w23-logic-2/inputs/lecture_11.tex

\lecture{11}{2023-11-23}{Cardinals}

\subsection{Cardinals}

\begin{definition}
    Let $a$ be any set.
    The \vocab{cardinality} of $a$
    denoted by $\overline{\overline{a}}$, $|a|$ or $\card(a)$,
    is the smallest ordinal $\alpha$
    such that there is some bijection $f\colon  \alpha \to a$.

    An ordinal $\alpha$ is called a \vocab{cardinal},
    iff \gist{%
        there is some set $a$ with $|a| = \alpha$
        (equivalently, $|\alpha| = \alpha$).%
    }{$|\alpha| = \alpha$.}
\end{definition}

We often write $\kappa, \lambda, \ldots$ for cardinals.

\begin{lemma}
    For every cardinal $\kappa$,
    there is come cardinal $\lambda > \kappa$.
\end{lemma}
\begin{proof}
    \gist{%
        Consider the powerset of  $\kappa$.
        We know that there is no surjection $\kappa \twoheadrightarrow \cP(\kappa)$.
        Hence $\kappa < |2^{\kappa}|$.
    }{$\kappa < |2^{\kappa}|$.}
\end{proof}

\begin{definition}
    For each cardinal $\kappa$,
    $\kappa^+$ denotes the least cardinal $\lambda > \kappa$.
\end{definition}
\gist{%
\begin{warning}
    This has nothing to do with the ordinal successor of $\kappa$.
\end{warning}
}{}
\begin{lemma}
    Let $X$ be any set of cardinals.
    Then $\sup X$ is a cardinal.
\end{lemma}
\begin{proof}
\gist{%
    If there is some $\kappa \in X$ with $\lambda \le \kappa$
    for all $\lambda \in X$,
    then $\kappa = \sup(X)$ is a cardinal.

    Let us now assume that for all $\kappa \in X$
    there is some $\lambda \in X$
    with $\lambda > \kappa$.
    Suppose that $\sup(X)$ is no a cardinal
    and write $\mu = |\sup(X)|$.
    Then $\mu \in \sup(X)$,
    since $\sup(X)$ is an ordinal.
    However $\sup(X)$ is the least ordinal
    larger than all $\alpha \in X$,
    so there is $\lambda \in X$
    with $\lambda > \mu$.
    However, there exists $\mu \twoheadrightarrow \sup(X)$,
    hence also $\mu \twoheadrightarrow \lambda$
    (which is in contradiction to $\lambda$ being a cardinal).
}{%
    \begin{itemize}
        \item If $\sup(X) \in X$ this is trivial.
        \item Let $\sup(X) \not\in X$, $\mu \coloneqq  |\sup(X)|$.
          \begin{itemize}
              \item Suppose that $\sup(X)$ is not a cardinal.
                  Then $\mu \in \sup(X)$, since $\sup(X) \in \OR$.
              \item $\exists  \lambda \in X.~\lambda > \mu$ $\lightning$.
          \end{itemize}
    \end{itemize}
}
\end{proof}
We may now use the \yaref{lem:recursion}
to define a sequence $\langle \aleph_\alpha : \alpha \in \OR \rangle$
with the following properties:
\begin{IEEEeqnarray*}{rCl}
    \aleph_0 &=& \omega,\\
    \aleph_{\alpha+1} &=& (\aleph_\alpha)^+,\\
    \aleph_{\lambda} &=& \sup \{\aleph_\alpha : \alpha < \lambda\}.
\end{IEEEeqnarray*}
Each $\aleph_\alpha$ is a cardinal.
Also, a trivial induction shows that $\alpha \le \aleph_\alpha$.
In particular $|\alpha| \le \aleph_{\alpha}$.
Therefore the $\aleph_\alpha$ are all the infinite cardinals:
If $a$ is any infinite set, then $|a| \le  \aleph_{|a|}$,
so $|a| = \aleph_\beta$ for some $\beta \le |\alpha|$.

\begin{notation}
    Sometimes we write $\omega_\alpha$ for $\aleph_\alpha$
    (when viewing it as an ordinal).
\end{notation}

\begin{notation}
    Let $\leftindex^a b \coloneqq  \{f : f \text{ is a function}, \dom(f) = \alpha, \ran(f) \subseteq b\}$.
\end{notation}

\begin{definition}[Cardinal arithmetic]
    Let $\kappa$, $\lambda$ be cardinals.
    Define
    \begin{IEEEeqnarray*}{rCl}
        \kappa + \lambda &\coloneqq & |\{0\} \times \kappa \cup \{1\} \times \lambda|,\\
        \kappa \cdot \lambda &\coloneqq & |\kappa \times  \lambda|,\\
        \kappa^{\lambda} &\coloneqq & |\leftindex^{\lambda}\kappa|.
    \end{IEEEeqnarray*}
\end{definition}
\gist{%
\begin{warning}
    This is very different from ordinal arithmetic!
\end{warning}
}{}

