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Josia Pietsch 2023-10-25 15:37:34 +02:00
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6
inputs/lecture_01.tex
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@ -26,7 +26,7 @@ Literature
\begin{definition} \begin{definition}
Let $A \neq \emptyset$, $B$ be arbitrary sets. Let $A \neq \emptyset$, $B$ be arbitrary sets.
We write $A \le B$ ($A$ is not bigger than $B$ ) We write $A \le B$ ($A$ is not bigger than $B$)
iff there is an injection $f\colon A \hookrightarrow B$. iff there is an injection $f\colon A \hookrightarrow B$.
\end{definition} \end{definition}
\begin{lemma} \begin{lemma}
@ -148,12 +148,12 @@ Literature
\end{proof} \end{proof}
\begin{definition} \begin{definition}
The \vocab{continuum hypothesis} ($\CH$ ) The \vocab{continuum hypothesis} ($\CH$)
says that there is no set $A$ such that says that there is no set $A$ such that
$\N < A < \R$. $\N < A < \R$.
$\CH$ is equivalent to the statement that there is no set $A \subset \R$ $\CH$ is equivalent to the statement that there is no set $A \subset \R$
which is uncountable ($\N < A$ ) which is uncountable ($\N < A$)
and there is no bijection $A \leftrightarrow \R$. and there is no bijection $A \leftrightarrow \R$.
\end{definition} \end{definition}

2
inputs/lecture_02.tex
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@ -9,7 +9,7 @@
\begin{remark} \begin{remark}
\begin{itemize} \begin{itemize}
\item If $\emptyset \neq O \R$ \item If $\emptyset \neq O \overset{\text{open}}{\subseteq} \R$
then $O \sim \R$. then $O \sim \R$.
\item If $O \subseteq \R$ \item If $O \subseteq \R$
is open, then $O$ is the union of open intervals is open, then $O$ is the union of open intervals

12
inputs/lecture_03.tex
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@ -1,8 +1,7 @@
\lecture{03}{2023-10–23}{Cantor-Bendixson} \lecture{03}{2023-10–23}{Cantor-Bendixson}
\begin{theorem}[Cantor-Bendixson] \begin{theorem}[Cantor-Bendixson]
\yaref{thm:cantorbendixson}{Cantor-Bendixson}{Cantor-Bendixson} \yalabel{Cantor-Bendixson}{Cantor-Bendixson}{thm:cantorbendixson}
If $A \subseteq \R$ is closed, If $A \subseteq \R$ is closed,
it is either at most countable or else it is either at most countable or else
$A$ contains a perfect set. $A$ contains a perfect set.
@ -30,9 +29,9 @@
\begin{definition} \begin{definition}
Let $A \subseteq \R$. Let $A \subseteq \R$.
We say that $x \in \R$ We say that $x \in \R$
is a \vocab{condensation point} of $A$ is a \vocab{condensation point} of $A$
iff for all $a < x < b$, $(a,b) \cap A$ iff for all $a < x < b$, $(a,b) \cap A$
is uncountable. is uncountable.
\end{definition} \end{definition}
By the fact we just proved, By the fact we just proved,
@ -128,6 +127,3 @@ all condensation points are accumulation points.
% is at most countable. % is at most countable.
% Also $A'$ is closed. % Also $A'$ is closed.
% \end{remark} % \end{remark}