better ref for axioms
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@ -32,7 +32,7 @@ $\ZFC$ stands for
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\end{notation}
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\end{notation}
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$\ZFC$ consists of the following axioms:
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$\ZFC$ consists of the following axioms:
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\begin{axiom}[\vocab{Extensionality}]
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\begin{axiom}[\vocab{Extensionality}]
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\yalabel{Axiom of Extensionality}{(AoE)}{ax:ext}
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\yalabel{Axiom of Extensionality}{(Ext)}{ax:ext} % (AoE)
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\[
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\[
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\forall x.~\forall y.~(x = y \iff \forall z.~(z \in x \iff z \in y)).
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\forall x.~\forall y.~(x = y \iff \forall z.~(z \in x \iff z \in y)).
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\]
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\]
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@ -55,7 +55,7 @@ $\ZFC$ consists of the following axioms:
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\]
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\]
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\end{axiom}
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\end{axiom}
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\begin{axiom}[\vocab{Pairing}]
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\begin{axiom}[\vocab{Pairing}]
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\yalabel{Axiom of Pairing}{(Pair)}{ax:aop} % AoP
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\yalabel{Axiom of Pairing}{(Pair)}{ax:pair} % AoP
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\[
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\[
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\forall x .~\forall y.~ \exists z.~(z = \{x,y\}).
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\forall x .~\forall y.~ \exists z.~(z = \{x,y\}).
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\]
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\]
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@ -75,18 +75,19 @@ $\ZFC$ consists of the following axioms:
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\end{remark}
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\end{remark}
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\begin{axiom}[\vocab{Union}]
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\begin{axiom}[\vocab{Union}]
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\yalabel{Axiom of Union}{(AoU)}{ax:union} % Union
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\yalabel{Axiom of Union}{(Union)}{ax:union} % Union (AoU)
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\[
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\[
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\forall x.~\exists y.~(y = \bigcup x).
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\forall x.~\exists y.~(y = \bigcup x).
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\]
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\]
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\end{axiom}
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\end{axiom}
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\begin{axiom}[\vocab{Powerset}]
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\begin{axiom}[\vocab{Power Set}]
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\yalabel{Powerset Axiom}{(Pow)}{ax:pow}
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\yalabel{Axiom of Power Set}{(Pow)}{ax:pow}
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% (PWA)
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We write $x = \cP(y)$
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We write $x = \cP(y)$
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for
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for
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$\forall z.~(z \in x \iff x \subseteq z)$.
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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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The power set axiom states
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\[
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\[
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\forall x.~\exists y.~y=\cP(x).
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\forall x.~\exists y.~y=\cP(x).
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\]
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\]
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@ -99,11 +100,11 @@ $\ZFC$ consists of the following axioms:
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The axiom of infinity says that there exists and inductive set.
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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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\end{axiom}
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\begin{axiomschema}[\vocab{Separation}]
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\begin{axiomschema}[\vocab{Separation}]
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\yalabel{Axiom Schema of Separation}{(Aus)}{ax:aus}
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\yalabel{Axiom of Separation}{(Aus)}{ax:aus}
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% TODO :(Aus)
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Let $\phi$ be some fixed
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Let $\phi$ be some fixed
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fist order formula in $\cL_\in$.
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fist order formula in $\cL_\in$.
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Then $\text{(Aus)}_{\phi}$
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Then $\AxAus_{\phi}$
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states
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states
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\[
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\[
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\forall v_1 .~\forall v_p .~\forall a .~\exists b .~\forall x.~
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\forall v_1 .~\forall v_p .~\forall a .~\exists b .~\forall x.~
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@ -112,7 +113,7 @@ $\ZFC$ consists of the following axioms:
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Let us write $b = \{x \in a | \phi(x)\}$
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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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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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Then \AxAus can be formulated as
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\[
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\[
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\forall a.~\exists b.~(b = \{x \in a; \phi(x)\}).
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\forall a.~\exists b.~(b = \{x \in a; \phi(x)\}).
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\]
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\]
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@ -125,7 +126,7 @@ $\ZFC$ consists of the following axioms:
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% $x = \bigcap y$ for ...
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% $x = \bigcap y$ for ...
