From 6936663173805ee9251581942ea501c4fa8e14de Mon Sep 17 00:00:00 2001 From: Josia Pietsch Date: Thu, 16 Nov 2023 15:38:46 +0100 Subject: [PATCH] lecture 09 --- inputs/lecture_08.tex | 31 ++++++++-- inputs/lecture_09.tex | 128 ++++++++++++++++++++++++++++++++++++++++++ logic.sty | 2 + logic2.tex | 1 + 4 files changed, 156 insertions(+), 6 deletions(-) create mode 100644 inputs/lecture_09.tex diff --git a/inputs/lecture_08.tex b/inputs/lecture_08.tex index 25e55c8..8a7c7e2 100644 --- a/inputs/lecture_08.tex +++ b/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 - - - - - - diff --git a/inputs/lecture_09.tex b/inputs/lecture_09.tex new file mode 100644 index 0000000..7274f99 --- /dev/null +++ b/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} + + + + diff --git a/logic.sty b/logic.sty index ec89433..a2cafdf 100644 --- a/logic.sty +++ b/logic.sty @@ -110,6 +110,8 @@ \DeclareMathOperator{\Zermelo}{Z} \DeclareSimpleMathOperator{ZF} \DeclareSimpleMathOperator{ZFC} +\DeclareSimpleMathOperator{BG} +\DeclareSimpleMathOperator{BGC} \DeclareSimpleMathOperator{HOD} \DeclareSimpleMathOperator{OD} \DeclareSimpleMathOperator{AC} diff --git a/logic2.tex b/logic2.tex index 47d53e4..871cebd 100644 --- a/logic2.tex +++ b/logic2.tex @@ -32,6 +32,7 @@ \input{inputs/lecture_06} \input{inputs/lecture_07} \input{inputs/lecture_08} +\input{inputs/lecture_09} \cleardoublepage