w23-logic-2/inputs/lecture_04.tex

\lecture{04}{}{ZFC}

% Model-theoretic concepts and ultraproducts

\section{$\ZFC$}

% 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}