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\end{notation}
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\end{notation}
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\begin{remark}
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\begin{remark}
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(Aus) proves that
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\AxAus proves that
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\begin{itemize}
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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 $\forall a.~\forall b.~\exists c.~(c = a \cap b)$,
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\item $\forall a.~\forall b.~\exists c.~(c = a \setminus b)$,
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\item $\forall a.~\forall b.~\exists c.~(c = a \setminus b)$,
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@ -133,6 +134,7 @@ $\ZFC$ consists of the following axioms:
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\end{itemize}
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\end{itemize}
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\end{remark}
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\end{remark}
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\begin{axiomschema}[\vocab{Replacement} (Fraenkel)]
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\begin{axiomschema}[\vocab{Replacement} (Fraenkel)]
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\yalabel{Axiom of Replacement}{(Rep)}{ax:rep}
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Let $\phi$ be some $\cL_{\in }$ formula.
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Let $\phi$ be some $\cL_{\in }$ formula.
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Then
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Then
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\[
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\[
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@ -141,6 +143,7 @@ $\ZFC$ consists of the following axioms:
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\end{axiomschema}
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\end{axiomschema}
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\begin{axiom}[\vocab{Choice}]
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\begin{axiom}[\vocab{Choice}]
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\yalabel{Axiom of Choice}{(C)}{ax:c}
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Every family of non-empty sets has a \vocab{choice set}:
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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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\begin{IEEEeqnarray*}{rCl}
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\forall x .~&(&\\
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\forall x .~&(&\\
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@ -2,24 +2,24 @@
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\begin{definition}
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\begin{definition}
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Zermelo:
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Zermelo:
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\[\Zermelo \coloneqq \AoE + \AoF + \AoP + \AoU + \Pow + \AoI + \Aus_{\phi}\]
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\[\Zermelo \coloneqq \AxExt + \AxFund + \AxPair + \AxUnion + \AxPow + \AxInf + \AxAus_{\phi}\]
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Zermelo and Fraenkl:
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Zermelo and Fraenkl:
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\[
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\[
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{\ZF} \coloneqq Z + (\Rep_{\phi})
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{\ZF} \coloneqq Z + \AxRep_{\phi} % TODO fix parenthesis
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\]
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\]
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\[
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\[
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{\ZFC} \coloneqq \ZF + \Choice
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{\ZFC} \coloneqq \ZF + \AxC
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\]
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\]
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Variants:
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Variants:
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\[
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\[
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{\ZFC^{-}} \coloneqq \ZFC \setminus \Pow.
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{\ZFC^{-}} \coloneqq \ZFC \setminus \AxPow.
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\]
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\]
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\[
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\[
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{\ZFC^{-\infty}} \coloneqq \ZFC \setminus \Inf
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{\ZFC^{-\infty}} \coloneqq \ZFC \setminus \AxInf
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\]
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\]
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\end{definition}
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\end{definition}
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@ -50,7 +50,7 @@
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$a \times b$ exists.
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$a \times b$ exists.
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\end{fact}
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\end{fact}
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\begin{proof}
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\begin{proof}
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Use $\Aus$ over $\cP(\cP(a \cup b))$.
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Use \AxAus over $\cP(\cP(a \cup b))$.
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\end{proof}
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\end{proof}
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\begin{definition}
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\begin{definition}
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@ -112,7 +112,7 @@
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${}^d b$ exists.
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${}^d b$ exists.
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\end{fact}
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\end{fact}
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\begin{proof}
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\begin{proof}
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Apply again $\Aus$ over $\cP(d \times b)$.
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Apply again \AxAus over $\cP(d \times b)$.
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\end{proof}
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\end{proof}
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\begin{definition}
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\begin{definition}
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@ -7,7 +7,7 @@
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and $ b$ linearly ordered, $b$ has an upper bound,
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and $ b$ linearly ordered, $b$ has an upper bound,
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Then $a$ has a maximal element.
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Then $a$ has a maximal element.
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\end{theorem}
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\end{theorem}
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\begin{proof}
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\begin{refproof}{thm:zorn}
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Fix $(a, \le )$ as in the hypothesis.
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Fix $(a, \le )$ as in the hypothesis.
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Let $A \coloneqq \{ \{(b,x) : x in b\} : b \le a, b \neq \emptyset\}$.
