Lecture 7

inputs/lecture_06.tex

@@ -223,13 +223,5 @@ Clearly, the $\in$-relation is a well-order on an ordinal $x$.
                 If $z+1 \in y$, then $z+1 \in y+1$ as well.
         \end{itemize}
     \end{itemize}
-        
 \end{proof}
 
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inputs/lecture_07.tex

@@ -0,0 +1,200 @@
+\lecture{07}{2023-11-09}{}
+
+\begin{lemma}+
+    By the axiom of foundation there cannot exist infinite
+    descending chains
+    $x_1 \ni x_2 \ni x_3 \ni \ldots$
+\end{lemma}
+\begin{proof}
+    \todo{TODO!}
+\end{proof}
+
+\begin{notation}
+    From now on, we will write $\alpha, \beta, \ldots$
+    for ordinals.
+\end{notation}
+\begin{lemma}
+    \begin{enumerate}[(a)]
+        \item $0$ is an ordinal, and if $\alpha$ is
+            an ordinal, so is $\alpha + 1$.
+        \item If $\alpha$ is an ordinal and $x \in a$,
+            then $x$ is an ordinal.
+        \item If $\alpha, \beta$ are ordinals
+            and $\alpha \subseteq \beta$,
+            then $\alpha = \beta$ or $\alpha \in \beta$.
+        \item If $\alpha$ and $\beta$ are ordinals,
+            then $\alpha \in \beta$, $\alpha = \beta$
+            or $\alpha \ni \beta$.
+    \end{enumerate}
+\end{lemma}
+\begin{proof}
+    We have already proved (a) before.
+
+    (b) Fix $x \in \alpha$. Then $x \subseteq \alpha$.
+    So if $y, z \in x$, then $y \in  z \lor y = z \lor y \ni z$.
+    Let $y \in x$. We need to see $y \subseteq x$.
+    Let $z \in y$.
+    \begin{claim}
+        $z \in x$
+    \end{claim}
+    \begin{subproof}
+    As $\alpha$ is transitive, we have that $z, y, x \in \alpha$.
+    Thus $z \in x \lor z = x \lor z \ni x$.
+
+    $z = x$ contradicts Fund:
+    Consider $\{x,y\}$. Then $x \cap \{x,y\}$ is non empty,
+    as it contains $y$.
+    Furthermore $x \in y \cap \{x,y\} $
+
+    $z \ni x$ also contradicts Fund:
+    If $x \in z$, then $z \ni x \ni y \ni z \ni x \ni \ldots$.
+    $\{x,y,z\}$ yields a contradiction,
+    as $y \in x \cap \{x,y,z\}$, $z \in  y \cap \{x,y,z\}$,
+    $x \in z \cap \{x,y,z\}$.
+
+    So $z \in x$ as desired.
+    \end{subproof}
+
+    (c) Say $\alpha \subsetneq \beta$.
+    Pick $\xi \in \beta \setminus \alpha$
+    such that $\eta \in \alpha$ for every $\eta \in\xi \cap \beta$.
+    (This exists by Fund).
+    We want to see that $\xi = \alpha$.
+    We have $\xi \subseteq \alpha$ by the choice of $\xi$.
+    On the other hand $\alpha \subseteq \xi$:
+    Let $\eta \in \alpha \subseteq \beta$.
+    We have that $\eta \in \xi \lor \eta = \xi \lor \eta \ni \xi$.
+    If $\xi \in \eta$, then since $\eta  \in \alpha$,
+    we get $\xi \in \alpha$ contradicting the choice of $\xi$.
+    If $\xi = \eta$, the $\xi = \eta \in \alpha$,
+    which also is a contradiction.
+    Thus $\eta \in \xi$.
+
+    This yields $\alpha \in \beta$, hence $\alpha$ is an ordinal.
+
+
+    (d) By (c) if $\alpha$ and $\beta$ are ordinals,
+    then $\alpha \subseteq  \beta \iff (\alpha = \beta \lor \alpha \in \beta)$.
+    We need tho see that if $ \alpha$, $\beta$ are ordinals,
+    then $\alpha \subseteq \beta$ or $\beta \subseteq \alpha$.
+    Suppose there are ordinals $\alpha$, $\beta$ such that
+    this is not the case.
+
+    Pick such an $\alpha$.
+
+    Let $\alpha_0 \in  \alpha \cup \{\alpha\}$
+    be such that there is some $\beta$
+    with $\lnot( \beta \subseteq \alpha_0 \lor \alpha_0 \subseteq \beta)$
+    for for all $\gamma \in \alpha_0$,
+    $\forall \beta. ( \beta \subseteq \gamma \lor \gamma \subseteq \beta)$.
+    Pick $\beta_0$ such that
+    \[
+    \lnot \left( \beta_0 \subseteq \alpha_0 \lor \alpha_0 \subseteq \beta_0 \right).
