240 lines
6.7 KiB
TeX
240 lines
6.7 KiB
TeX
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\lecture{08}{2023-11-13}{Induction and recursion}
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\subsection{Classes}
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It is often very handy to work in a class theory rather than
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in set theory.
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To formulate a class theory,
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we start out with a first order language
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with two types of variables,
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sets (denoted by lower case letters)
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and classes (denoted by capital letters),
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as well as one binary relation symbol $\in$
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for membership.
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\vocab{Bernays-Gödel class theory} (\vocab{BG})
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has the following axioms:
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\begin{axiom}[Extensionality]
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\yalabel{Axiom of Extensionality}{(Ext)}{ax:bg:ext}
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\[
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\forall x, y. \left( x = y \iff \left( \forall z.~(z \in x \iff z \in y \right) \right).
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\]
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\end{axiom}
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\begin{axiom}[Foundation]
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\yalabel{Axiom of Foundation}{(Fund)}{ax:bg:fund}
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\[
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\forall x .(x \neq \emptyset \implies \exists y \in x . y \cap x = \emptyset).
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\]
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\end{axiom}
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\begin{axiom}[Pairing]
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\yalabel{Axiom of Pairing}{(Pair)}{ax:bg:pair}
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\[
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\forall x \forall y \exists z . z = \{x,y\}.
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\]
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\end{axiom}
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\begin{axiom}[Union]
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\yalabel{Axiom of Union}{(Union)}{ax:bg:union}
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\[
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\forall x \exists y .~ y = \bigcup x.
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\]
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\end{axiom}
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\begin{axiom}[Power]
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\yalabel{Powerset Axiom}{(Pow)}{ax:bg:pow}
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\[
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\forall x \exists y .~ y = \cP(x).
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\]
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\end{axiom}
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\begin{axiom}[Infinity]
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\yalabel{Axiom of Infinity}{(Infinity)}{ax:bg:inf}
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\[
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\exists x . ~(\emptyset \in x \land \left( \forall y \in x .~y \cup \{y\} \in x \right)).
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\]
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\end{axiom}
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Together with the following axioms for classes:
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\begin{axiom}[Extensionality for classes]
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\[
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\forall X \forall Y \left( \forall x(x \in X \iff y \in X) \implies X = Y).
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\]
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\end{axiom}
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\begin{axiom}
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Every set is a class:
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\[
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\forall x \exists X. x = X.
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\]
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\end{axiom}
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\begin{axiom}
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Every element of a class is a set:
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\[
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\forall X \exists Y.~(X \in Y \to \exists x.~x = X).
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\]
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\end{axiom}
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\begin{axiom}[Replacement]
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\yalabel{Axiom of Replacement}{(Rep)}{ax:bg:rep}
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If $F$ is a function and inf $a $ is a set,
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then $F"a$ is a set.
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\end{axiom}
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Here a \vocab[Class function]{(class) function} is a class
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consisting of pairs $(x,y)$,
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such that for every $x$ there is at most one $y$
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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}
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Note that we didn't need to use an axiom schema,
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\yarefs{ax:bg:rep} is a single axiom.
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\end{remark}
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\begin{axiom}[Comprehension]
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\yalabel{Axiom of Comprehension}{(Comp)}{ax:bg:comp}
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\[
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\forall X_1 \ldots \forall X_k .~\exists Y ( \forall x .~x \in Y \iff \phi(x,X_1,\ldots, X_k))
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\]
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where $\phi(x, X_1, \ldots, X_k)$
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is a formula which contains exactly $X_1, \ldots, X_k, x$
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as free variables,
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and $\phi$ does not have quantifiers
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ranging over classes.%
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\footnote{If one removes the restriction regarding
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quantifiers another theory, called
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\vocab{Morse-Kelly} set theory, is obtained.}
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\end{axiom}
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\todo{notation: $\emptyset, \cap$}
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\subsection{Induction and Recursion}
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\begin{definition}
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A binary relation $R$ on a set $X$,
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i.e.~$R \subseteq X \times X$,
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is called \vocab{well-founded}
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iff for all $\emptyset \neq Y \subseteq X$
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there is some $x \in Y$
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such that for no $y \in Y. (y,x) \in R$.
