This commit is contained in:
parent
ab3a4b7556
commit
340e034fdf
1 changed files with 147 additions and 0 deletions
147
inputs/lecture_04.tex
Normal file
147
inputs/lecture_04.tex
Normal file
|
@ -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
|
Loading…
Reference in a new issue