w23-logic-2/inputs/lecture_08.tex

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\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}
\[
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\forall x.~\forall y.~ \left( x = y \iff \left( \forall z.~(z \in x \iff z \in y \right) \right).
\]
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\end{axiom}
\begin{axiom}[Foundation]
\yalabel{Axiom of Foundation}{(Fund)}{ax:bg:fund}
\[
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\forall x .~(x \neq \emptyset \implies \exists y \in x .~ y \cap x = \emptyset).
\]
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\end{axiom}
\begin{axiom}[Pairing]
\yalabel{Axiom of Pairing}{(Pair)}{ax:bg:pair}
\[
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\forall x.~\forall y . ~\exists z .~ z = \{x,y\}.
\]
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\end{axiom}
\begin{axiom}[Union]
\yalabel{Axiom of Union}{(Union)}{ax:bg:union}
\[
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\forall x .~\exists y .~ y = \bigcup x.
\]
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\end{axiom}
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\begin{axiom}[Power Set]
\yalabel{Power Set Axiom}{(Pow)}{ax:bg:pow}
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\[
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\forall x .~\exists y .~ y = \cP(x).
\]
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\end{axiom}
\begin{axiom}[Infinity]
\yalabel{Axiom of Infinity}{(Infinity)}{ax:bg:inf}
\[
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\exists x .~(\emptyset \in x \land \left( \forall y \in x .~y \cup \{y\} \in x \right)).
\]
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\end{axiom}
Together with the following axioms for classes:
\begin{axiom}[Extensionality for classes]
\[
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\forall X .~\forall Y.~ \left( \forall x.~(x \in X \iff x \in Y) \implies X = Y\right).
\]
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\end{axiom}
\begin{axiom}
Every set is a class:
\[
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\forall x.~ \exists X.~ x = X.
\]
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\end{axiom}
\begin{axiom}
Every element of a class is a set:
\[
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\forall X .~\exists Y.~(X \in Y \to \exists x.~x = X).
\]
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\end{axiom}
\begin{axiom}[Replacement]
\yalabel{Axiom of Replacement}{(Rep)}{ax:bg:rep}
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If $F$ is a function and $a$ is a set,
then $F\,''a$ is a set.
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\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 $(x,y) \in F$.
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Furthermore $F\,''a \coloneqq \{y : \exists x \in a .~(x,y) \in F\}$.
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\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}
\[
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\forall X_1 .~\ldots \forall X_k .~\exists Y.~ ( \forall x .~x \in Y \iff \phi(x,X_1,\ldots, X_k))
\]
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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: $\emptyset, \cap$}
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\todo{the following was actually done in lecture 9}
$\BGC$ (in German often NBG) is defined to be $\BG$
together with the additional axiom:
\begin{axiom}[Choice]
\[
\exists F.~(F \text{ is a function} \land \forall x \neq \emptyset. F(x) \in x).
\]
\end{axiom}
\begin{fact}
$\BGC$ is conservative over $\ZFC$,
i.e.~for all formulae $\phi$ in the language
of set theory (only set variables):
if $\BGC \vdash \phi$ then $\ZFC \vdash \phi$.
\end{fact}
We cannot prove this fact at this point,
as the proof requires forcing.
The converse is easy however, i.e.~if
$\ZFC \vdash \phi$ then $\BGC \vdash \phi$.
\begin{notation}
From now on, objects denoted by capital letters
are (potentially proper) classes.
\end{notation}
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\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 $\in\defon{M} \coloneqq \{(x,y) : x,y \in M \land x \in y\}$.
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\AxFund is equivalent to saying that
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this is a well-founded relation for every $M$.
\end{enumerate}
\end{example}
\begin{lemma}
\label{lem:fundseq}
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In $\ZFC - \AxFund$,
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the following are equivalent:
\begin{itemize}
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\item \AxFund,
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\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)
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violates \AxFund.
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For the other direction let $M \neq \emptyset$ be some set.
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Suppose that \AxFund does not hold for $M$.
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Using \AxC,
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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,
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a weaker form of the \yaref{ax:c}.
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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*}
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$G$ exists as it can be obtained by \AxAus
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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}
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In $\ZF$ this is a weaker form of \AxC.
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\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