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\end{definition}
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\begin{example}
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\begin{enumerate}[(a)]
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\item $(\N, <)$ is well-founded.
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\item Let $M$ be a set,
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and let $\in\defon{M} \coloneqq \{(x,y) : x,y \in M \land x \in y\}$.
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Fund is equivalent to saying that
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this is a well-founded relation for every $M$.
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\end{enumerate}
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\end{example}
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\begin{lemma}
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\label{lem:fundseq}
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In $\ZFC - \Fund$,
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the following are equivalent:
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\begin{itemize}
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\item \Fund,
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\item There is no sequence $\langle x_n : n < \omega \rangle$
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such that $x_{n+1} \in x_n$ for all $n < \omega$.
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\end{itemize}
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\end{lemma}
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\begin{proof}
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Suppose such sequence exists.
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Then $\{x_n : n < \omega\}$
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(this exists as by definition sequence of the $x_n$ is a function
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and this set is the range of that function)
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violates \Fund.
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For the other direction let $M \neq \emptyset$ be some set.
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Suppose that \Fund does not hold for $M$.
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Using \Choice,
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we construct an infinite sequence $x_0 \ni x_1 \ni x_2 \ni \ldots$
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of elements of $M$.
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More formally,
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for each $x \in M$
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let $A_x \coloneqq \{y \in M: y \in x\}$.
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Suppose that $A_x \neq \emptyset$ for all $x \in M$.
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Using \AxC
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we get a function for $\langle A_x : x \in M \rangle$,%
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\footnote{Actually we only need the axiom of dependent choice,
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a weaker form of the axiom of choice.
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We'll discuss this later.% TODO REF
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}
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i.e.~a function $f\colon M \to M$
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such that $f(x) \in A_x$ for $x \in M$.
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Now fix $x \in M$.
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We want to produce
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a function
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$g\colon \omega \to M$
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such that
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\begin{itemize}
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\item $g(0) = x$,
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\item $g(n+1) = f(g(n)) \in A_{g(n)}$.
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\end{itemize}
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Let
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\begin{IEEEeqnarray*}{rCl}
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G &=& \{\overline{g} : \exists n \in \omega . \\
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&&~ ~\overline{g} \text{ is a function with domain $n$ and range $\subseteq M$, such that}\\
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&&~ ~\overline{g}(0) = x \land \forall m \in \omega.~(m+1 \in \dom(\overline{g}) \implies \overline{g}(m+1) = f(\overline{g}(m)))\}.
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\end{IEEEeqnarray*}
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$G$ exists as it can be obtained by \AxSep
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from ${}^{< \omega}M$.
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By induction,
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for every $n \in \omega$,
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there is a $\overline{g} \in G$
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with $\dom(\overline{g}) \in n+1$:
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This holds for $n = 0$,
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as $\{(0,x)\} \in G$.
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If $\overline{g} \in G$ with $\dom(\overline{g}) = n+ 1$,
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then $\overline{g} \cup \{(n+1, f(\overline{g}(n)))\} \in G$.
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Also by induction,
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for every $n \in \omega$,
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there is a \emph{unique}
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$\overline{g}$ with $\dom(\overline{g}) = n+1$.
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Now let $g = \bigcup \overline{G}$.
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Also let $g(0) = x$ and $g(n+1) = f(g(n))$
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for all $n \in \omega$.
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\end{proof}
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\begin{lemma}[Dependent Choice]
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Suppose that $M \neq \emptyset$
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and $R$ is a binary relation on $M$
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such that for all $x \in M$,
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$A_x \coloneqq \{y \in M : (y,x) \in R\}$
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is not empty.
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Then for every $x \in M$ there exists a function
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$g\colon \omega \to M$
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such that $g(0) = x$
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and $g(n+1) \in A_{g(n)}$
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for all $n < \omega$.
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\end{lemma}
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\begin{proof}
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We showed a special case of this in the proof of
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\yaref{lem:fundseq}.
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\end{proof}
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\begin{remark}
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In $\ZF$ this is a weaker form of \Choice.
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\end{remark}
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The construction of $g$ in the previous proof was a special case of
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a construction on the proof of the recursion theorem: % TODO REF
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