lecture 1

inputs/lecture_01.tex

@@ -0,0 +1,164 @@
+\lecture{01}{2023-10-16}{}
+
+Literature
+
+\begin{itemize}
+    \item Schindler, Set theory
+    \item K. Kunen
+    \item T.~Jech
+    \item A.~Kanamori, The higher infinite
+\end{itemize}
+
+\section*{Outline}
+\begin{itemize}
+    \item Set theory
+        \begin{itemize}
+            \item Naive set theory
+            \item $\ZFC$
+            \item Ordinals and Cardinals
+            \item Models of set theory (in particular forcing)
+            \item Independence of $\CH$.
+        \end{itemize}
+\end{itemize}
+
+
+\section{Naive set theory}
+
+\begin{definition}
+    Let $A \neq \emptyset$, $B$ be arbitrary sets.
+    We write $A \le  B$ ($A$ is not bigger than $B$ )
+    iff there is an injection $f\colon A \hookrightarrow B$.
+\end{definition}
+\begin{lemma}
+    If $A \le B$,
+    then there is a surjection $g\colon B \twoheadrightarrow B$.
+\end{lemma}
+\begin{proof}
+    Fix $f\colon A \hookrightarrow B$.
+    If $f$ is also surjective, then $f^{-1}\colon B \to A$
+    is also a bijection.
+    Otherwise define $g$ by choosing an arbitrary $x_0 \in B$
+    and let
+    \[
+    g(y) \coloneqq  \begin{cases}
+        x &: f(x) = y,\\
+        x_0 &: \text{if there is no such $x$}.
+    \end{cases}.
+    \]
+\end{proof}
+\begin{lemma}
+    If there is a surjection $f\colon A \twoheadrightarrow B$,
+    then $B \le A$.
+\end{lemma}
+\begin{proof}
+    For every $x \in B$ choose  one of its preimages under $f$.
+    This is basically equivalent to $\AC$.
+\end{proof}
+
+\begin{definition}
+    For sets $A$, $B$
+    write $A < B$ iff $A \le B \land B \not\le A$.
+\end{definition}
+\begin{theorem}[Cantor]
+    $\N < \R$.
+\end{theorem}
+\begin{proof}[Cantor's original proof]
+    Clearly $\N \le \R$.
+    Take some function $f\colon \N \to \R$
+
+    Define a sequence $([a_n, b_n], n \in \N)$
+    of nonempty closed nested intervals,
+    i.e.~$a_n \le a_{n+1} < b_{n+1} \le b_n$
+    as follows:
+    Set $a_0 \coloneqq 0$, $b_0 \coloneqq 1$,
+    and $a_{n+1}$, $b_{n+1}$ such that $x_n \notin [a_{n+1}, b_{n+1}]$.
+    Then $\bigcap_{n \in \N} [a_n, b_n] \neq \emptyset$
+    since $\R$ is complete.
+    Thus $f$ is not surjective.
+\end{proof}
+
+\begin{notation}
+    For a set $A$,
+    $\bP(A)$ denotes the \vocab{power set}  of $A$,
+    i.e.~the set of all subsets of $A$.
+\end{notation}
+\begin{theorem}
+    For all sets $A$, $A < \bP(A)$.
+\end{theorem}
+\begin{proof}
+    Clearly $A \le \bP(A)$
+    since $A \ni a \mapsto \{a\} \in \bP(A)$ is an injection.
+
+    Let $f\colon A \to \bP(A)$, we want to show that this is not
+    surjective.
+    Let $c \coloneqq \{x \in A | x \not\in f(x)\} \in \bP(A)$.
+    Suppose that $f(x_0) = c$.
+    Then both $x_0 \in c$ and $x_0 \not\in c$
+    lead to a contradiction.
+\end{proof}
+
+\begin{definition}
+    For sets $A$, $B$
+    write $A \sim B$
+    for $A \le B$ and $B \le A$.
+\end{definition}
+
+\begin{theorem}[Schröder-Bernstein]
+    Let $A$, $B$ be any sets.
+    If $A \sim B$, there is a bijection $h\colon A \to B$.
+\end{theorem}
+\begin{proof}
+    Let $f\colon A \hookrightarrow B$ and $g\colon B \hookrightarrow A$
+    be injective.
+    We need to define a bijection $h\colon A \to  B$.
+    For each $x \in A$ we define $N(x) \in \N \cup \{\infty\}$
+    and the maximal ``preimage sequence'' $(x_n : n < N(x))$
+    as follows:
+    $x_0 \coloneqq x$,
+    if $n + 1 < N$ and $n$ is even, then $x_n\coloneqq g(x_{n+1})$,
+    if it is odd, $x_n \coloneqq  f(x_{n+1})$
+    and either $N =  \infty$
+    or $x_{N-1}$ has no preimage under $f$ if $N-1$ is even,
+    resp.~$g$ if $N-1$ is odd.
+
+
+    Similarly for each $y \in B$
+    an $M = M(y) \in  \N \cup \{\infty\}$
+    and the maximal preimage sequence $(y_n : n < M)$
+    can be defined.
+
+    Let  $A^{\text{odd}} \coloneqq  \{x \in A : N(x) \text{ is an odd natural number}\}$,
+    $A^{\text{even}} \coloneqq \{x \in  A : N(x) \text{ is an even natural number}\}$,
+    $A^{\infty} \coloneqq \{x \in A : N(x) = \infty\}$
+    and similarly for $B$.
+
+    Now define
+    \begin{IEEEeqnarray*}{rCl}
+        h\colon A &\longrightarrow & B \\
+        x &\longmapsto &
+        \begin{cases}
+            f(x) &: x \in A^{\text{odd}} \cup A^{\infty},\\
+            g^{-1}(x) &: x \in A^{\text{even}}.
+        \end{cases}
+    \end{IEEEeqnarray*}
+
+    It is clear that this is bijective.
+
+    \todo{missing picture $f(A^{\text{odd}}) \subseteq B^{\text{even}}$, $f(A^\infty) = B^\infty$}.
+\end{proof}
+
+\begin{definition}
+    The \vocab{continuum hypothesis} ($\CH$ )
+    says that there is no set $A$ such that
+    $\N < A < \R$.
+
+    $\CH$ is equivalent to the statement that there is no set $A \subset \R$
+    which is uncountable ($\N < A$ )
+    and there is no bijection $A \leftrightarrow \R$.
+\end{definition}
+
+What we'll do next:
+Define open and closed subsets of $\R$.
+Show $\CH$ for open and closed sets.
+
+

logic2.tex

@@ -1,8 +1,8 @@
-\documentclass[10pt,ngerman,a4paper,fancyfoot,git]{mkessler-script}
+\documentclass[10pt,a4paper,fancyfoot,git]{mkessler-script}
 
 \course{Logic II}
 \lecturer{Ralf Schindler}
-\assistant{}
+\assistant{Mirko Bartsch}
 \author{Josia Pietsch}
 
 \usepackage{logic}
@@ -24,6 +24,7 @@
 
 \newpage
 
+\input{inputs/lecture_01}
 
 
 \cleardoublepage