151 lines
4.5 KiB
TeX
151 lines
4.5 KiB
TeX
\lecture{04}{}{ZFC}
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% Model-theoretic concepts and ultraproducts
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\section{$\ZFC$}
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% 1900, Russel's paradox
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\todo{Russel's Paradox}
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$\ZFC$ stands for
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\begin{itemize}
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\item \textsc{Zermelo}’s axioms (1905), % crises around 19000
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\item \vocab{Fraenkel}'s axioms,
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\item the axiom of choice.
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\end{itemize}
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\begin{notation}
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We write $x \subseteq y$ as a shorthand
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for $\forall z.~(z \in x \implies z \in y)$.
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We write $x = \emptyset$ for $\lnot \exists y . y \in x$
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and $x \cap y = \emptyset$ for $\lnot \exists z . ~(z \in x \land z \in y)$.
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We use $x = \{y,z\}$
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for
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\[
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y \in x \land z \in x \land \forall a .~(a \in x \implies a = y \lor a = z).
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\]
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Let $x = \bigcup y$ denote
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\[
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\forall z.~(z \in x \iff \exists v.(v \in y \land z \in v)).
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\]
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\end{notation}
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$\ZFC$ consists of the following axioms:
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\begin{axiom}[\vocab{Extensionality}]
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\yalabel{Axiom of Extensionality}{(AoE)}{ax:ext}
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\[
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\forall x.~\forall y.~(x = y \iff \forall z.~(z \in x \iff z \in y)).
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\]
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Equivalent statements using $\subseteq$:
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\[
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\forall x.~\forall y.~(x = y \iff (x \subseteq y \land y \subseteq x)).
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\]
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\end{axiom}
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\begin{axiom}[\vocab{Foundation}]
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\yalabel{Axiom of Foundation}{(Fund)}{ax:fund}
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Every set has an $\in$-minimal member:
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\[
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\forall x .~ \left(\exists a .~(a\in x) \implies
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\exists y .~ y \in x \land \lnot \exists z.~(z \in y \land z \in x)\right).
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\]
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Shorter:
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\[
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\forall x.~(x \neq \emptyset \implies \exists y \in x .~ x \cap y = \emptyset).
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\]
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\end{axiom}
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\begin{axiom}[\vocab{Pairing}]
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\yalabel{Axiom of Pairing}{(Pair)}{ax:aop} % AoP
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\[
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\forall x .~\forall y.~ \exists z.~(z = \{x,y\}).
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\]
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\end{axiom}
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\begin{remark}
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Together with the axiom of pairing,
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the axiom of foundation implies
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that there can not be a set $x$ such that
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$x \in x$:
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Suppose that $x \in x$.
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Then $x$ is the only element of $\{x\}$,
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but $x \cap \{x\} \neq \emptyset$.
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A similar argument shows that chains like
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$x_0 \in x_1 \in x_2 \in x_0$
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are ruled out as well.
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\end{remark}
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\begin{axiom}[\vocab{Union}]
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\yalabel{Axiom of Union}{(AoU)}{ax:union} % Union
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\[
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\forall x.~\exists y.~(y = \bigcup x).
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\]
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\end{axiom}
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\begin{axiom}[\vocab{Powerset}]
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\yalabel{Powerset Axiom}{(Pow)}{ax:pow}
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We write $x = \cP(y)$
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for
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$\forall z.~(z \in x \iff x \subseteq z)$.
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The powerset axiom (PWA) states
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\[
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\forall x.~\exists y.~y=\cP(x).
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\]
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\end{axiom}
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\begin{axiom}[\vocab{Infinity}]
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\yalabel{Axiom of Infinity}{(Inf)}{ax:inf}
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A set $x$ is called \vocab{inductive},
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iff $\emptyset \in x \land \forall y.~(y \in x \implies y \cup \{y\} \in x)$.
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The axiom of infinity says that there exists and inductive set.
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\end{axiom}
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\begin{axiomschema}[\vocab{Separation}]
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\yalabel{Axiom Schema of Separation}{(Aus)}{ax:aus}
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% TODO :(Aus)
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Let $\phi$ be some fixed
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fist order formula in $\cL_\in$.
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Then $\text{(Aus)}_{\phi}$
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states
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\[
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\forall v_1 .~\forall v_p .~\forall a .~\exists b .~\forall x.~
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(x \in b \implies x \in a \land \phi(x,v_1,v_p))
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\]
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Let us write $b = \{x \in a | \phi(x)\}$
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for $\forall x.~(x \in b \iff x \in a \land f(x))$.
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Then (Aus) can be formulated as
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\[
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\forall a.~\exists b.~(b = \{x \in a; \phi(x)\}).
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\]
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\end{axiomschema}
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\begin{notation}
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\todo{$\cap, \setminus, \bigcap$}
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% We write $z = x \cap y$ for $\forall u.~((u \in z) \implies u \in x \land u \in y)$,
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% $Z = x \setminus y$ for ...
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% $x = \bigcap y$ for ...
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\end{notation}
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\begin{remark}
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(Aus) proves that
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\begin{itemize}
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\item $\forall a.~\forall b.~\exists c.~(c = a \cap b)$,
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\item $\forall a.~\forall b.~\exists c.~(c = a \setminus b)$,
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\item $\forall a.~\exists b.~(b = \bigcap a)$.
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\end{itemize}
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\end{remark}
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\begin{axiomschema}[\vocab{Replacement} (Fraenkel)]
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Let $\phi$ be some $\cL_{\in }$ formula.
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Then
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\[
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\forall v_1 .~\exists b.~\forall y.~(y \in b \iff \exists x .~(x \in a \land \phi(x,y,v_1,v_p))).
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\]
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\end{axiomschema}
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\begin{axiom}[\vocab{Choice}]
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Every family of non-empty sets has a \vocab{choice set}:
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\begin{IEEEeqnarray*}{rCl}
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\forall x .~&(&\\
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&& ((\forall y \in x.~y \neq \emptyset) \land (\forall y \in x .~\forall y' \in x .~(y \neq y' \implies y \cap y' = \emptyset))\\
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&& \implies\exists z.~\forall y \in x.~\exists u.~(z \cap y = \{u\})\\
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&)&
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\end{IEEEeqnarray*}
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\end{axiom}
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