w23-logic-2/inputs/lecture_03.tex
Josia Pietsch 68280c4b70
Some checks failed
Build latex and deploy / checkout (push) Failing after 15m24s
typo
2023-12-04 16:07:22 +01:00

130 lines
4.6 KiB
TeX
Raw Blame History

This file contains ambiguous Unicode characters

This file contains Unicode characters that might be confused with other characters. If you think that this is intentional, you can safely ignore this warning. Use the Escape button to reveal them.

\lecture{03}{2023-1023}{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{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 &=& \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}
$P \neq \emptyset$: $\checkmark$
$P \subseteq P'$ (i.e. $P$ is closed):
% \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$.
$P' \subseteq P$ :
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}