230 lines
7.6 KiB
TeX
230 lines
7.6 KiB
TeX
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\lecture{12}{2023-11-27}{}
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\subsection{Ordinal arithmetic}
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We define $+$, $\cdot $ and exponentiation
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for ordinals as follows:
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Fix an ordinal $\beta$.
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We recursively define
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\begin{IEEEeqnarray*}{rClClr}
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\beta &+& 0 &\coloneqq & \beta\\
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\beta &+& (\alpha + 1)&\coloneqq &(\beta + \alpha),\\
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\beta &+& \lambda &\coloneqq & \sup_{\alpha < \lambda} \beta + \alpha &~ ~\text{for limit ordinals $\lambda$}
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\end{IEEEeqnarray*}
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(Recall that $\alpha + 1 = \alpha \cup \{\alpha\}$
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was already defined.)
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\begin{IEEEeqnarray*}{rClClr}
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\beta &\cdot& 0 &\coloneqq& 0,\\
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\beta &\cdot& (\alpha+1) &\coloneqq & \beta\cdot \alpha + \beta,\\
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\beta &\cdot& \lambda &\coloneqq &\sup_{\alpha < \lambda} \beta\cdot \alpha &~ ~\text{for limit ordinals $\lambda$}
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\end{IEEEeqnarray*}
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and
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\begin{IEEEeqnarray*}{rClr}
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\beta^0 &\coloneqq & 1,\\
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\beta^{\alpha+1} &\coloneqq & \beta^{\alpha} \cdot \beta,\\
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\beta^{\lambda} &\coloneqq & \sup_{\alpha < \lambda} \beta^{\alpha} &~ ~\text{for limit ordinals $\lambda$}.
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\end{IEEEeqnarray*}
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\begin{example}
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\leavevmode
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\begin{itemize}
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\item $2+ 2 =4$,
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\item $1 + \omega = \sup_{n < \omega} 1 + n = \omega \neq \omega +1$,
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\item $2 \cdot \omega = \sup_{n < \omega} 2\cdot n = \omega$,
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\item $\omega \cdot 2 =\omega \cdot 1 + \omega = \omega + \omega$..
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\end{itemize}
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\end{example}
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\begin{warning}
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Cardinal arithmetic and ordinal arithmetic are very different!
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The symbols are the same, but usually we will distinguish
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between the two by the symbols used for variables
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(i.e.~$\alpha,\beta, \omega, \omega_1$ are viewed primarily as ordinals
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and $\kappa,\lambda, \aleph_\alpha$ as cardinals).
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\end{warning}
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We will very rarely use ordinal arithmetic.
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\subsection{Cofinality}
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\begin{definition}
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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 $\eta < \alpha$
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such that $f(\eta) \ge \xi$.
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\end{definition}
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\begin{remark}
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If $\beta$ is a limit ordinal,
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this is equivalent to
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\[
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\forall \xi < \beta .~\exists \eta \alpha.~f(\eta) > \xi.
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\]
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\end{remark}
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\begin{example}
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\begin{enumerate}[(a)]
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\item Look at $\omega + \omega$.
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\begin{IEEEeqnarray*}{rCl}
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f\colon \omega &\longrightarrow & \omega + \omega \\
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n &\longmapsto & \omega + n
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\end{IEEEeqnarray*}
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is cofinal.
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\item Look at $\aleph_\omega$.
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Then
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\begin{IEEEeqnarray*}{rCl}
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f\colon \omega &\longrightarrow & \aleph_{\omega} \\
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n &\longmapsto & \aleph_n
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\end{IEEEeqnarray*}
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is cofinal.
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\end{enumerate}
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\end{example}
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\begin{definition}
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Let $\beta$ be an ordinal.
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The \vocab{cofinality} of $\beta$,
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denoted $\cf(\beta)$,
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is the least ordinal $\alpha$
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such that there exists a cofinal
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$f\colon \alpha \to \beta$.
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\end{definition}k
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\begin{example}
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\begin{itemize}
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\item $\cf(\aleph_\omega) = \omega$.
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In fact $\cf(\aleph_{\lambda}) \le \lambda$
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for limit ordinals $\lambda$
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(consider $\alpha \mapsto \aleph_\alpha$).
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\item $\cf(\aleph_{\omega + \omega}) = \omega$.
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\end{itemize}
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\end{example}
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\begin{lemma}
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For any ordinal $\beta$,
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$\cf(\beta)$ is a cardinal.
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\end{lemma}
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\begin{proof}
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Let $f\colon \alpha \to \beta$ be cofinal.
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Then $\tilde{f}\colon |\alpha| \to \beta$,
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the composition with $\alpha \leftrightarrow|\alpha|$
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is cofinal as well and $|\alpha| \le \alpha$.
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\end{proof}
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\begin{question}
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How does one imagine ordinals with
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cofinality $> \omega$?
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\end{question}
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No idea.
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\begin{definition}
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An ordinal $\beta$ is \vocab{regular}
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iff $\cf(\beta) = \beta$.
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Otherwise $\beta$ is called \vocab{singular}.
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\end{definition}
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In particular, a regular ordinal is always a cardinal.
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\begin{lemma}
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Let $\beta$ be an ordinal
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Then $\cf(\beta)$ is a regular cardinal,
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i.e.
