w23-logic-2/inputs/lecture_12.tex

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\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),\\
\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*}
\begin{example}
\leavevmode
\begin{itemize}
\item $2+ 2 =4$,
\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}
\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.
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We say that $f\colon \alpha \to \beta$ is \vocab{cofinal}
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iff for all $\xi < \beta$, there is some $\eta < \alpha$
such that $f(\eta) \ge \xi$.
\end{definition}
\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}k
\begin{example}
\begin{itemize}
\item $\cf(\aleph_\omega) = \omega$.
In fact $\cf(\aleph_{\lambda}) \le \lambda$
for limit ordinals $\lambda$
(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}
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 $|\alpha| \le \alpha$.
\end{proof}
\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 map
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}
\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 $\ZFC$.
\begin{theorem}[Hausdorff]
\[
\aleph_{\alpha+1}^{\aleph_\beta} = \aleph_\alpha^{\aleph_\beta} \cdot \aleph_{\alpha+1}.
\]
\end{theorem}
\begin{proof}
Recall that
\begin{IEEEeqnarray*}{rCl}
\aleph_{\alpha+1}^{\aleph_\beta} &=& |{}^{\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,
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
\[
{}^{\aleph_{\beta}}\aleph_{\alpha+1} = \bigcup_{\xi < \aleph_{\alpha+1}} {}^{\aleph_\beta} \xi
\]
for each $\xi < \aleph_{\alpha+1}$, $|\xi| \le \aleph_\alpha$,
hence
\[
|{}^{\aleph_{\beta}}\xi| \le \aleph_\alpha^{\aleph_\beta}
\]
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}
\end{proof}