Josia Pietsch
cac1563244
Some checks failed
Build latex and deploy / checkout (push) Failing after 17m37s
234 lines
8.6 KiB
TeX
234 lines
8.6 KiB
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 not 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{thm: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 $\leftindex^a b \coloneqq \{f : f \text{ is a function}, \dom(f) = a, \ran(f) \subseteq b\}$.
|
|
\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 $2^{\aleph_0} = \aleph_1$.
|
|
|