Josia Pietsch
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130 lines
4.6 KiB
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130 lines
4.6 KiB
TeX
\lecture{03}{2023-10–23}{Cantor-Bendixson}
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\begin{theorem}[Cantor-Bendixson]
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\yalabel{Cantor-Bendixson}{Cantor-Bendixson}{thm:cantorbendixson}
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If $A \subseteq \R$ is closed,
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it is either at most countable or else
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$A$ contains a perfect set.
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\end{theorem}
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\begin{corollary}
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If $A \subseteq \R$ is closed,
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then either $A \le \N$ or $A \sim \R$.
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\end{corollary}
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\begin{fact}
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$A' = \{x \in \R | \forall a < x < b.~ (a,b) \cap A \text{ is at least countable}\}$.
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\end{fact}
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\begin{proof}
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$\supseteq$ is clear.
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For $\subseteq $, fix $a < x < b$
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and let us define $(y_n: n \in \omega)$
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as well as $((a_n, b_n): n \in \omega)$.
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Set $a_0 \coloneqq a$, $b_0 \coloneqq b$.
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Having defined $(a_n, b_n)$,
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pick $x \neq y_n \in A \cap (a_n, b_n)$,
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Then pick $a_n < a_{n+1} < x < b_{n+1} < b_n$
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such that $y_n \not\in (a_{n+1}, b_{n+1})$.
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Clearly $y_n \neq y_{n+1}$,
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hence $\{y_n : n \in \N\}$ is a countable subset of $A \cap (a,b)$.
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\end{proof}
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\begin{definition}
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Let $A \subseteq \R$.
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We say that $x \in \R$
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is a \vocab{condensation point} of $A$
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iff for all $a < x < b$, $(a,b) \cap A$
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is uncountable.
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\end{definition}
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By the fact we just proved,
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all condensation points are accumulation points.
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\begin{refproof}{thm:cantorbendixson}
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Fix $A \subseteq \R$ closed.
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We want to see that $A$ is at most countable
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or there is some perfect $P \subseteq A$.
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Let
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\[P \coloneqq \{x \in \R | x \text{ is a condensation point of $A$}\}.\]
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Since $A $ is closed, $P \subseteq A$.
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\begin{claim}
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$A \setminus P$ is at most countable.
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\end{claim}
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\begin{subproof}
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For each $x \in A \setminus P$,
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there is $a_x < x < b_x$
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such that $(a_x, b_x) \cap A$
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is at most countable.
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Since $\Q$ is dense in $\R$,
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we may assume that $a_x, b_x \in \Q$.
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Then
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\begin{IEEEeqnarray*}{rCl}
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A \setminus P &=& \bigcup_{x \in A \setminus P} (a_x, b_x) \cap A.
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\end{IEEEeqnarray*}
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$\subseteq $ holds by the choice of $a_x$ and $b_x$.
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For $\supseteq$ let $y$ be an element of the RHS.
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Then $y \in (a_{x_0}, b_{x_0}) \cap A$ for some $x_0$.
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As $(a_{x_0}, b_{x_0}) \cap A$ is at most countable,
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$y \not\in P$.
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Now we have that $A \setminus P$ is a union
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of at most countably many sets,
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each of which is at most countable.
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\end{subproof}
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\begin{claim}
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If $P \neq \emptyset$, the $P$ is perfect.
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\end{claim}
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\begin{subproof}
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$P \neq \emptyset$: $\checkmark$
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$P \subseteq P'$ (i.e. $P$ is closed):
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% \begin{IEEEeqnarray*}{rCl}
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% P &=& \{x \in A | \text{every open neighbourhood of $x$ is uncountable}\}\\
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% &\subseteq & \{x \in A | \text{every open neighbourhood of $x$ is at least countable}\} = P'.
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% \end{IEEEeqnarray*}
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Let $x \in P$.
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Let $a < x < b$.
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We need to show that there is some $y \in (a,b) \cap P \setminus \{x\}$.
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Suppose that for all $y \in (a,b) \setminus \{x\}$
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there is some $a_y < y < b_y$
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with $(a_y, b_y) \cap A$ being at most countable.
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Wlog.~$a_y, b_y \in \Q$.
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Then
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\[
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(a,b) \cap A = \{x \} \cup \bigcup_{\substack{y \in (a,b)\\y \neq x}} [(a_y, b_y) \cap A].
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\]
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But then $(a,b) \cap A$ is at most countable
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contradicting $ x \in P$.
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$P' \subseteq P$ :
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Let $x \in P'$.
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Then for $a < x < b$ the set
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$(a,b) \cap P$
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always has a member $y$ such that $y \neq x$.
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Since $y \in P$, we get that $(a,b) \cap A$
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in uncountable, hence $x \in P$.
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\end{subproof}
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But now
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\[
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A = \overbrace{P}^{\mathclap{\text{perfect, unless $= \emptyset$}}} \cup \underbrace{(A \setminus P)}_{\mathclap{\text{at most countable}}}.
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\]
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\end{refproof}
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\todo{Alternative proof of Cantor-Bendixson}
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% \begin{remark}
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% There is an alternative proof of Cantor-Bendixson, going as follows:
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% Fix $A \subseteq \R$ closed.
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% Define a sequence
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% \[
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% A \supseteq A' \supseteq A'' \supseteq \ldots \supseteq \bigcap_{n} A^{(n)}
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% \supseteq \left( \bigcap_{n} A^{(n)} \right)' \supseteq \ldots
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% \]
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% Then $A \setminus A'$ has at most countably many points.
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% For all $a \in A \setminus A'$
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% pick $\Q\ni a_x < x < b_x \in \Q$
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% such that $(a_x, b_x) \cap A = \{x\}$.
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% Then $A \setminus A' = \bigcup_{x \in A \setminus A'} [(a_x, b_x) \cap A]$
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% is at most countable.
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% Also $A'$ is closed.
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% \end{remark}
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