\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.