\begin{theorem}[Hessenberg]
    \label{thm:hessenberg}
    For all $\alpha$ we have
    \[
    \aleph_\alpha \cdot \aleph_\alpha = \aleph_\alpha.
    \]
\end{theorem}
\begin{corollary}
    For all $\alpha, \beta$ it is
    \[
    \aleph_\alpha + \aleph_\beta = \aleph_\alpha \cdot  \aleph_\beta = \max \{\aleph_\alpha, \aleph_\beta\}.
    \]
\end{corollary}
\begin{proof}
    Wlog.~$\alpha \le \beta$.
    \gist{%
        Trivially $\aleph_\alpha \le \aleph_\beta$.
        It is also clear that
        \[
            \aleph_{\beta} \le  \aleph_\alpha + \aleph_{\beta} \le  \aleph_\alpha \cdot \aleph_\beta \le \aleph_\beta \cdot \aleph_\beta = \aleph_\beta.
        \]
    }{Then $\aleph_{\beta} \le  \aleph_\alpha + \aleph_{\beta} \le  \aleph_\alpha \cdot \aleph_\beta \le \aleph_\beta \cdot \aleph_\beta = \aleph_\beta$.}
\end{proof}
\begin{refproof}{thm:hessenberg}
\gist{%
    Define a well-order $<^\ast$ on $\OR \times \OR$ by setting
    \[
    (\alpha,\beta) <^\ast (\gamma,\delta)
    \]
    iff
    \begin{itemize}
        \item $\max(\alpha,\beta) < \max(\gamma,\delta)$ or
        \item $\max(\alpha,\beta) = \max(\gamma,\delta)$ and $\alpha < \gamma$ or
        \item $\max(\alpha,\beta) = \max(\gamma,\delta)$ and $\alpha = \gamma$ and $\beta < \delta$.
    \end{itemize}

    It is clear that this is a well-order.

    There is an isomorphism
    \[
        (\OR, <) \cong^{\Gamma^{-1}} (\OR \times  \OR, <^\ast).
    \]
    $\Gamma$ is called the \vocab{Gödel pairing function}.
    \begin{claim}
        For all $\alpha$ it is
        $\ran(\Gamma\defon{\aleph_\alpha \times \aleph_\alpha}) = \aleph_\alpha$,
        i.e.
        \[
        \aleph_\alpha = \{\xi: \exists  \eta, \eta' < \aleph_\alpha.~\xi = \Gamma((\eta,\eta'))\}.
        \]
    \end{claim}
    \begin{subproof}
        We use induction of $\alpha$.
        The claim is trivial for $\alpha = 0$.
        Now let $\alpha > 0$ and suppose the claim
        to be true for all $\beta < \alpha$.
        It is easy to see that
        \[
        \ran(\Gamma\defon{\aleph_\alpha \times  \aleph_\alpha}) \supseteq \aleph_\alpha,
        \] 
        as otherwise $\Gamma\defon{\aleph_\alpha \times  \aleph_\alpha}: \aleph_{\alpha} \times  \aleph_\alpha \to  \eta$
        would be a bijection for some $\eta < \aleph_\alpha$,
        but $\aleph_\alpha$ is a cardinal.

        Suppose that $\ran(\Gamma\defon{\aleph_\alpha \times  \aleph_\alpha}) \supsetneq \aleph_\alpha$.
        Then there exist $\eta, \eta' < \aleph_\alpha$ with
        \[
        \Gamma((\eta, \eta')) = \aleph_\alpha.
        \] 

        So $\Gamma\defon{\{(\gamma,\delta) : (\gamma,\delta) <^\ast (\eta, \eta'\}}$
        is bijective onto $\aleph_\alpha$.
        If $(\gamma,\delta) <^\ast (\eta, \eta')$,
        then $\max \{\gamma,\delta\} \le \max \{\eta, \eta'\}$.
        Say $\eta \le  \eta' < \aleph_\alpha$
        and let $\aleph_\beta = |\eta'|$.
        There is a surjection
        \[f\colon  \underbrace{(\eta +1)}_{ \le \aleph_\beta} \times \underbrace{(\eta' + 1)}_{\sim  \aleph_\beta} \twoheadrightarrow \aleph_\alpha.\]

        This gives rise to a surjection $f^\ast \colon  \aleph_\beta \times \aleph_\beta \to \aleph_\alpha$.
        The inductive hypothesis then produces a surjection
        $f^\ast\colon  \aleph_\beta \to  \aleph_\alpha \lightning$.
    \end{subproof}
}{
    \begin{itemize}
        \item Define wellorder $<^\ast \subseteq \OR \times \OR$,
            $(\alpha,\beta) <^\ast (\gamma,\delta)$ iff
            \begin{itemize}
                \item $\max(\alpha,\beta) < \max(\gamma,\delta)$ or
                \item $\max(\alpha,\beta) = \max(\gamma,\delta)$ and $\alpha < \gamma$ or
                \item $\max(\alpha,\beta) = \max(\gamma,\delta)$ and $\alpha = \gamma$ and $\beta < \delta$.
            \end{itemize}
        \item Gödel pairing function $\Gamma : (\OR \times \OR, <^\ast) \xrightarrow{\cong} (\OR, <)$.
        \item $\forall \alpha.~\ran(\Gamma\defon{\aleph_\alpha \times \aleph_\alpha}) = \aleph_\alpha$.
            \begin{itemize}
                \item Induction on $\alpha$.
                \item Clearly $\aleph_\alpha \subseteq \ran(\Gamma\defon{\aleph_\alpha \times \aleph_\alpha})$
                    ($\aleph_\alpha$ is a cardinal).
                \item Suppose $\aleph_\alpha \subsetneq \ran(\Gamma\defon{\aleph_\alpha \times  \aleph_\alpha})$,
                    We get a surjection $\aleph_\beta \xrightarrow{\text{induction}} \aleph_\beta \times  \aleph_\beta \to \aleph_\alpha$
                    for some $\beta < \alpha$.
            \end{itemize}
    \end{itemize}
}
\end{refproof}
\gist{%
However, exponentiation of cardinals is far from trivial:
\begin{observe}
    $2^{\kappa} = |\cP(\kappa)|$,
    since $\leftindex^{\kappa} \{0,1\} \leftrightarrow\cP(\kappa)$.

    Hence by Cantor $2^{\kappa}\ge \kappa^+$.
\end{observe}
This is basically all we can say.
}{}

The \vocab{continuum hypothesis} states that $2^{\aleph_0} = \aleph_1$.