273 lines
9.1 KiB
TeX
273 lines
9.1 KiB
TeX
\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: $\beta \ge \alpha+1$.
|
|
Note that for all $\gamma \le \beta$
|
|
we have
|
|
\[
|
|
\aleph_{\gamma}^{\aleph_\beta} \le \aleph_\beta^{\aleph_\beta}
|
|
\le \left( 2^{\aleph_\beta} \right)^{\aleph_\beta}
|
|
= 2^{\aleph_\beta \cdot \aleph_\beta} = 2^{\aleph_{\beta}}
|
|
\le \aleph_\gamma^{\aleph_\beta}.
|
|
\]
|
|
So in this case $\aleph_\alpha^{\aleph_{\beta}} = 2^{\aleph_\beta}$
|
|
and $\aleph_{\alpha+1}^{\aleph_\beta} = 2^{\aleph_{\beta}}$.
|
|
Thus
|
|
\[
|
|
\aleph_{\alpha+1}^{\aleph_{\beta}} = 2^{\aleph_{\beta}}
|
|
= \aleph_{\alpha}^{\aleph_\beta} = \aleph_{\alpha}^{\aleph_\beta} \cdot \aleph_{\alpha+1}.
|
|
\]
|
|
\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}
|