From 340e034fdfd1ab6e6116f846d2ae603f11080175 Mon Sep 17 00:00:00 2001 From: Josia Pietsch Date: Sat, 4 Nov 2023 23:28:39 +0100 Subject: [PATCH] lecture 4 --- inputs/lecture_04.tex | 147 ++++++++++++++++++++++++++++++++++++++++++ 1 file changed, 147 insertions(+) create mode 100644 inputs/lecture_04.tex diff --git a/inputs/lecture_04.tex b/inputs/lecture_04.tex new file mode 100644 index 0000000..3bdf433 --- /dev/null +++ b/inputs/lecture_04.tex @@ -0,0 +1,147 @@ +\lecture{04}{}{ZFC} + +% Model-theoretic concepts and ultraproducts + +\section{$\ZFC$} + +% 1900, Russel's paradox +Russel's Paradox: +$\ZFC$ stands for +\begin{itemize} + \item \textsc{Zermelo}’s axioms (1905), % crises around 19000 + \item \vocab{Fraenkel}'s axioms, + \item the axiom of choice. +\end{itemize} +\begin{notation} + We write $x \subseteq y$ as a shorthand + for $\forall z.~(z \in x \implies z \in y)$. + + We write $x = \emptyset$ for $\lnot \exists y . y \in x$ + and $x \cap y = \emptyset$ for $\lnot \exists z . ~(z \in x \land z \in y)$. + + We use $x = \{y,z\}$ + for + \[ + y \in x \land z \in x \land \forall a .~(a \in x \implies a = y \lor a = z). + \] + + Let $x = \bigcup y$ denote + \[ + \forall z.~(z \in x \iff \exists v.(v \in y \land z \in v)). + \] +\end{notation} +$\ZFC$ consists of the following axioms: +\begin{axiom}[\vocab{Extensionality}] + \[ + \forall x.~\forall y.~(x = y \iff \forall z.~(z \in x \iff z \in y)). + \] + Equivalent statements using $\subseteq$: + \[ + \forall x.~\forall y.~(x = y \iff (x \subseteq y \land y \subseteq x)). + \] +\end{axiom} + +\begin{axiom}[\vocab{Foundation}] + Every set has an $\in$-minimal member: + \[ + \forall x .~ \left(\exists a .~(a\in x) \implies + \exists y .~ y \in x \land \lnot \exists z.~(z \in y \land z \in x)\right). + \] + Shorter: + \[ + \forall x.~(x \neq \emptyset \implies \exists y \in x .~ x \cap y = \emptyset). + \] +\end{axiom} +\begin{axiom}[\vocab{Pairing}] + \[ + \forall x .~\forall y.~ \exists z.~(z = \{x,y\}). + \] +\end{axiom} +\begin{remark} + Together with the axiom of pairing, + the axiom of foundation implies + that there can not be a set $x$ such that + $x \in x$: + Suppose that $x \in x$. + Then $x$ is the only element of $\{x\}$, + but $x \cap \{x\} \neq \emptyset$. + + A similar argument shows that chains like + $x_0 \in x_1 \in x_2 \in x_0$ + are ruled out as well. +\end{remark} + +\begin{axiom}[\vocab{Union}] + \[ + \forall x.~\exists y.~(y = \bigcup x). + \] +\end{axiom} + +\begin{axiom}[\vocab{Powerset}] + We write $x = \cP(y)$ + for + $\forall z.~(z \in x \iff x \subseteq z)$. + The powerset axiom (PWA) states + \[ + \forall x.~\exists y.~y=\cP(x). + \] +\end{axiom} +\begin{axiom}[\vocab{Infinity}] + A set $x$ is called \vocab{inductive}, + iff $\emptyset \in x \land \forall y.~(y \in x \implies y \cup \{y\} \in x)$. + + The axiom of infinity says that there exists and inductive set. +\end{axiom} + +\begin{axiomscheme}[Separation] + % TODO :(Aus) + Let $\phi$ be some fixed + fist order formula in $\cL_\in$. + Then $\text{(Aus)}_{\phi}$ + states + \[ + \forall v_1 .~\forall v_p .~\forall a .~\exists b .~\forall x.~ + (x \in b \implies x \in a \land \phi(x,v_1,v_p)) + \] + + Let us write $b = \{x \in a | \phi(x)\}$ + for $\forall x.~(x \in b \iff x \in a \land f(x))$. + Then (Aus) can be formulated as + \[ + \forall a.~\exists b.~(b = \{x \in a; \phi(x)\}). + \] +\end{axiomscheme} + +\begin{notation} + \todo{$\cap, \setminus, \bigcap$} + % We write $z = x \cap y$ for $\forall u.~((u \in z) \implies u \in x \land u \in y)$, + % $Z = x \setminus y$ for ... + % $x = \bigcap y$ for ... +\end{notation} +\begin{remark} + (Aus) proves that + \begin{itemize} + \item $\forall a.~\forall b.~\exists c.~(c = a \cap b)$, + \item $\forall a.~\forall b.~\exists c.~(c = a \setminus b)$, + \item $\forall a.~\exists b.~(b = \bigcap a)$. + \end{itemize} +\end{remark} +\begin{axiomscheme}[\vocab{Replacement} (Fraenkel)] + Let $\phi$ be some $\cL_{\in }$ formula. + Then + \[ + \forall v_1 .~\exists b.~\forall y.~(y \in b \iff \exists x .~(x \in a \land \phi(x,y,v_1,v_p))). + \] +\end{axiomscheme} + +\begin{axiom}[\vocab{Choice}] + Every family of non-empty sets has a \vocab{choice set}: + \todo{TODO} + \[ + \forall x.~(\forall y \in x.~x \neq \emptyset \land + \forall y \in x \forall y' \in x .(y \neq y' \implies x \cap y' = \emptyset) \implies \exists z .~\forall y \in x .~\exists u.~(z \cap y = \{u\})). + \] +\end{axiom} + + +% TODO Hier weiter