lecture 09

inputs/lecture_08.tex

@@ -111,6 +111,31 @@ Furthermore $F\,''a \coloneqq  \{y : \exists  x \in  a .~(x,y) \in F\}$.
 \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}
 
@@ -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
 a construction on the proof of the recursion theorem: % TODO REF
 
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inputs/lecture_09.tex

@@ -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}
+
+
+
+

logic.sty

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

logic2.tex

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