lecture 12

inputs/lecture_12.tex

@@ -0,0 +1,229 @@
+\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.
+    We say that $f\colon  \alpha \to \beta$ is \vocab{cofinal} 
+    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}

logic.sty

@@ -139,5 +139,7 @@
 \DeclareSimpleMathOperator{rk}
 \DeclareSimpleMathOperator{otp}
 
+\DeclareSimpleMathOperator{cf}
+
 \newcommand\lecture[3]{\hrule{\color{darkgray}\hfill{\tiny[Lecture #1, #2]}}}
 

logic2.tex

@@ -35,6 +35,7 @@
 \input{inputs/lecture_09}
 \input{inputs/lecture_10}
 \input{inputs/lecture_11}
+\input{inputs/lecture_12}
 
 
 \cleardoublepage