diff --git a/inputs/lecture_06.tex b/inputs/lecture_06.tex index 8bfebf0..5abfdab 100644 --- a/inputs/lecture_06.tex +++ b/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} - - - - - - - diff --git a/inputs/lecture_07.tex b/inputs/lecture_07.tex new file mode 100644 index 0000000..1595c4e --- /dev/null +++ b/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} + diff --git a/inputs/tutorial_02.tex b/inputs/tutorial_02.tex new file mode 100644 index 0000000..24c2fd3 --- /dev/null +++ b/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. + + + + + + diff --git a/logic2.tex b/logic2.tex index bf32004..9f99611 100644 --- a/logic2.tex +++ b/logic2.tex @@ -30,6 +30,7 @@ \input{inputs/lecture_04} \input{inputs/lecture_05} \input{inputs/lecture_06} +\input{inputs/lecture_07} \cleardoublepage