Josia Pietsch
ab14be172f
Some checks failed
Build latex and deploy / checkout (push) Failing after 14m16s
193 lines
6.4 KiB
TeX
193 lines
6.4 KiB
TeX
\lecture{13}{2023-11-30}{}
|
|
\gist{%
|
|
\begin{remark}[``Constructive'' approach to $\omega_1$ ]
|
|
There are many well-orders on $\omega$.
|
|
Let $W$ be the set of all such well-orders.
|
|
For $R, S \in W$,
|
|
write $R \le S$ iff $R$ is isomorphic to
|
|
an initial segment of $S$.
|
|
Consider $\faktor{W}{\sim}$,
|
|
where $R \sim S :\iff R \le S \land S \le R$.
|
|
Define $\le $ on $\faktor{W}{\sim }$
|
|
by $[R] \le [S] :\iff R \le S$.
|
|
Clearly this is well-defined
|
|
and $<$ is a well-order on $\faktor{W}{\sim}$:
|
|
Suppose that $\{R_n : n \in \omega\} \subseteq W$
|
|
is such that $R_{n+1} < R_n$.
|
|
Then there exist $n_i \in \omega$
|
|
such that $R_i \cong R_0\defon{\{x : x <_{R_0} n_i\}}$
|
|
and these form a $<_{R_0}$ strictly decreasing sequence.
|
|
|
|
So $(\faktor{W}{\sim})$
|
|
is a well-ordered set.
|
|
Every well-order on a countable set is isomorphic
|
|
to $(\omega, R)$ for some $[R] \in \faktor{W}{\sim}$.
|
|
|
|
Moreover if $R \in W$,
|
|
then
|
|
\[
|
|
(\omega; R) \cong (\underbrace{\{[S] \in \faktor{W}{\sim} : [S] < [R]\}}_{I}; <\defon{I}),
|
|
\]
|
|
where the isomorphism is given by
|
|
\[
|
|
n \longmapsto [R \defon{\{ m : (m,n) \in R\} }].
|
|
\]
|
|
|
|
This also shows that every $[R] \in \faktor{W}{\sim }$
|
|
has only countably many $<$-predecessors.
|
|
This then also shows that $(\faktor{W}{\sim}, <)$ itself
|
|
is not a well-order on a countable set.
|
|
|
|
Thus $\otp(\faktor{W}{\sim}, <) = \omega_1$.
|
|
\end{remark}
|
|
}{}
|
|
|
|
\gist{%
|
|
\begin{notation}
|
|
Let $I \neq \emptyset$
|
|
and let $\{\kappa_i : i \in I\}$
|
|
be a set of cardinals.
|
|
|
|
Then
|
|
\[
|
|
\sum_{i \in I} \kappa_i \coloneqq \left|\bigcup_{ i \in I} (\kappa_i \times \{i\})\right|
|
|
\]
|
|
and
|
|
\[
|
|
\prod_{i \in I} \coloneqq \left| \bigtimes_{i \in I} \kappa_i\right|,
|
|
\]
|
|
where
|
|
\[
|
|
\bigtimes_{i \in I} A_i \coloneqq \{f : f \text{ is a function}, \dom(f) = I, \forall i.~f(i) \in A_i\}.
|
|
\]
|
|
\end{notation}
|
|
\begin{remark}
|
|
\AxC is equivalent to $\forall i \in I.~A_i \neq \emptyset \implies \bigtimes_{i \in I} A_i \neq \emptyset$.
|
|
\end{remark}
|
|
}{}
|
|
\begin{theorem}[K\H{o}nig]
|
|
\yalabel{K\H{o}nig's Theorem}{K\H{o}nig}{thm:koenig}
|
|
Let $I \neq \emptyset$.
|
|
Let $\{\kappa_i : i \in I\}$,
|
|
$\{\lambda_i : i \in I\}$
|
|
be sets of cardinals
|
|
such that $\kappa_i < \lambda_i$ for all $i \in I$.
|
|
|
|
Then
|
|
\[
|
|
\sum_{i \in I} \kappa_i < \prod_{i \in I} \lambda_i.
|
|
\]
|
|
\end{theorem}
|
|
\begin{proof}
|
|
\gist{%
|
|
Consider a function $F\colon \bigcup_{i \in I} (\kappa_i \times \{i\}) \to \bigtimes_{i \in I}\lambda_i$.
|
|
We want to show that $F$ is not surjective.
|
|
|
|
For $i \in I$, let $\xi_i$ be the least $\xi < \lambda_i$
|
|
such that for all $\eta < \kappa_i$
|
|
\[
|
|
\underbrace{F((\eta, i))(i)}_{\in \lambda_i} \neq \xi.
|
|
\]
|
|
Such $\xi$ exists, since $\kappa_i < \lambda_i$.
|
|
|
|
Let $f \in \bigtimes_{i \in I} \lambda_i$
|
|
be defined by $i \mapsto \xi_i$.
|
|
|
|
Then $f \not\in \ran(F)$.
|
|
}{%
|
|
\begin{itemize}
|
|
\item Consider $F\colon \bigcup_{i \in I}(\kappa_i \times \{i\}) \to \bigtimes_{i \in I} \lambda_i$. Want: $F$ is not surjective.
