w23-logic-2/inputs/lecture_01.tex

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2023-10-16 16:51:42 +02:00
\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.