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Let $A \coloneqq \{ \{(b,x) : x in b\} : b \le a, b \neq \emptyset\}$.
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Note that $A$ is a set (use separation on $\cP(\cP(a) \times \bigcup \cP(a))$).
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Note that $A$ is a set (use separation on $\cP(\cP(a) \times \bigcup \cP(a))$).
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@ -65,7 +65,7 @@
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Then $B = B_{u_0}^{\le^{\ast\ast}}$.
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Then $B = B_{u_0}^{\le^{\ast\ast}}$.
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So $\le^{\ast\ast} \in W$, but now $n_0 \in b$.
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So $\le^{\ast\ast} \in W$, but now $n_0 \in b$.
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So $b$ must have a maximum.
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So $b$ must have a maximum.
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\end{proof}
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\end{refproof}
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\begin{remark}
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\begin{remark}
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Over $\ZF$ the axiom of choice and \yaref{thm:zorn}
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Over $\ZF$ the axiom of choice and \yaref{thm:zorn}
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@ -107,20 +107,20 @@
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of well-orders.
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of well-orders.
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% TODO theorem
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% TODO theorem
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\end{goal}
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\end{goal}
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Recall that (AoI) states the existence of an inductive set $x$.
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Recall that \AxInf states the existence of an inductive set $x$.
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We can hence form the smallest inductive set
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We can hence form the smallest inductive set
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\[
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\[
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\omega \coloneqq \bigcap \{ x : x \text{ is inductive}\}
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\omega \coloneqq \bigcap \{ x : x \text{ is inductive}\}
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\]
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\]
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Note that $\omega$ exists, as it is a subset of the inductive
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Note that $\omega$ exists, as it is a subset of the inductive
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set given by AoI.
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set given by \AxInf.
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We call $\omega$ the set of \vocab{natural numbers}.
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We call $\omega$ the set of \vocab{natural numbers}.
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\begin{notation}
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\begin{notation}
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We write $0$ for $\emptyset$,
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We write $0$ for $\emptyset$,
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and $y + 1$ for $y \cup \{y\}$.
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and $y + 1$ for $y \cup \{y\}$.
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\end{notation}
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\end{notation}
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With this notation the AoI is equivalent to
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With this notation the \AxInf is equivalent to
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\[
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\[
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\exists x_0.~(0 \in x_0 \land \forall n. ~(n \in x_0 \implies n+1 \in x_0)).
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\exists x_0.~(0 \in x_0 \land \forall n. ~(n \in x_0 \implies n+1 \in x_0)).
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\]
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\]
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@ -218,7 +218,7 @@ Clearly, the $\in$-relation is a well-order on an ordinal $x$.
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\item In the first case, $z+1 = y+1$.
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\item In the first case, $z+1 = y+1$.
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\item Suppose that $z \in y$.
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\item Suppose that $z \in y$.
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Then by the induction hypothesis $\phi(y, z+1)$ holds.
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Then by the induction hypothesis $\phi(y, z+1)$ holds.
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If $y \in z+1$, then $\{y,z\}$ would violate AoF.
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If $y \in z+1$, then $\{y,z\}$ would violate \AxFund.
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If $y = z+1$, then $z + 1 \in y + 1$.
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If $y = z+1$, then $z + 1 \in y + 1$.
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If $z+1 \in y$, then $z+1 \in y+1$ as well.
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If $z+1 \in y$, then $z+1 \in y+1$ as well.
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\end{itemize}
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\end{itemize}
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@ -14,6 +14,7 @@
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for ordinals.
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for ordinals.
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\end{notation}
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\end{notation}
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\begin{lemma}
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\begin{lemma}
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\label{lem:7:ordinalfacts}
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\begin{enumerate}[(a)]
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\begin{enumerate}[(a)]
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\item $0$ is an ordinal, and if $\alpha$ is
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\item $0$ is an ordinal, and if $\alpha$ is
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an ordinal, so is $\alpha + 1$.
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an ordinal, so is $\alpha + 1$.
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@ -27,7 +28,7 @@
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or $\alpha \ni \beta$.
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or $\alpha \ni \beta$.
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\end{enumerate}
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\end{enumerate}
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\end{lemma}
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\end{lemma}
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\begin{proof}
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\begin{refproof}{lem:7:ordinalfacts}
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We have already proved (a) before.