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\[\cf(\cf(\beta)) = \cf(\beta).\]
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\end{lemma}
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\begin{proof}
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Suppose not.
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Let $f\colon \cf(\beta) \to \beta$ be cofinal
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and $g\colon \cf(\cf(\beta)) \to \cf(\beta)$.
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Consider
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\begin{IEEEeqnarray*}{rCl}
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h\colon \cf(\cf(\beta)) &\longrightarrow & \beta \\
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\eta &\longmapsto & \sup \{f(\xi): \xi \le g(\eta)\} < \beta.
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\end{IEEEeqnarray*}
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Clearly this is cofinal.
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\end{proof}
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\begin{warning}
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Note that in general, a composition of cofinal map
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is not necessarily cofinal.
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\end{warning}
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\begin{theorem}
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Let $\kappa > \aleph_0$.
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Then $\kappa^+$ is regular.
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\end{theorem}
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\begin{proof}
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Suppose that $\cf(\kappa^+) < \kappa^+$.
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Then $\cf(\kappa^+) \le \kappa$,
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i.e.~there is a cofinal function $f\colon \kappa \to \kappa^+$.
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By the axiom of choice,
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there is a function $g$ with domain $\kappa$,
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such that $g(\eta)\colon \kappa \twoheadrightarrow f(\eta)$ is onto.
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Now define
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\begin{IEEEeqnarray*}{rCl}
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h\colon \kappa\times \kappa &\longrightarrow & \kappa^+ \\
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(\eta, \xi) &\longmapsto & g(\eta)(\xi).
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\end{IEEEeqnarray*}
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Clearly this is surjective,
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but $|\kappa \times \kappa| < \kappa^+$,
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by \yaref{thm:hessenberg}.
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\end{proof}
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\begin{itemize}
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\item $\aleph_0, \aleph_1, \aleph_2, \ldots$ are regular,
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\item $\aleph_\omega$ is singular,
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\item $\aleph_{\omega + 1}, \aleph_{\omega + 2}, \ldots$ are regular,
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\item $\aleph_{\omega + \omega}$ is singular,
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\item $\aleph_{\omega + \omega + 1}, \ldots$ are regular,
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\item $\aleph_{\omega + \omega + \omega}$ is singular,
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\item $\ldots$
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\item $\aleph_{\omega_1}$ is singular,
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\item $\aleph_{\omega_1 + 1}, \ldots$ is regular,
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\item $\aleph_{\omega_2}$ is singular.
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\end{itemize}
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\begin{question}[Hausdorff]
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Is there a regular limit cardinal?
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\end{question}
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Maybe. This is independent of $\ZFC$.
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\begin{theorem}[Hausdorff]
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\[
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\aleph_{\alpha+1}^{\aleph_\beta} = \aleph_\alpha^{\aleph_\beta} \cdot \aleph_{\alpha+1}.
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\]
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\end{theorem}
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\begin{proof}
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Recall that
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\begin{IEEEeqnarray*}{rCl}
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\aleph_{\alpha+1}^{\aleph_\beta} &=& |{}^{\aleph_\beta} \aleph_{\alpha+1}|.
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\end{IEEEeqnarray*}
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\begin{itemize}
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\item First case: $\beta \ge \alpha+1$.
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Then
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\[
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\aleph_{a+1}^{\aleph_\beta} \le \aleph_{\beta}^{\aleph_\beta}
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\le \left( 2^{\aleph_{\beta}} \right)^{\aleph_\beta}
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= 2^{\aleph_{\beta} \cdot \aleph_\beta} = 2^{\aleph_\beta} \le \aleph_{\alpha+1}^{\aleph_\beta}.
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\]
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Also $\aleph_\alpha^{\aleph_{\beta} = 2^{\aleph_\beta}}$
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in this case,
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so
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\[
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\aleph_{\alpha+1}^{\aleph_{\beta}} = 2^{\aleph_{\beta}}
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= \aleph_{\alpha}^{\aleph_\beta} = \aleph_{\alpha}^{\aleph_\beta} \cdot \aleph_{\alpha+1}.
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\]
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\item Second case:
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Suppose $\beta < \alpha+1$.
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By case hypothesis and because $\aleph_{\alpha+1}$
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is regular,
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no $f\colon \aleph_{\beta} \to \aleph_{\alpha+1}$
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is unbounded.
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So
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\[
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{}^{\aleph_{\beta}}\aleph_{\alpha+1} = \bigcup_{\xi < \aleph_{\alpha+1}} {}^{\aleph_\beta} \xi
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\]
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for each $\xi < \aleph_{\alpha+1}$, $|\xi| \le \aleph_\alpha$,
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hence
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\[
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|{}^{\aleph_{\beta}}\xi| \le \aleph_\alpha^{\aleph_\beta}
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\]
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for each $\xi < \aleph_{\alpha+1}$.
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Therefore, \[\aleph_{\alpha+1}^{\aleph_\beta} \le \aleph_{\alpha+1} \cdot \aleph_{\alpha}^{\aleph_{\beta}} \le \aleph_{\alpha+1}^{\aleph_\beta}.\]
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\end{itemize}
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\end{proof}
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