+    \]
+
+    Consider $\alpha_0 \cup \beta_0$.
+    \begin{claim}
+        $\alpha_0 \cup \beta_0$
+        is an ordinal.
+    \end{claim}
+    \begin{subproof}
+        $\alpha_0 \cup \beta_0$ is clearly transitive.
+        Let $\gamma,\delta \in \alpha_0 \cup \beta_0$.
+        We claim that $\gamma \in \delta \lor \gamma = \delta \lor \gamma \in \delta$.
+        This can only fail if $\gamma \in \alpha_0$ and $\delta \in \beta_0$
+        (or the other way around).
+        But then $\gamma \in \delta \lor \gamma = \delta \lor \delta \in \gamma$
+        by the choice of $\alpha_0$.
+    \end{subproof}
+
+    \begin{claim}
+        $\alpha_0 = \alpha_0 \cup \beta_0$
+        or $\beta_0 = \alpha_0 \cup \beta_0$.
+    \end{claim}
+    \begin{subproof}
+        If that is not the case,
+        then $\alpha_0 \in \alpha_0 \cup \beta_0$
+        and $\beta_0 \in  \alpha_0 \cup \beta_0$.
+        $\alpha_0 \in  \alpha_0$ violates Fund.
+        Hence $\alpha_0 \in \beta_0$.
+        By the same argument, $\beta_0 \in \alpha_0$.
+        But this violates Fund,
+        as $\alpha_0 \in  \beta_0 \in \alpha_0$.
+    \end{subproof}
+\end{proof}
+
+\begin{lemma}
+    Let $X$ be a set of ordinals,
+    $X \neq \emptyset$.
+    Then $\bigcap X$ and $\bigcup X$ are ordinals.
+\end{lemma}
+\begin{proof}
+    Easy.
+\end{proof}
+It is actually the case that $\bigcap X \in X$:
+Pick $\alpha \in X$ such that $\alpha \subseteq \beta$
+for all $\beta \in X$. This exists
+by Fund and since all ordinals are comparable.
+Then $\alpha = \bigcap X$.
+
+\begin{notation}
+    We write $\min(X)$ for  $\bigcap X$
+    and $\sup(X)$ for $\bigcup X$.
+\end{notation}
+
+It need not be the case that $\bigcup X \in X$,
+for example $\bigcup \omega = \omega$.
+% but  $\bigcup 2 = 1$.
+
+\begin{definition}
+    An ordinal $\alpha$ is called
+    a \vocab[Ordinal!successor]{successor ordinal},
+    iff $\alpha = \beta \cup \{\beta\}$
+    for some $\beta \in \alpha$.
+    Otherwise $\alpha$ is called
+    a \vocab[Ordinal!limit]{limit ordinal}.
+\end{definition}
+\begin{observe}
+    Note that $\alpha$ is a limit ordinal iff for all
+    $\beta \in \alpha$, $\beta + 1 \in \alpha$:
+    If there is $\beta \in \alpha$ such that $\beta+1 \not\in \alpha$,
+    then either $\alpha = \beta+1$ (i.e.~$\alpha$ is a successor)
+    or $\alpha \in \beta+1$,
+    in which case $\beta \in \alpha \in \beta \cup \{\beta\} \lightning$.
+
+    Also if $\alpha$ is a successor,
+    then by definition there is some $\beta \in \alpha$,
+    with $\beta + 1 = \alpha$, so $\beta + 1 \not\in \alpha$.
+\end{observe}
+
+\begin{notation}
+    If $\alpha, \beta$ are ordinals,
+    we write $\alpha < \beta$ for $\alpha \in \beta$
+    (equivalently $\alpha \subsetneq \beta$).
+    We also write $\alpha \le \beta$
+    for $\alpha \in \beta \lor \alpha = \beta$
+    (i.e.~$\alpha \subseteq \beta$).
+\end{notation}
+
+\begin{example}
+    Limit ordinals:
+    \begin{itemize}
+        \item $0$,
+        \item $\omega$,
+        \item $\omega + \omega = \sup(\omega \cup \{\omega, \omega + 1, \ldots\})$,%
+            \footnote{To show that this exists, we need the recursion theorem
+                and replacement.}
+            $\omega + \omega + \omega, \omega + \omega + \omega + \omega, \ldots$
+    \end{itemize}
+
+    Successor ordinals:
+    \begin{itemize}
+        \item $1 = \{0\}, 2 = \{0,1\}, 3, \ldots$
+        \item $\omega +1 = \omega \cup \{\omega\} , \omega + 2, \ldots$,
+    \end{itemize}
+    
+    
+\end{example}
+
+\subsection{Induction and Recursion}
+

