lecture 1
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inputs/lecture_01.tex
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\lecture{01}{2023-10-16}{}
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Literature
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\begin{itemize}
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\item Schindler, Set theory
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\item K. Kunen
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\item T.~Jech
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\item A.~Kanamori, The higher infinite
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\end{itemize}
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\section*{Outline}
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\begin{itemize}
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\item Set theory
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\begin{itemize}
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\item Naive set theory
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\item $\ZFC$
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\item Ordinals and Cardinals
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\item Models of set theory (in particular forcing)
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\item Independence of $\CH$.
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\end{itemize}
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\end{itemize}
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\section{Naive set theory}
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\begin{definition}
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Let $A \neq \emptyset$, $B$ be arbitrary sets.
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We write $A \le B$ ($A$ is not bigger than $B$ )
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iff there is an injection $f\colon A \hookrightarrow B$.
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\end{definition}
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\begin{lemma}
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If $A \le B$,
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then there is a surjection $g\colon B \twoheadrightarrow B$.
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\end{lemma}
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\begin{proof}
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Fix $f\colon A \hookrightarrow B$.
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If $f$ is also surjective, then $f^{-1}\colon B \to A$
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is also a bijection.
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Otherwise define $g$ by choosing an arbitrary $x_0 \in B$
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and let
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\[
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g(y) \coloneqq \begin{cases}
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x &: f(x) = y,\\
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x_0 &: \text{if there is no such $x$}.
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\end{cases}.
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\]
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\end{proof}
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\begin{lemma}
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If there is a surjection $f\colon A \twoheadrightarrow B$,
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then $B \le A$.
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\end{lemma}
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\begin{proof}
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For every $x \in B$ choose one of its preimages under $f$.
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This is basically equivalent to $\AC$.
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\end{proof}
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\begin{definition}
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For sets $A$, $B$
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write $A < B$ iff $A \le B \land B \not\le A$.
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\end{definition}
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\begin{theorem}[Cantor]
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$\N < \R$.
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\end{theorem}
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\begin{proof}[Cantor's original proof]
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Clearly $\N \le \R$.
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Take some function $f\colon \N \to \R$
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Define a sequence $([a_n, b_n], n \in \N)$
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of nonempty closed nested intervals,
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i.e.~$a_n \le a_{n+1} < b_{n+1} \le b_n$
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as follows:
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Set $a_0 \coloneqq 0$, $b_0 \coloneqq 1$,
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and $a_{n+1}$, $b_{n+1}$ such that $x_n \notin [a_{n+1}, b_{n+1}]$.
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Then $\bigcap_{n \in \N} [a_n, b_n] \neq \emptyset$
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since $\R$ is complete.
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Thus $f$ is not surjective.
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\end{proof}
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\begin{notation}
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For a set $A$,
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$\bP(A)$ denotes the \vocab{power set} of $A$,
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i.e.~the set of all subsets of $A$.
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\end{notation}
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\begin{theorem}
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For all sets $A$, $A < \bP(A)$.
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\end{theorem}
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\begin{proof}
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Clearly $A \le \bP(A)$
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since $A \ni a \mapsto \{a\} \in \bP(A)$ is an injection.
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Let $f\colon A \to \bP(A)$, we want to show that this is not
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surjective.
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Let $c \coloneqq \{x \in A | x \not\in f(x)\} \in \bP(A)$.
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Suppose that $f(x_0) = c$.
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Then both $x_0 \in c$ and $x_0 \not\in c$
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lead to a contradiction.
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\end{proof}
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\begin{definition}
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For sets $A$, $B$
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write $A \sim B$
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for $A \le B$ and $B \le A$.
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\end{definition}
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\begin{theorem}[Schröder-Bernstein]
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Let $A$, $B$ be any sets.
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If $A \sim B$, there is a bijection $h\colon A \to B$.
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\end{theorem}
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\begin{proof}
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Let $f\colon A \hookrightarrow B$ and $g\colon B \hookrightarrow A$
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be injective.
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We need to define a bijection $h\colon A \to B$.
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For each $x \in A$ we define $N(x) \in \N \cup \{\infty\}$
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and the maximal ``preimage sequence'' $(x_n : n < N(x))$
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as follows:
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$x_0 \coloneqq x$,
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if $n + 1 < N$ and $n$ is even, then $x_n\coloneqq g(x_{n+1})$,
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if it is odd, $x_n \coloneqq f(x_{n+1})$
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and either $N = \infty$
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or $x_{N-1}$ has no preimage under $f$ if $N-1$ is even,
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resp.~$g$ if $N-1$ is odd.
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Similarly for each $y \in B$
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an $M = M(y) \in \N \cup \{\infty\}$
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and the maximal preimage sequence $(y_n : n < M)$
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can be defined.
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Let $A^{\text{odd}} \coloneqq \{x \in A : N(x) \text{ is an odd natural number}\}$,
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$A^{\text{even}} \coloneqq \{x \in A : N(x) \text{ is an even natural number}\}$,
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$A^{\infty} \coloneqq \{x \in A : N(x) = \infty\}$
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and similarly for $B$.
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Now define
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\begin{IEEEeqnarray*}{rCl}
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h\colon A &\longrightarrow & B \\
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x &\longmapsto &
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\begin{cases}
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f(x) &: x \in A^{\text{odd}} \cup A^{\infty},\\
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g^{-1}(x) &: x \in A^{\text{even}}.
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\end{cases}
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\end{IEEEeqnarray*}
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It is clear that this is bijective.
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\todo{missing picture $f(A^{\text{odd}}) \subseteq B^{\text{even}}$, $f(A^\infty) = B^\infty$}.
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\end{proof}
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\begin{definition}
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The \vocab{continuum hypothesis} ($\CH$ )
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says that there is no set $A$ such that
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$\N < A < \R$.
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$\CH$ is equivalent to the statement that there is no set $A \subset \R$
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which is uncountable ($\N < A$ )
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and there is no bijection $A \leftrightarrow \R$.
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\end{definition}
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What we'll do next:
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Define open and closed subsets of $\R$.
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Show $\CH$ for open and closed sets.
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@ -1,8 +1,8 @@
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\documentclass[10pt,ngerman,a4paper,fancyfoot,git]{mkessler-script}
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\documentclass[10pt,a4paper,fancyfoot,git]{mkessler-script}
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\course{Logic II}
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\lecturer{Ralf Schindler}
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\assistant{}
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\assistant{Mirko Bartsch}
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\author{Josia Pietsch}
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\usepackage{logic}
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\newpage
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\input{inputs/lecture_01}
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\cleardoublepage
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