Josia Pietsch
9956de5277
Some checks failed
Build latex and deploy / checkout (push) Failing after 15m58s
131 lines
4.6 KiB
TeX
131 lines
4.6 KiB
TeX
\lecture{03}{2023-10–23}{Cantor-Bendixson}
|
||
|
||
\begin{theorem}[Cantor-Bendixson]
|
||
\yalabel{Cantor-Bendixson}{Cantor-Bendixson}{thm:cantorbendixson}
|
||
If $A \subseteq \R$ is closed,
|
||
it is either at most countable or else
|
||
$A$ contains a perfect set.
|
||
\end{theorem}
|
||
\begin{corollary}
|
||
If $A \subseteq \R$ is closed,
|
||
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.
|
||
For $\subseteq $, fix $a < x < b$
|
||
and let us define $(y_n: n \in \omega)$
|
||
as well as $((a_n, b_n): n \in \omega)$.
|
||
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 $A \cap (a,b)$.
|
||
\end{proof}
|
||
|
||
\begin{definition}
|
||
Let $A \subseteq \R$.
|
||
We say that $x \in \R$
|
||
is a \vocab{condensation point} of $A$
|
||
iff for all $a < x < b$, $(a,b) \cap A$
|
||
is uncountable.
|
||
\end{definition}
|
||
By the fact we just proved,
|
||
all condensation points are accumulation points.
|
||
|
||
\begin{yarefproof}{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 &=& \bigcup_{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,
|
||
$y \not\in P$.
|
||
|
||
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}
|
||
$P \neq \emptyset$: $\checkmark$
|
||
|
||
$P \subseteq P'$:
|
||
% \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*}
|
||
|
||
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$.
|
||
|
||
$P' \subseteq P$ (i.e.~$P$ is closed):
|
||
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{yarefproof}
|
||
|
||
\gist{\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}
|