Lecture 7
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@ -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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\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}

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\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.

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