w23-logic-2/inputs/lecture_03.tex

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\lecture{03}{2023-1023}{Cantor-Bendixson}
\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,
it is either at most countable or else
$A$ contains a perfect set.
\end{theorem}
\begin{corollary}
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If $A \subseteq \R$ is closed,
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then either $A \le \N$ or $A \sim \R$.
\end{corollary}
\begin{fact}
$A' = \{x \in \R | \forall a < x < b.~ (a,b) \cap A \text{ is at least countable}\}$.
\end{fact}
\begin{proof}
$\supseteq$ is clear.
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For $\subseteq $, fix $a < x < b$
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and let us define $(y_n: n \in \omega)$
as well as $((a_n, b_n): n \in \omega)$.
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Set $a_0 \coloneqq a$, $b_0 \coloneqq b$.
Having defined $(a_n, b_n)$,
pick $x \neq y_n \in A \cap (a_n, b_n)$,
Then pick $a_n < a_{n+1} < x < b_{n+1} < b_n$
such that $y_n \not\in (a_{n+1}, b_{n+1})$.
Clearly $y_n \neq y_{n+1}$,
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}
Let $A \subseteq \R$.
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We say that $x \in \R$
is a \vocab{condensation point} of $A$
iff for all $a < x < b$, $(a,b) \cap A$
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is uncountable.
\end{definition}
By the fact we just proved,
all condensation points are accumulation points.
\begin{refproof}{thm:cantorbendixson}
Fix $A \subseteq \R$ closed.
We want to see that $A$ is at most countable
or there is some perfect $P \subseteq A$.
Let $P \coloneqq \{x \in \R | x \text{ is a condensation point of $A$}\}$.
Since $A $ is closed, $P \subseteq A$.
\begin{claim}
$A \setminus P$ is at most countable.
\end{claim}
\begin{subproof}
For each $x \in A \setminus P$,
there is $a_x < x < b_x$
such that $(a_x, b_x) \cap A$
is at most countable.
Since $\Q$ is dense in $\R$,
we may assume that $a_x, b_x \in \Q$.
Then
\begin{IEEEeqnarray*}{rCl}
A \setminus P &=& \bigcap_{x \in A \setminus P} (a_x, b_x) \cap A.
\end{IEEEeqnarray*}
$\subseteq $ holds by the choice of $a_x$ and $b_x$.
For $\supseteq$ let $y$ be an element of the RHS.
Then $y \in (a_{x_0}, b_{x_0}) \cap A$ for some $x_0$.
As $(a_{x_0}, b_{x_0}) \cap A$ is at most countable,
$y \in P$.
Now we have that $A \setminus P$ is a union
of at most countably many sets,
each of which is at most countable.
\end{subproof}
\begin{claim}
If $P \neq \emptyset$, the $P$ is perfect.
\end{claim}
\begin{subproof}
$P \neq \emptyset$: $\checkmark$
$P \subseteq P'$ (i.e. $P$ is closed):
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% \begin{IEEEeqnarray*}{rCl}
% P &=& \{x \in A | \text{every open neighbourhood of $x$ is uncountable}\}\\
% &\subseteq & \{x \in A | \text{every open neighbourhood of $x$ is at least countable}\} = P'.
% \end{IEEEeqnarray*}
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Let $x \in P$.
Let $a < x < b$.
We need to show that there is some $y \in (a,b) \cap P \setminus \{x\}$.
Suppose that for all $y \in (a,b) \setminus \{x\}$
there is some $a_y < y < b_y$
with $(a_y, b_y) \cap A$ being at most countable.
Wlog.~$a_y, b_y \in \Q$.
Then
\[
(a,b) \cap A = \{x \} \cup \bigcup_{\substack{y \in (a,b)\\y \neq x}} [(a_y, b_y) \cap A].
\]
But then $(a,b) \cap A$ is at most countable
contradicting $ x \in P$.
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$P' \subseteq P$ :
Let $x \in P'$.
Then for $a < x < b$ the set
$(a,b) \cap P$
always has a member $y$ such that $y \neq x$.
Since $y \in P$, we get that $(a,b) \cap A$
in uncountable, hence $x \in P$.
\end{subproof}
But now
\[
A = \overbrace{P}^{\mathclap{\text{perfect, unless $= \emptyset$}}} \cup \underbrace{(A \setminus P)}_{\mathclap{\text{at most countable}}}.
\]
\end{refproof}
\todo{Alternative proof of Cantor-Bendixson}
% \begin{remark}
% There is an alternative proof of Cantor-Bendixson, going as follows:
% Fix $A \subseteq \R$ closed.
% Define a sequence
% \[
% A \supseteq A' \supseteq A'' \supseteq \ldots \supseteq \bigcap_{n} A^{(n)}
% \supseteq \left( \bigcap_{n} A^{(n)} \right)' \supseteq \ldots
% \]
% Then $A \setminus A'$ has at most countably many points.
% For all $a \in A \setminus A'$
% pick $\Q\ni a_x < x < b_x \in \Q$
% such that $(a_x, b_x) \cap A = \{x\}$.
% Then $A \setminus A' = \bigcup_{x \in A \setminus A'} [(a_x, b_x) \cap A]$
% is at most countable.
% Also $A'$ is closed.
% \end{remark}