lecture 09
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@ -111,6 +111,31 @@ Furthermore $F\,''a \coloneqq \{y : \exists x \in a .~(x,y) \in F\}$.
\todo{notation: $\emptyset, \cap$} \todo{notation: $\emptyset, \cap$}
\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}
\subsection{Induction and Recursion} \subsection{Induction and Recursion}
@ -230,9 +255,3 @@ Furthermore $F\,''a \coloneqq \{y : \exists x \in a .~(x,y) \in F\}$.
The construction of $g$ in the previous proof was a special case of The construction of $g$ in the previous proof was a special case of
a construction on the proof of the recursion theorem: % TODO REF a construction on the proof of the recursion theorem: % TODO REF

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inputs/lecture_09.tex Normal file
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@ -0,0 +1,128 @@
\lecture{09}{2023-11-16}{}
\begin{definition}
Let $R$ be a binary relation.
$R$ is called \vocab{well-founded}
if for all classes $X$,
there is an $R$-least $y$ such that
there is no $z \in X$ with $(z,y) \in R$.
\end{definition}
\begin{theorem}[Induction (again, but now with classes)]
Suppose that $R$ is a well-founded relation.
Let $X$ be a class such that for all sets $x$,
\[
\{y : (y,x) \in R\} \subseteq X \implies x \in X.
\]
Then $X$ contains all sets.
\end{theorem}
\begin{proof}
Assume otherwise.
Consider $Y = \{x : x \not\in X\} \neq \emptyset$.
By hypothesis, there is some
$x \in Y$ such that $(y,x) \not\in R$ for all $y \in Y$.
In other words, if $(y,x) \not\in R$,
then $x \not\in Y$, i.e.~$x \in X$.
Thus $\{y: (y,x) \in R\} \subseteq X$.
Hence $x \in X \lightning$.
\end{proof}
An alternative way of formulating this is
\begin{theorem}
Suppose $R$ is a well-founded binary relation on $A$,
i.e.~$R \subseteq A \times A$.
Suppose for all $\overline{A} \subseteq A$
is such that for all $x \in X$,
\[
\{y \in A: (y,x) \in R\} \subseteq \overline{A} \implies x \in \overline{A}.
\]
Then $\overline{A} = A$.
\end{theorem}
\begin{definition}
Let $R$ be a binary relation.
$R$ is called \vocab{set-like}
iff for all $x$,
$\{y : (y,x) \in R\}$ is a set.
\end{definition}
\begin{theorem}
Let $R$ be a well-founded
and set-like relation on $A$ (i.e.~$R \subseteq A \times A$).
Let $D$ be a class of triples
such that for all $u,x$ there is exactly
one $y$ with $(u,x,y) \in D$
(basicalls $(u,x) \mapsto y$ is a function).
Then there is a unique function $f$ on $A$
such that for all $x \in A$,
\[
(F\defon{ \{y \in A : (y,x) \in R\}}, x, F(x)) \in D.
\]
I.e.~$F(x)$ is computed from $F\defon{\{y \in A: (y,x) \in R\}}$.
\end{theorem}
\begin{proof}
Uniqueness:
Let $F, F'$ be two such functions.
Suppose that $\overline{A} = \{ x \in A : F(x) \neq F'(x)\} \neq \emptyset$.
As $R$ is well-founded,
there is some $x \in \overline{A}$
such that $y \not\in \overline{A}$
for all $y \in A, (y,x) \in R$.
I.e. $F(y) = F'(y)$
for all $y \in A$, $(y,x) \in R$.
But then $F(x)$
is the unique $y$
with $(F\defon{\{z\colon (z,x) \in R\}}, x, y) \in D$,
in particular it is the same as $F'(x) \lightning$
Existence:
Let us call a (set) function $f$
\emph{good},
if
\begin{itemize}
\item $\dom(f) \subseteq A$,
\item if $x \in \dom(f)$ and $y \in A, (y,x) \in R$,
then $y \in \dom(f)$ and
\item for all $x \in \dom(f)$ :
\[
(f\defon{\{y \in A: (y,x) \in R\}}, x, f(x)) \in D.
\]
\end{itemize}
By the proof of uniqueness,
we have that all good functions are coherent,
i.e.~$f(x) = f'(x)$ for good functions
$f,f'$ and all $x \in \dom(f) \cap \dom(f')$.
We may now let $F = \bigcup \{f: f \text{ is good}\}$,
this exists by comprehension.
If $x \in \dom(F)$ and $y \in A$ with $(y,x) \in R$,
then $y \in \dom(F)$
and $(F\defon{\{y : (y,x) \in R\}}, x,F(x)) \in D$.
We need to show that $\dom(F) = A$.
This holds by induction:
Suppose for a contradiction that $A \setminus \dom(F) \neq \emptyset$.
Then there exists an $R$-least element $x$ in this set,
i.e.$x \not\in \dom(F)$,
but $y \in \dom(F)$ for all $(y,x) \in R$.
For each $y \in A$ with $(y,x) \in R$,
pick some good function $f_y$ with $y \in \dom(f_y)$
Since $R$ is set-like,
we have that $f = \bigcup_y f_y$ is a good function.
But then $f \cup (x,z)$,
where $z$ is unique such that $(f\defon{\{y : (y,x) \in R\}}, x, z) \in D$,
is good $\lightning$.
\end{proof}

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@ -110,6 +110,8 @@
\DeclareMathOperator{\Zermelo}{Z} \DeclareMathOperator{\Zermelo}{Z}
\DeclareSimpleMathOperator{ZF} \DeclareSimpleMathOperator{ZF}
\DeclareSimpleMathOperator{ZFC} \DeclareSimpleMathOperator{ZFC}
\DeclareSimpleMathOperator{BG}
\DeclareSimpleMathOperator{BGC}
\DeclareSimpleMathOperator{HOD} \DeclareSimpleMathOperator{HOD}
\DeclareSimpleMathOperator{OD} \DeclareSimpleMathOperator{OD}
\DeclareSimpleMathOperator{AC} \DeclareSimpleMathOperator{AC}

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@ -32,6 +32,7 @@
\input{inputs/lecture_06} \input{inputs/lecture_06}
\input{inputs/lecture_07} \input{inputs/lecture_07}
\input{inputs/lecture_08} \input{inputs/lecture_08}
\input{inputs/lecture_09}
\cleardoublepage \cleardoublepage