w23-logic-2/inputs/lecture_08.tex

\lecture{08}{2023-11-13}{Induction and recursion}

\subsection{Classes}
It is often very handy to work in a class theory rather than
in set theory.

To formulate a class theory,
we start out with a first order language
with two types of variables,
sets (denoted by lower case letters)
and classes (denoted by capital letters),
as well as one binary relation symbol $\in$
for membership.

\vocab{Bernays-Gödel class theory} (\vocab{BG})
has the following axioms:

\begin{axiom}[Extensionality]
    \yalabel{Axiom of Extensionality}{(Ext)}{ax:bg:ext}
    \[
    \forall x, y. \left( x = y \iff \left( \forall z.~(z \in x \iff z \in y \right)  \right).
    \] 
\end{axiom}
\begin{axiom}[Foundation]
    \yalabel{Axiom of Foundation}{(Fund)}{ax:bg:fund}
    \[
    \forall x .(x \neq \emptyset \implies \exists  y \in x . y \cap x = \emptyset).
    \] 
\end{axiom}

\begin{axiom}[Pairing]
    \yalabel{Axiom of Pairing}{(Pair)}{ax:bg:pair}
    \[
    \forall x \forall  y \exists  z . z = \{x,y\}.
    \] 
\end{axiom}

\begin{axiom}[Union]
    \yalabel{Axiom of Union}{(Union)}{ax:bg:union}
    \[
    \forall  x \exists  y .~ y = \bigcup x.
    \] 
\end{axiom}
\begin{axiom}[Power]
    \yalabel{Powerset Axiom}{(Pow)}{ax:bg:pow}
    \[
    \forall x \exists  y .~ y = \cP(x).
    \] 
\end{axiom}

\begin{axiom}[Infinity]
    \yalabel{Axiom of Infinity}{(Infinity)}{ax:bg:inf}
    \[
    \exists  x . ~(\emptyset \in x \land \left( \forall y \in x .~y \cup \{y\}  \in x \right)).
    \] 
\end{axiom}

Together with the following axioms for classes:

\begin{axiom}[Extensionality for classes]
    \[
        \forall X \forall  Y \left( \forall  x(x \in X \iff y \in X) \implies X = Y).
    \] 
\end{axiom}

\begin{axiom}
    Every set is a class:
    \[
    \forall  x \exists X. x = X.
    \] 
\end{axiom}
\begin{axiom}
    Every element of a class is a set:
    \[
        \forall X \exists Y.~(X \in Y \to  \exists x.~x = X).
    \] 
\end{axiom}
\begin{axiom}[Replacement]
    \yalabel{Axiom of Replacement}{(Rep)}{ax:bg:rep}

    If $F$ is a function and inf $a $ is a set,
    then $F"a$ is a set.
\end{axiom}
Here a \vocab[Class function]{(class) function} is a class
consisting of pairs $(x,y)$,
such that for every  $x$ there is at most one $y$
with $(x,y) \in F$.
Furthermore $F"a \coloneqq  \{y : \exists  x \in  a .~(x,y) \in F\}$.

\begin{remark}
    Note that we didn't need to use an axiom schema,
    \yarefs{ax:bg:rep} is a single axiom.
\end{remark}

\begin{axiom}[Comprehension]
    \yalabel{Axiom of Comprehension}{(Comp)}{ax:bg:comp}

    \[
    \forall X_1 \ldots \forall X_k .~\exists Y ( \forall  x .~x \in Y \iff \phi(x,X_1,\ldots, X_k))
    \] 
    where $\phi(x, X_1, \ldots, X_k)$
    is a formula which contains exactly $X_1, \ldots, X_k, x$
    as free variables,
    and $\phi$ does not have quantifiers
    ranging over classes.%
    \footnote{If one removes the restriction regarding
        quantifiers another theory, called
        \vocab{Morse-Kelly} set theory, is obtained.}
\end{axiom}


