From f7bd6359bd1524821c40b4d2902eab7e328245f1 Mon Sep 17 00:00:00 2001
From: Josia Pietsch <git@jrpie.de>
Date: Mon, 23 Oct 2023 11:44:15 +0200
Subject: [PATCH] lecture 02

---
 inputs/lecture_02.tex | 128 ++++++++++++++++++++++++++++++++++++++++++
 logic2.tex            |   3 +-
 2 files changed, 130 insertions(+), 1 deletion(-)
 create mode 100644 inputs/lecture_02.tex

diff --git a/inputs/lecture_02.tex b/inputs/lecture_02.tex
new file mode 100644
index 0000000..331efb9
--- /dev/null
+++ b/inputs/lecture_02.tex
@@ -0,0 +1,128 @@
+\lecture{02}{2023-10-19}{Topology on $\R$}
+\begin{definition}
+    A set $O \subseteq \R$ is called \vocab{open} in $\R$ iff it is the union
+    of a set of open intervals.
+
+    A set $A \subseteq \R$ is called \vocab{closed} in $\R$
+    iff it is the complement of an open set.
+\end{definition}
+
+\begin{remark}
+    \begin{itemize}
+        \item If $\emptyset \neq  O   \R$
+            then $O \sim \R$.
+        \item If $O \subseteq \R$
+            is open, then $O$ is the union of open intervals
+            with rational endpoints, since $\Q$ is dense.
+    \end{itemize}
+\end{remark}
+
+\begin{remark}+
+    $\{O \subseteq \R\} \sim 2^{\aleph_0} < \cP(\R)$.
+\end{remark}
+
+\begin{definition}
+    We call $x \in \R$
+    an \vocab{accumulation point}
+    of $A$ iff for all $a < x < b$
+    there is some $y \in A$, $y \in (a,b)$, $y \neq x$.
+    We write \vocab{$A'$} for the set of all accumulation points of $A$.
+\end{definition}
+\begin{example}
+    $\{\frac{1}{n+1} | n \in \N\}' = \{0\}$.
+\end{example}
+
+\begin{lemma}
+    \label{lem:closedaccumulation}
+    A set $A \subseteq  \R$ is closed iff $A' \subseteq  A$.
+\end{lemma}
+\begin{refproof}{lem:closedaccumulation}
+    ``$\implies$''
+    Let $A$ be closed. Suppose that $x \in A' \setminus A$.
+    Then there exists $(a,b) \ni x$
+    disjoint from $A$. Hence $x \not\in A' \lightning$
+
+    ``$\impliedby$''
+    Suppose $A' \subseteq  A$.
+    \begin{claim}
+        $A \subseteq  \R$ is closed iff all Cauchy sequences
+        in $A$ converge in $A$.
+    \end{claim}
+    \begin{subproof}
+        Let $A$ be closed and $\langle x_n : n \in \omega \rangle$
+        a Cauchy sequence in $A$.
+        Suppose that $x = \lim_{n \to \infty} x_n \not\in A$.
+        Then there is $(a,b) \ni x$ disjoint from $A$.
+        However $x_n \in (a,b)$ for almost all $n \in \omega$ $\lightning$
+
+        On the other hand let  $A$ not be closed.
+        Then there exists a witness $x \in \R \setminus A$
+        such that $A \cap (a,b) \neq  \emptyset$ for all $(a,b) \ni x$.
+        In particular,
+        we may pick $x_n \in (x - \frac{1}{n+1}, x + \frac{1}{n+1}) \cap A$
+        for all $n < \omega$.
+    \end{subproof}
+
+    Now if $A' \subseteq  A$ and $A$ were not closed,
+    there would be some Cauchy-sequence $(x_n)$
+    in $A$ such that $\lim_{n \to  \infty} x_n \not\in A$.
+    But then $x \in A' \subseteq A \lightning$.
+\end{refproof}
+\begin{definition}
+    $P \subseteq  \R$ is called \vocab{perfect}
+    iff $P \neq \emptyset$ and $P = P'$.
+\end{definition}
+
+We want to prove two things:
+\begin{itemize}
+    \item If $P$ is perfect, then $P \sim \R$.
+    \item If $A$ is closed and uncountable then $A$ has a perfect subset.
+        In particular $A \sim \R$.
+\end{itemize}
+
+\begin{lemma}
+    Let $P \subseteq  \R$ be perfect.
+    Then $P \sim \R$.
+\end{lemma}
+\begin{proof}
+    It suffices to find an injection $f\colon \R \hookrightarrow P$.
+    We have $\underbrace{\{0,1\}^{\omega}}_{\text{infinite 0-1-sequences}} \sim \R$,
+    hence it suffices to construct $f\colon \{0,1\}^\omega\hookrightarrow P$.
+
+    In order to do that, we are going to construct some
+    $g\colon \underbrace{\{0,1\}^{<\omega}}_{\text{finite 0-1-sequences}} \to P$
+    with certain properties
+    by recursion on the length of $s \in \{0,1\}^{<\omega}$.
+
+    Let $g(\emptyset)$ be any point in $P$.
+    Suppose that $g(s) \in P$ has been chosen for all $s$ of
+    length $\le n$.
+    For each $s \in \{0,1\}^{n}$ pick
+    $g(s) \in  (a_s, b_s)$
+    such that $(a_s, b_s) \cap (a_{s'}, b_{s'}) = \emptyset$
+    for all $s, s'$ of length $n$,
+    $b_s - a_s \le  \frac{1}{n^3}$
+    and
+    $(a_{s\defon{n-1}}, b_{s\defon{n-1}}) \subseteq (a_s, b_s)$.
+
+    For each such $s$ pick $x_s \in (a_s, b_s) \cap P$
+    with $x_s \neq f(s)$.
+    This is possible since $P \subseteq P'$.
+    Now set $g(s\concat 0) \coloneqq g(s)$
+    and $g(s \concat 1) \coloneqq  g(x_s)$.
+    This finishes the construction.
+
+    If $t \in \{0,1\}^{\omega}$,
+    then $(g(t\defon{n}), n < \omega)$ is a Cauchy sequence.
+
+    By $P' \subseteq P$ we get that this sequence
+    converges to a point in $P$.
+    Define $f(t)$ to be this point.
+
+    If $t \neq t' \in \{0,1\}^{\omega}$,
+    then there is some $n$ such that $t\defon{n} \neq  t'\defon{n}$,
+    hence $f(t) \in [a_{t\defon{n}}, b_{t\defon{n}}]$
+    and $f(t') \in [a_{t\defon{n}}, b_{t\defon{n}}]$
+    which are disjoint.
+    Thus $f(t) \neq f(t')$, i.e.~$f$ is injective.
+\end{proof}
diff --git a/logic2.tex b/logic2.tex
index 15cf247..c430f4a 100644
--- a/logic2.tex
+++ b/logic2.tex
@@ -3,7 +3,7 @@
 \course{Logic II}
 \lecturer{Ralf Schindler}
 \assistant{Mirko Bartsch}
-\author{Josia Pietsch}
+\author{Mirko Bartsch, Josia Pietsch}
 
 \usepackage{logic}
 
@@ -25,6 +25,7 @@
 \newpage
 
 \input{inputs/lecture_01}
+\input{inputs/lecture_02}
 
 
 \cleardoublepage