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We have already proved (a) before.
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(b) Fix $x \in \alpha$. Then $x \subseteq \alpha$.
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(b) Fix $x \in \alpha$. Then $x \subseteq \alpha$.
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@ -41,12 +42,12 @@
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As $\alpha$ is transitive, we have that $z, y, x \in \alpha$.
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As $\alpha$ is transitive, we have that $z, y, x \in \alpha$.
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Thus $z \in x \lor z = x \lor z \ni x$.
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Thus $z \in x \lor z = x \lor z \ni x$.
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$z = x$ contradicts Fund:
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$z = x$ contradicts \AxFund:
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Consider $\{x,y\}$. Then $x \cap \{x,y\}$ is non empty,
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Consider $\{x,y\}$. Then $x \cap \{x,y\}$ is non empty,
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as it contains $y$.
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as it contains $y$.
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Furthermore $x \in y \cap \{x,y\} $
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Furthermore $x \in y \cap \{x,y\} $
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$z \ni x$ also contradicts Fund:
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$z \ni x$ also contradicts \AxFund:
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If $x \in z$, then $z \ni x \ni y \ni z \ni x \ni \ldots$.
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If $x \in z$, then $z \ni x \ni y \ni z \ni x \ni \ldots$.
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$\{x,y,z\}$ yields a contradiction,
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$\{x,y,z\}$ yields a contradiction,
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as $y \in x \cap \{x,y,z\}$, $z \in y \cap \{x,y,z\}$,
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as $y \in x \cap \{x,y,z\}$, $z \in y \cap \{x,y,z\}$,
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(c) Say $\alpha \subsetneq \beta$.
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(c) Say $\alpha \subsetneq \beta$.
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Pick $\xi \in \beta \setminus \alpha$
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Pick $\xi \in \beta \setminus \alpha$
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such that $\eta \in \alpha$ for every $\eta \in\xi \cap \beta$.
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such that $\eta \in \alpha$ for every $\eta \in\xi \cap \beta$.
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(This exists by Fund).
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(This exists by \AxFund).
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We want to see that $\xi = \alpha$.
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We want to see that $\xi = \alpha$.
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We have $\xi \subseteq \alpha$ by the choice of $\xi$.
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We have $\xi \subseteq \alpha$ by the choice of $\xi$.
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On the other hand $\alpha \subseteq \xi$:
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On the other hand $\alpha \subseteq \xi$:
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If that is not the case,
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If that is not the case,
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then $\alpha_0 \in \alpha_0 \cup \beta_0$
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then $\alpha_0 \in \alpha_0 \cup \beta_0$
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and $\beta_0 \in \alpha_0 \cup \beta_0$.
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and $\beta_0 \in \alpha_0 \cup \beta_0$.
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$\alpha_0 \in \alpha_0$ violates Fund.
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$\alpha_0 \in \alpha_0$ violates \AxFund.
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Hence $\alpha_0 \in \beta_0$.
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Hence $\alpha_0 \in \beta_0$.
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By the same argument, $\beta_0 \in \alpha_0$.
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By the same argument, $\beta_0 \in \alpha_0$.
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But this violates Fund,
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But this violates \AxFund,
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as $\alpha_0 \in \beta_0 \in \alpha_0$.
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as $\alpha_0 \in \beta_0 \in \alpha_0$.
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\end{subproof}
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\end{subproof}
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\end{proof}
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\end{refproof}
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\begin{lemma}
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\begin{lemma}
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Let $X$ be a set of ordinals,
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Let $X$ be a set of ordinals,
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It is actually the case that $\bigcap X \in X$:
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It is actually the case that $\bigcap X \in X$:
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Pick $\alpha \in X$ such that $\alpha \subseteq \beta$
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Pick $\alpha \in X$ such that $\alpha \subseteq \beta$
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for all $\beta \in X$. This exists
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for all $\beta \in X$. This exists
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by Fund and since all ordinals are comparable.
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by \AxFund and since all ordinals are comparable.
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Then $\alpha = \bigcap X$.
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Then $\alpha = \bigcap X$.
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\begin{notation}
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\begin{notation}
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@ -41,8 +41,8 @@ has the following axioms:
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\forall x .~\exists y .~ y = \bigcup x.