\todo{notation: $\emptyset, \cap$}



\subsection{Induction and Recursion}

\begin{definition}
    A binary relation $R$ on a set $X$,
    i.e.~$R \subseteq X \times X$,
    is called \vocab{well-founded}
    iff for all $\emptyset \neq Y \subseteq X$
    there is some $x \in Y$
    such that for no $y \in Y. (y,x) \in R$.
\end{definition}

\begin{example}
    \begin{enumerate}[(a)]
        \item $(\N, <)$ is well-founded.
        \item Let $M$ be a set,
            and let $\in\defon{M} \coloneqq  \{(x,y) : x,y \in M \land x \in y\}$.
            Fund is equivalent to saying that
            this is a well-founded relation for every $M$.
    \end{enumerate}
\end{example}


\begin{lemma}
    \label{lem:fundseq}
    In $\ZFC - \Fund$,
    the following are equivalent:
    \begin{itemize}
        \item \Fund,
        \item There is no sequence $\langle x_n : n < \omega \rangle$
             such that $x_{n+1} \in x_n$ for all $n < \omega$.
    \end{itemize}
\end{lemma}
\begin{proof}
    Suppose such sequence exists.
    Then $\{x_n : n < \omega\}$
    (this exists as by definition sequence of the $x_n$ is a function
    and this set is the range of that function)
    violates \Fund.

    For the other direction let $M \neq \emptyset$ be some set.
    Suppose that \Fund does not hold for $M$.

    Using \Choice,
    we construct an infinite sequence $x_0 \ni x_1 \ni x_2 \ni \ldots$
    of elements of $M$.

    More formally,
    for each $x \in M$
    let $A_x \coloneqq  \{y \in M: y \in x\}$.
    Suppose that $A_x \neq \emptyset$ for all $x \in M$.
    Using \AxC
    we get a function for $\langle A_x : x \in M \rangle$,%
    \footnote{Actually we only need the axiom of dependent choice,
        a weaker form of the axiom of choice.
        We'll discuss this later.% TODO REF
    }
    i.e.~a function $f\colon M \to  M$
    such that $f(x) \in A_x$ for $x \in M$.
    Now fix $x \in M$.
    We want to produce
    a function
    $g\colon \omega \to  M$
    such that
    \begin{itemize}
        \item $g(0) = x$,
        \item $g(n+1) = f(g(n)) \in A_{g(n)}$.
    \end{itemize}

    Let
    \begin{IEEEeqnarray*}{rCl}
        G &=& \{\overline{g} : \exists  n \in  \omega . \\
          &&~ ~\overline{g} \text{ is a function with domain $n$ and range $\subseteq  M$, such that}\\
          &&~ ~\overline{g}(0) = x \land \forall m \in  \omega.~(m+1 \in \dom(\overline{g}) \implies \overline{g}(m+1) = f(\overline{g}(m)))\}.
    \end{IEEEeqnarray*}
    $G$ exists as it can be obtained by \AxSep
    from ${}^{< \omega}M$.
    By induction,
    for every $n \in  \omega$,
    there is a $\overline{g} \in G$
    with $\dom(\overline{g}) \in n+1$:
    This holds for $n = 0$,
    as $\{(0,x)\} \in  G$.
    If $\overline{g} \in  G$ with $\dom(\overline{g}) = n+ 1$,
    then $\overline{g} \cup  \{(n+1, f(\overline{g}(n)))\} \in G$.
    Also by induction,
    for every $n \in \omega$,
    there is a \emph{unique}
    $\overline{g}$ with $\dom(\overline{g}) = n+1$.

    Now let $g = \bigcup \overline{G}$.
    Also let $g(0) = x$ and $g(n+1) = f(g(n))$
    for all  $n \in \omega$.
\end{proof}

\begin{lemma}[Dependent Choice]
    Suppose that $M \neq  \emptyset$
    and $R$ is a binary relation on $M$
    such that for all $x \in M$,
    $A_x \coloneqq  \{y \in M : (y,x) \in R\}$
    is not empty.

    Then for every $x \in M$ there exists a function
    $g\colon \omega \to M$
    such that $g(0) = x$
    and $g(n+1) \in A_{g(n)}$
    for all $n < \omega$.
\end{lemma}
\begin{proof}
    We showed a special case of this in the proof of
    \yaref{lem:fundseq}.
\end{proof}
\begin{remark}
    In $\ZF$ this is a weaker form of \Choice.
\end{remark}

The construction of $g$ in the previous proof was a special case of
a construction on the proof of the recursion theorem: % TODO REF