|
|
\item Let $\xi_i$ minimal such that $\underbrace{F((\eta,i))(i)}_{\in \lambda_i} \neq \xi$
|
|
for all $\eta < \kappa_i$.
|
|
\item $(i \mapsto \xi_i) \not\in \ran(F)$.
|
|
\end{itemize}
|
|
}
|
|
\end{proof}
|
|
|
|
\begin{corollary}
|
|
\label{cor:cfpow}
|
|
For infinite cardinals $\kappa$,
|
|
it is $\cf(2^{\kappa}) > \kappa$.
|
|
\end{corollary}
|
|
\begin{proof}
|
|
\gist{%
|
|
If $2^{\kappa}$ is a successor cardinal,
|
|
then $\cf(2^{\kappa}) = 2^{\kappa} > \kappa$,
|
|
since successor cardinals are regular.
|
|
|
|
Suppose $\cf(2^{\kappa}) \le \kappa$ is a limit cardinal.
|
|
Then there is some cofinal $f\colon \kappa\to 2^{\kappa}$.
|
|
Write $\kappa_i = f(i)$
|
|
(replacing $f(i)$ by $|f(i)|^{+}$ we may assume
|
|
that every $\kappa_i$ is a cardinal).
|
|
|
|
For $i \in \kappa$, write $\lambda_i = 2^{\kappa}$.
|
|
By \yaref{thm:koenig},
|
|
\[
|
|
\sup \{\kappa_i : i < \kappa\} \le \sum_{i \in \kappa} \kappa_i < \prod_{i \in \kappa} \lambda_i = \left( 2^{\kappa} \right)^{\kappa} = 2^{\kappa \cdot \kappa} = 2^{\kappa}
|
|
\]
|
|
and $f$ is not cofinal.
|
|
}{%
|
|
\begin{itemize}
|
|
\item Trivial for successors ($\cf(2^{\kappa}) \overset{\text{regular}}{=} 2^{\kappa} > \kappa$).
|
|
\item Suppose $2^\kappa$ is a limit, $\exists f\colon \kappa \to 2^{\kappa}$ cofinal.
|
|
Wlog.~$f(i) \in \Card$.
|
|
But
|
|
\[
|
|
\sup \{f(i)) : i < \kappa\} \le \sum_{i \in \kappa} f(i)
|
|
\overset{\yaref{thm:koenig}}{<} \prod_{i \in \kappa} 2^{\kappa}
|
|
= (2^{\kappa})^{\kappa} = 2^{\kappa}. ~ ~\qedhere
|
|
\]
|
|
\end{itemize}
|
|
}
|
|
\end{proof}
|
|
|
|
\begin{fact}
|
|
Properties of the function $\kappa \mapsto 2^{\kappa}$.
|
|
\begin{itemize}
|
|
\item $\mu < \kappa \implies 2^{\mu} \le 2^{\kappa}$
|
|
(it is independent of $\ZFC$ whether or not this is strictly increasing).
|
|
\item $\cf(2^{\kappa}) \ge \kappa^+$.
|
|
\end{itemize}
|
|
This is ``all'' you can prove in $\ZFC$.
|
|
\end{fact}
|
|
|
|
The next goal is to show the following:
|
|
(However the method might be more interesting than the result)
|
|
\begin{theorem}[Silver]
|
|
\yalabel{Silver's Theorem (case of $\aleph_{\omega_1}$)}{Silver ($\aleph_{\omega_{1}}$)}{thm:silver1}
|
|
If $2^{\aleph_\alpha} = \aleph_{\alpha + 1}$
|
|
for all $\alpha < \omega_1$,
|
|
then $2^{\aleph_{\omega_1}} = \aleph_{\omega_1 + 1}$.
|
|
\end{theorem}
|
|
|
|
|
|
Relevant concepts to prove this theorem:
|
|
|
|
\begin{definition}
|
|
Let $\alpha$ be a limit ordinal.
|
|
\begin{itemize}
|
|
\item We say that $A \subseteq \alpha$
|
|
is \vocab{unbounded} (in $\alpha$),
|
|
iff for all $\beta < \alpha$,
|
|
there is some $\gamma \in A$
|
|
such that $\beta < \gamma$.
|
|
\item We say that $A \subseteq \alpha$
|
|
is \vocab{closed},
|
|
iff it is closed with respect to the order topology on $\alpha$,
|
|
i.e.~for all $\beta < \alpha$,
|
|
\[
|
|
\sup(A \cap \beta) \in A \cup \{0\} .
|
|
\]
|
|
\item $A$ is \vocab{club} (\emph{cl}osed \emph{un}bounded)
|
|
iff it is closed and unbounded.
|
|
\end{itemize}
|
|
\end{definition}
|
|
The interesting case is that $\alpha$ is a regular uncountable cardinal.
|
|
\begin{fact}
|
|
$A \subseteq \alpha$ being unbounded
|
|
is equivalent to $f\colon \beta\to \alpha$ being cofinal,
|
|
where
|
|
$(\beta, \in ) \overset{f}{\cong} (A, \in )$.
|
|
\end{fact}
|