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