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\forall x .~\exists y .~ y = \bigcup x.
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\]
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\]
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\end{axiom}
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\end{axiom}
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\begin{axiom}[Power]
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\begin{axiom}[Power Set]
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\yalabel{Powerset Axiom}{(Pow)}{ax:bg:pow}
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\yalabel{Power Set Axiom}{(Pow)}{ax:bg:pow}
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\[
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\[
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\forall x .~\exists y .~ y = \cP(x).
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\forall x .~\exists y .~ y = \cP(x).
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\]
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\]
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@ -128,7 +128,7 @@ Furthermore $F\,''a \coloneqq \{y : \exists x \in a .~(x,y) \in F\}$.
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\item $(\N, <)$ is well-founded.
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\item $(\N, <)$ is well-founded.
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\item Let $M$ be a set,
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\item Let $M$ be a set,
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and let $\in\defon{M} \coloneqq \{(x,y) : x,y \in M \land x \in y\}$.
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and let $\in\defon{M} \coloneqq \{(x,y) : x,y \in M \land x \in y\}$.
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Fund is equivalent to saying that
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\AxFund is equivalent to saying that
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this is a well-founded relation for every $M$.
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this is a well-founded relation for every $M$.
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\end{enumerate}
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\end{enumerate}
|
||||||
\end{example}
|
\end{example}
|
||||||
|
@ -136,10 +136,10 @@ Furthermore $F\,''a \coloneqq \{y : \exists x \in a .~(x,y) \in F\}$.
|
||||||
|
|
||||||
\begin{lemma}
|
\begin{lemma}
|
||||||
\label{lem:fundseq}
|
\label{lem:fundseq}
|
||||||
In $\ZFC - \Fund$,
|
In $\ZFC - \AxFund$,
|
||||||
the following are equivalent:
|
the following are equivalent:
|
||||||
\begin{itemize}
|
\begin{itemize}
|
||||||
\item \Fund,
|
\item \AxFund,
|
||||||
\item There is no sequence $\langle x_n : n < \omega \rangle$
|
\item There is no sequence $\langle x_n : n < \omega \rangle$
|
||||||
such that $x_{n+1} \in x_n$ for all $n < \omega$.
|
such that $x_{n+1} \in x_n$ for all $n < \omega$.
|
||||||
\end{itemize}
|
\end{itemize}
|
||||||
|
@ -149,12 +149,12 @@ Furthermore $F\,''a \coloneqq \{y : \exists x \in a .~(x,y) \in F\}$.
|
||||||
Then $\{x_n : n < \omega\}$
|
Then $\{x_n : n < \omega\}$
|
||||||
(this exists as by definition sequence of the $x_n$ is a function
|
(this exists as by definition sequence of the $x_n$ is a function
|
||||||
and this set is the range of that function)
|
and this set is the range of that function)
|
||||||
violates \Fund.
|
violates \AxFund.
|
||||||
|
|
||||||
For the other direction let $M \neq \emptyset$ be some set.
|
For the other direction let $M \neq \emptyset$ be some set.
|
||||||
Suppose that \Fund does not hold for $M$.
|
Suppose that \AxFund does not hold for $M$.
|
||||||
|
|
||||||
Using \Choice,
|
Using \AxC,
|
||||||
we construct an infinite sequence $x_0 \ni x_1 \ni x_2 \ni \ldots$
|
we construct an infinite sequence $x_0 \ni x_1 \ni x_2 \ni \ldots$
|
||||||
of elements of $M$.
|
of elements of $M$.
|
||||||
|
|
||||||
|
@ -186,7 +186,7 @@ Furthermore $F\,''a \coloneqq \{y : \exists x \in a .~(x,y) \in F\}$.
|
||||||
&&~ ~\overline{g} \text{ is a function with domain $n$ and range $\subseteq M$, such that}\\
|
&&~ ~\overline{g} \text{ is a function with domain $n$ and range $\subseteq M$, such that}\\
|
||||||
&&~ ~\overline{g}(0) = x \land \forall m \in \omega.~(m+1 \in \dom(\overline{g}) \implies \overline{g}(m+1) = f(\overline{g}(m)))\}.