inputs/tutorial_02.tex

@@ -0,0 +1,42 @@
+\tutorial{02}{}{}
+
+\subsection{Exercise 1}
+
+(Cantor-Bendixson Derrivative)
+
+For $A \subseteq \R$
+Let $A^{(0)} \coloneqq A$,
+$A^{(\alpha+1)} \coloneqq \left( A^{(\alpha)}' \right)$
+and $A^{(\lambda)} \coloneqq  \bigcap_{\alpha < \lambda} A^{(\alpha)}$.
+
+We want to find $A_i$ such that $A_i^{(i)} = \emptyset$,
+but $A_i^{(j)} \neq \emptyset$ for all $j < i$.
+
+Let $A_1 = \{0\}$,
+$A_2 = \{0\} \cup \{\frac{1}{n+1} | n < \omega\}$
+and so on
+(in every step add sequences contained in a gap of the previous set converging to the points of the previous set):
+Define intervals:
+For all  $x \in A_n \setminus \{\max A_n\}$
+let $\epsilon_{x,n} \coloneqq \min \{y \in A_n | y > x\} - x$
+and
+let $I_x^{n} \coloneqq  (x, \epsilon_{x,n} + x)$.
+Construct a sequence converging to $x$,
+$(x_n)_{n < \omega}$,
+where $x +\frac{\epsilon_{x,n}}{n+2}$.
+This yields $A_n$ for every $n < \omega$,
+such that $A_n^\left( 1 \right) = A_{n-1}$,
+by setting
+\[
+A_{n+1} \coloneqq  A_n \cup \bigcup_{x \in A_n \setminus \left( \bigcup_{i=1}^n I^1_{x_i} \righ))} \{(x_n)_{n < \omega} | x_n + x+\frac{\epsilon_{x,n}}{n+1}\}
+\]
+
+Let $A_\omega \coloneqq \bigcup_{n < \omega} A_n$.
+Let $A_{\omega + 1} \coloneqq \{0\} \cup  \{ \frac{1}{n}A_\omega + \frac{1}{n} | n < \omega\}$
+and so on.
+
+
+
+
+
+

logic2.tex

@@ -30,6 +30,7 @@
 \input{inputs/lecture_04}
 \input{inputs/lecture_05}
 \input{inputs/lecture_06}
+\input{inputs/lecture_07}
 
 
 \cleardoublepage