w23-logic-2/inputs/lecture_01.tex

\lecture{01}{2023-10-16}{}
\gist{%
\paragraph{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 $g\colon B \twoheadrightarrow A$.
\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}
    \gist{%
    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 $f(A^{\text{odd}}) \subseteq B^{\text{even}}$, $f(A^\infty) = B^\infty$}.
}{Preimage sequence}
\end{proof}

\begin{definition}
    The \vocab{continuum hypothesis} ($\CH$)
    says that there is no set $A$ such that
    $\N < A < \R$,
    i.e.~every uncountable subset $A \subseteq \R$ is in bijection with $\R$.

    $\CH$ 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.