeqnarray*}{rCl}
        h\colon \kappa\times \kappa &\longrightarrow & \kappa^+ \\
        (\eta, \xi) &\longmapsto & g(\eta)(\xi).
    \end{IEEEeqnarray*}
    Clearly this is surjective,
    but $|\kappa \times \kappa| < \kappa^+$,
    by \yaref{thm:hessenberg}.
\end{proof}

\gist{%
\begin{itemize}
    \item $\aleph_0, \aleph_1, \aleph_2, \ldots$ are regular,
    \item $\aleph_\omega$ is singular,
    \item $\aleph_{\omega + 1}, \aleph_{\omega + 2}, \ldots$ are regular,
    \item $\aleph_{\omega + \omega}$ is singular,
    \item $\aleph_{\omega + \omega + 1}, \ldots$ are regular,
    \item $\aleph_{\omega + \omega + \omega}$ is singular,
    \item $\ldots$
    \item $\aleph_{\omega_1}$ is singular,
    \item $\aleph_{\omega_1 + 1}, \ldots$ is regular,
    \item $\aleph_{\omega_2}$ is singular.
\end{itemize}
}{}

\begin{question}[Hausdorff]
    Is there a regular limit cardinal?
\end{question}
Maybe. This is independent of $\ZFC$,
cf.~\yaref{def:inaccessible}.


\begin{theorem}[Hausdorff]
    \[
    \aleph_{\alpha+1}^{\aleph_\beta} = \aleph_\alpha^{\aleph_\beta} \cdot \aleph_{\alpha+1}.
    \]
\end{theorem}
\begin{proof}
\gist{%
    Recall that
    \begin{IEEEeqnarray*}{rCl}
        \aleph_{\alpha+1}^{\aleph_\beta} &=& |\leftindex^{\aleph_\beta} \aleph_{\alpha+1}|.
    \end{IEEEeqnarray*}
    \begin{itemize}
        \item First case: $\beta \ge \alpha+1$.
            Then
            \[
            \aleph_{a+1}^{\aleph_\beta} \le  \aleph_{\beta}^{\aleph_\beta}
            \le \left( 2^{\aleph_{\beta}} \right)^{\aleph_\beta}
            = 2^{\aleph_{\beta} \cdot \aleph_\beta} = 2^{\aleph_\beta} \le  \aleph_{\alpha+1}^{\aleph_\beta}.
            \]
            Also $\aleph_\alpha^{\aleph_{\beta}} = 2^{\aleph_\beta}$
            in this case (by the same argument),
            so
            \[
            \aleph_{\alpha+1}^{\aleph_{\beta}} = 2^{\aleph_{\beta}}
            = \aleph_{\alpha}^{\aleph_\beta} = \aleph_{\alpha}^{\aleph_\beta} \cdot  \aleph_{\alpha+1}.
            \]
        \item Second case:
            Suppose $\beta < \alpha+1$.
            By case hypothesis and because $\aleph_{\alpha+1}$
            is regular,
            no $f\colon \aleph_{\beta} \to  \aleph_{\alpha+1}$
            is unbounded.
            So
            \[
            \leftindex^{\aleph_{\beta}}\aleph_{\alpha+1} = \bigcup_{\xi < \aleph_{\alpha+1}} \leftindex^{\aleph_\beta} \xi
            \]
            for each $\xi < \aleph_{\alpha+1}$, $|\xi| \le \aleph_\alpha$,
            hence
            \[
            |\leftindex^{\aleph_{\beta}}\xi| \le \aleph_\alpha^{\aleph_\beta}
            \]
            for each $\xi < \aleph_{\alpha+1}$.
            Therefore, \[\aleph_{\alpha+1}^{\aleph_\beta} \le \aleph_{\alpha+1} \cdot \aleph_{\alpha}^{\aleph_{\beta}} \le \aleph_{\alpha+1}^{\aleph_\beta}.\]
    \end{itemize}
}{%
    $\ge$ is trivial.
    \begin{itemize}
        \item First case: $\beta \ge \alpha + 1$.
            Note that $\gamma \le \beta \implies \aleph_{\gamma}^{\aleph_\beta} = 2^{\aleph_\beta}$,
            so
            \[
                \aleph_{\alpha+1}^{\aleph_\beta} \overset{\alpha+1 \le \beta}{=}
                2^{\aleph_\beta}
                \overset{\alpha < \beta}{=} \aleph_\alpha^{\aleph_\beta}
                \le \aleph_\alpha^{\aleph_\beta} \cdot \aleph_{\alpha+1}.
            \]
        \item Second case: $\beta < \alpha+1$:
            \begin{itemize}
                \item $\aleph_{\alpha+1}$ is regular, so all $f\colon \aleph_\beta \to  \aleph_{\alpha+1}$ are bounded.
                \item Thus $\leftindex^{\aleph_\beta}\aleph_{\alpha+1} = \bigcup_{\xi < \aleph_{\alpha+1}} \leftindex^{\aleph_\beta} \xi$ for all $\xi < \aleph_{\alpha+1}$.
                \item So $\aleph_{\alpha+1}^{\aleph_\beta} \le \aleph_{\alpha+1} \aleph_\alpha^{\aleph_\beta}$.
            \end{itemize}
            
    \end{itemize}
}
\end{proof}