|
&&~ ~\overline{g}(0) = x \land \forall m \in \omega.~(m+1 \in \dom(\overline{g}) \implies \overline{g}(m+1) = f(\overline{g}(m)))\}.
|
||||||
\end{IEEEeqnarray*}
|
\end{IEEEeqnarray*}
|
||||||
$G$ exists as it can be obtained by \AxSep
|
$G$ exists as it can be obtained by \AxAus
|
||||||
from ${}^{< \omega}M$.
|
from ${}^{< \omega}M$.
|
||||||
By induction,
|
By induction,
|
||||||
for every $n \in \omega$,
|
for every $n \in \omega$,
|
||||||
|
@ -224,7 +224,7 @@ Furthermore $F\,''a \coloneqq \{y : \exists x \in a .~(x,y) \in F\}$.
|
||||||
\yaref{lem:fundseq}.
|
\yaref{lem:fundseq}.
|
||||||
\end{proof}
|
\end{proof}
|
||||||
\begin{remark}
|
\begin{remark}
|
||||||
In $\ZF$ this is a weaker form of \Choice.
|
In $\ZF$ this is a weaker form of \AxC.
|
||||||
\end{remark}
|
\end{remark}
|
||||||
|
|
||||||
The construction of $g$ in the previous proof was a special case of
|
The construction of $g$ in the previous proof was a special case of
|
||||||
|
|
28
logic.sty
28
logic.sty
|
@ -114,27 +114,15 @@
|
||||||
\DeclareSimpleMathOperator{OD}
|
\DeclareSimpleMathOperator{OD}
|
||||||
\DeclareSimpleMathOperator{AC}
|
\DeclareSimpleMathOperator{AC}
|
||||||
\newcommand{\AxC}{\yarefs{ax:c}}
|
\newcommand{\AxC}{\yarefs{ax:c}}
|
||||||
\newcommand{\AxSep}{\yarefs{ax:sep}} % Separation
|
\newcommand{\AxExt}{\yarefs{ax:ext}} % AoE
|
||||||
\newcommand{\Choice}{\yarefs{ax:c}}
|
\newcommand{\AxFund}{\yarefs{ax:fund}} % AoF
|
||||||
% \DeclareSimpleMathOperator{Choice}
|
\newcommand{\AxPair}{\yarefs{ax:pair}} % AoP
|
||||||
% \DeclareSimpleMathOperator{Fund}
|
\newcommand{\AxUnion}{\yarefs{ax:union}} % AoU
|
||||||
\newcommand{\Fund}{\yarefs{ax:fund}}
|
\newcommand{\AxPow}{\yarefs{ax:pow}}
|
||||||
\DeclareSimpleMathOperator{Pair}
|
\newcommand{\AxRep}{\yarefs{ax:rep}}
|
||||||
\DeclareSimpleMathOperator{Union}
|
\newcommand{\AxInf}{\yarefs{ax:inf}} % AoI
|
||||||
\DeclareSimpleMathOperator{Rep}
|
\newcommand{\AxAus}{\yarefs{ax:aus}} % Separation
|
||||||
|
|
||||||
\DeclareSimpleMathOperator{Pow}
|
|
||||||
\DeclareSimpleMathOperator{AoE}
|
|
||||||
\DeclareSimpleMathOperator{AoF}
|
|
||||||
\DeclareSimpleMathOperator{AoP}
|
|
||||||
\DeclareSimpleMathOperator{AoU}
|
|
||||||
\DeclareSimpleMathOperator{AoI}
|
|
||||||
\DeclareSimpleMathOperator{Inf}
|
|
||||||
|
|
||||||
|
|
||||||
\renewcommand{\Aus}{\text{Aus}}
|
|
||||||
% \DeclareSimpleMathOperator{Aus}
|
|
||||||
\DeclareSimpleMathOperator{Infinity}
|
|
||||||
|
|
||||||
\DeclareSimpleMathOperator{CH}
|
\DeclareSimpleMathOperator{CH}
|
||||||
\DeclareSimpleMathOperator{DC}
|
\DeclareSimpleMathOperator{DC}
|
||||||
|
|
1
todo
1
todo
|
@ -1 +0,0 @@
|
||||||
Better REF for axioms
|
|
Loading…
Reference in a new issue