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@ -111,6 +111,31 @@ Furthermore $F\,''a \coloneqq \{y : \exists x \in a .~(x,y) \in F\}$.
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\todo{notation: $\emptyset, \cap$}
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\todo{notation: $\emptyset, \cap$}
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\todo{the following was actually done in lecture 9}
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$\BGC$ (in German often NBG) is defined to be $\BG$
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together with the additional axiom:
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\begin{axiom}[Choice]
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\[
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\exists F.~(F \text{ is a function} \land \forall x \neq \emptyset. F(x) \in x).
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\]
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\end{axiom}
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\begin{fact}
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$\BGC$ is conservative over $\ZFC$,
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i.e.~for all formulae $\phi$ in the language
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of set theory (only set variables):
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if $\BGC \vdash \phi$ then $\ZFC \vdash \phi$.
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\end{fact}
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We cannot prove this fact at this point,
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as the proof requires forcing.
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The converse is easy however, i.e.~if
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$\ZFC \vdash \phi$ then $\BGC \vdash \phi$.
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\begin{notation}
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From now on, objects denoted by capital letters
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are (potentially proper) classes.
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\end{notation}
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\subsection{Induction and Recursion}
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\subsection{Induction and Recursion}
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@ -230,9 +255,3 @@ Furthermore $F\,''a \coloneqq \{y : \exists x \in a .~(x,y) \in F\}$.
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The construction of $g$ in the previous proof was a special case of
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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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a construction on the proof of the recursion theorem: % TODO REF
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128
inputs/lecture_09.tex
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128
inputs/lecture_09.tex
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@ -0,0 +1,128 @@
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\lecture{09}{2023-11-16}{}
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\begin{definition}
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Let $R$ be a binary relation.
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$R$ is called \vocab{well-founded}
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if for all classes $X$,
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there is an $R$-least $y$ such that
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there is no $z \in X$ with $(z,y) \in R$.
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\end{definition}
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\begin{theorem}[Induction (again, but now with classes)]
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Suppose that $R$ is a well-founded relation.
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Let $X$ be a class such that for all sets $x$,
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\[
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\{y : (y,x) \in R\} \subseteq X \implies x \in X.
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\]
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Then $X$ contains all sets.
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\end{theorem}
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\begin{proof}
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Assume otherwise.
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Consider $Y = \{x : x \not\in X\} \neq \emptyset$.
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By hypothesis, there is some
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$x \in Y$ such that $(y,x) \not\in R$ for all $y \in Y$.
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In other words, if $(y,x) \not\in R$,
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then $x \not\in Y$, i.e.~$x \in X$.
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Thus $\{y: (y,x) \in R\} \subseteq X$.
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Hence $x \in X \lightning$.
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\end{proof}
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An alternative way of formulating this is
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\begin{theorem}
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Suppose $R$ is a well-founded binary relation on $A$,
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i.e.~$R \subseteq A \times A$.
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Suppose for all $\overline{A} \subseteq A$
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is such that for all $x \in X$,
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\[
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\{y \in A: (y,x) \in R\} \subseteq \overline{A} \implies x \in \overline{A}.
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\]
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Then $\overline{A} = A$.
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\end{theorem}
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\begin{definition}
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Let $R$ be a binary relation.
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$R$ is called \vocab{set-like}
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iff for all $x$,
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$\{y : (y,x) \in R\}$ is a set.
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\end{definition}
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\begin{theorem}
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Let $R$ be a well-founded
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and set-like relation on $A$ (i.e.~$R \subseteq A \times A$).
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Let $D$ be a class of triples
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such that for all $u,x$ there is exactly
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one $y$ with $(u,x,y) \in D$
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(basicalls $(u,x) \mapsto y$ is a function).
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Then there is a unique function $f$ on $A$
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such that for all $x \in A$,
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\[
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(F\defon{ \{y \in A : (y,x) \in R\}}, x, F(x)) \in D.
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\]
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I.e.~$F(x)$ is computed from $F\defon{\{y \in A: (y,x) \in R\}}$.
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\end{theorem}
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\begin{proof}
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Uniqueness:
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Let $F, F'$ be two such functions.
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Suppose that $\overline{A} = \{ x \in A : F(x) \neq F'(x)\} \neq \emptyset$.
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As $R$ is well-founded,
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there is some $x \in \overline{A}$
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such that $y \not\in \overline{A}$
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for all $y \in A, (y,x) \in R$.
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I.e. $F(y) = F'(y)$
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for all $y \in A$, $(y,x) \in R$.
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But then $F(x)$
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is the unique $y$
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with $(F\defon{\{z\colon (z,x) \in R\}}, x, y) \in D$,
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in particular it is the same as $F'(x) \lightning$
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Existence:
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Let us call a (set) function $f$
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\emph{good},
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if
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\begin{itemize}
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\item $\dom(f) \subseteq A$,
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\item if $x \in \dom(f)$ and $y \in A, (y,x) \in R$,
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then $y \in \dom(f)$ and
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\item for all $x \in \dom(f)$ :
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\[
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(f\defon{\{y \in A: (y,x) \in R\}}, x, f(x)) \in D.
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\]
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\end{itemize}
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By the proof of uniqueness,
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we have that all good functions are coherent,
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i.e.~$f(x) = f'(x)$ for good functions
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$f,f'$ and all $x \in \dom(f) \cap \dom(f')$.
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We may now let $F = \bigcup \{f: f \text{ is good}\}$,
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this exists by comprehension.
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If $x \in \dom(F)$ and $y \in A$ with $(y,x) \in R$,
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then $y \in \dom(F)$
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and $(F\defon{\{y : (y,x) \in R\}}, x,F(x)) \in D$.
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We need to show that $\dom(F) = A$.
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This holds by induction:
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Suppose for a contradiction that $A \setminus \dom(F) \neq \emptyset$.
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Then there exists an $R$-least element $x$ in this set,
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i.e.$x \not\in \dom(F)$,
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but $y \in \dom(F)$ for all $(y,x) \in R$.
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For each $y \in A$ with $(y,x) \in R$,
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pick some good function $f_y$ with $y \in \dom(f_y)$
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Since $R$ is set-like,
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we have that $f = \bigcup_y f_y$ is a good function.
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But then $f \cup (x,z)$,
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where $z$ is unique such that $(f\defon{\{y : (y,x) \in R\}}, x, z) \in D$,
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is good $\lightning$.
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\end{proof}
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@ -110,6 +110,8 @@
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\DeclareMathOperator{\Zermelo}{Z}
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\DeclareMathOperator{\Zermelo}{Z}
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\DeclareSimpleMathOperator{ZF}
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\DeclareSimpleMathOperator{ZF}
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\DeclareSimpleMathOperator{ZFC}
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\DeclareSimpleMathOperator{ZFC}
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\DeclareSimpleMathOperator{BG}
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\DeclareSimpleMathOperator{BGC}
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\DeclareSimpleMathOperator{HOD}
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\DeclareSimpleMathOperator{HOD}
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\DeclareSimpleMathOperator{OD}
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\DeclareSimpleMathOperator{OD}
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\DeclareSimpleMathOperator{AC}
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\DeclareSimpleMathOperator{AC}
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@ -32,6 +32,7 @@
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\input{inputs/lecture_06}
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\input{inputs/lecture_06}
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\input{inputs/lecture_07}
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\input{inputs/lecture_07}
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\input{inputs/lecture_08}
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\input{inputs/lecture_08}
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\input{inputs/lecture_09}
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\cleardoublepage
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\cleardoublepage
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