w23-logic-2/inputs/lecture_04.tex
Josia Pietsch 858e017a4a
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\lecture{04}{}{ZFC}
% Model-theoretic concepts and ultraproducts
\section{\texorpdfstring{$\ZFC$}{ZFC}}
% 1900, Russel's paradox
% \todo{Russel's Paradox}
$\ZFC$ stands for
\begin{itemize}
\item \textsc{Zermelo}s axioms (1905), % crises around 19000
\item \textsc{Fraenkel}'s axioms,
\item the \yaref{ax:c}.
\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
\[
y \in x \land z \in x \land \forall a .~(a \in x \implies a = y \lor a = z).
\]
We write $z = x \cap y$ for
\[\forall u.~((u \in z) \implies u \in x \land u \in y),\]
$z = x \cup y$ for
\[
\forall u.~((u \in z) \iff (u \in x \lor u \in y)),
\]
$z = \bigcap x$ for
\[
\forall u.~((u \in z) \iff (\forall v.~(v \in x \implies u \in v))),
\]
$z = \bigcup x$ for
\[
\forall u.~((u \in z) \iff \exists v.~(v \in x \land u \in v))
\]
and
$z = x \setminus y$ for
\[
\forall u.~((u \in z) \iff (u \in x \land u \not\in y)).
\]
\end{notation}
$\ZFC$ consists of the following axioms:
\begin{axiom}[\vocab{Extensionality}]
\yalabel{Axiom of Extensionality}{(Ext)}{ax:ext} % (AoE)
\[
\forall x.~\forall y.~(x = y \iff \forall z.~(z \in x \iff z \in y)).
\]
Equivalent statements using $\subseteq$:
\[
\forall x.~\forall y.~(x = y \iff (x \subseteq y \land y \subseteq x)).
\]
\end{axiom}
\begin{axiom}[\vocab{Foundation}]
\yalabel{Axiom of Foundation}{(Fund)}{ax:fund}
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}]
\yalabel{Axiom of Pairing}{(Pair)}{ax:pair} % AoP
\[
\forall x .~\forall y.~ \exists z.~(z = \{x,y\}).
\]
\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}]
\yalabel{Axiom of Union}{(Union)}{ax:union} % Union (AoU)
\[
\forall x.~\exists y.~(y = \bigcup x).
\]
\end{axiom}
\begin{axiom}[\vocab{Power Set}]
\yalabel{Axiom of Power Set}{(Pow)}{ax:pow}
% (PWA)
We write $x = \cP(y)$
for
$\forall z.~(z \in x \iff z \subseteq x)$.
The power set axiom states
\[
\forall x.~\exists y.~y=\cP(x).
\]
\end{axiom}
\begin{axiom}[\vocab{Infinity}]
\yalabel{Axiom of Infinity}{(Inf)}{ax:inf}
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{axiomschema}[\vocab{Separation}]
\yalabel{Axiom of Separation}{(Aus)}{ax:aus}
Let $\phi$ be some fixed
fist order formula in $\cL_\in$
with free variables $x, v_1, \ldots, v_p$.
Let $b$ be a variable that is not free in $\phi$.
Then $\AxAus_{\phi}$
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 \AxAus can be formulated as
\[
\forall a.~\exists b.~(b = \{x \in a | \phi(x)\}).
\]
\end{axiomschema}
\begin{remark}
\AxAus 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{axiomschema}[\vocab{Replacement} (Fraenkel)]
\yalabel{Axiom of Replacement}{(Rep)}{ax:rep}
Let $\phi$ be some $\cL_{\in }$ formula
with free variables $x, y$.\todo{Allow more variables}
Then
\[
\forall x \in a.~\exists !y \phi(x,y) \implies \exists b.~\forall x \in a.~\exists y \in b.~\phi(x,y).
% \forall v_1 \ldots \forall v_p .~\forall x.~ \exists y'.~\forall y.~(y = y' \iff \phi(x))
% \implies \forall a.~\exists b.~\forall y.~(y \in b \iff \exists x.~(x \in a \land \phi(x))
\]
\end{axiomschema}
\begin{axiom}[\vocab{Choice}]
\yalabel{Axiom of Choice}{(C)}{ax:c}
Every family of pairwise disjoint non-empty sets has a \vocab{choice set}:
\begin{IEEEeqnarray*}{rCl}
\forall x .~&(&\\
&& ((\forall y \in x.~y \neq \emptyset) \land (\forall y \in x .~\forall y' \in x .~(y \neq y' \implies y \cap y' = \emptyset)))\\
&& \implies\exists z.~\forall y \in x.~\exists u.~(z \cap y = \{u\})\\
&)&
\end{IEEEeqnarray*}
\end{axiom}