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4 changed files with 28 additions and 26 deletions
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@ -41,7 +41,8 @@ all condensation points are accumulation points.
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Fix $A \subseteq \R$ closed.
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We want to see that $A$ is at most countable
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or there is some perfect $P \subseteq A$.
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Let $P \coloneqq \{x \in \R | x \text{ is a condensation point of $A$}\}$.
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Let
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\[P \coloneqq \{x \in \R | x \text{ is a condensation point of $A$}\}.\]
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Since $A $ is closed, $P \subseteq A$.
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\begin{claim}
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@ -9,7 +9,7 @@
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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 \textsc{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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@ -18,74 +18,74 @@ has the following axioms:
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\begin{axiom}[Extensionality]
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\yalabel{Axiom of Extensionality}{(Ext)}{ax:bg:ext}
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\[
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\forall x, y. \left( x = y \iff \left( \forall z.~(z \in x \iff z \in y \right) \right).
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\]
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\forall x.~\forall y.~ \left( x = y \iff \left( \forall z.~(z \in x \iff z \in y \right) \right).
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\]
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\end{axiom}
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\begin{axiom}[Foundation]
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\yalabel{Axiom of Foundation}{(Fund)}{ax:bg:fund}
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\[
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\forall x .(x \neq \emptyset \implies \exists y \in x . y \cap x = \emptyset).
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\]
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\forall x .~(x \neq \emptyset \implies \exists y \in x .~ y \cap x = \emptyset).
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\]
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\end{axiom}
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\begin{axiom}[Pairing]
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\yalabel{Axiom of Pairing}{(Pair)}{ax:bg:pair}
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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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\forall x.~\forall y . ~\exists z .~ z = \{x,y\}.
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\]
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\end{axiom}
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\begin{axiom}[Union]
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\yalabel{Axiom of Union}{(Union)}{ax:bg:union}
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\[
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\forall x \exists y .~ y = \bigcup x.
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\]
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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}[Power]
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\yalabel{Powerset Axiom}{(Pow)}{ax:bg:pow}
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\[
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\forall x \exists y .~ y = \cP(x).
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\]
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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}[Infinity]
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\yalabel{Axiom of Infinity}{(Infinity)}{ax:bg:inf}
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\[
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\exists x . ~(\emptyset \in x \land \left( \forall y \in x .~y \cup \{y\} \in x \right)).
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\]
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\exists x .~(\emptyset \in x \land \left( \forall y \in x .~y \cup \{y\} \in x \right)).
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\]
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\end{axiom}
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Together with the following axioms for classes:
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\begin{axiom}[Extensionality for classes]
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\[
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\forall X \forall Y \left( \forall x(x \in X \iff y \in X) \implies X = Y).
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\]
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\forall X .~\forall Y.~ \left( \forall x.~(x \in X \iff x \in Y) \implies X = Y\right).
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\]
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\end{axiom}
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\begin{axiom}
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Every set is a class:
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\[
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\forall x \exists X. x = X.
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\]
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\forall x.~ \exists X.~ x = X.
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\]
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\end{axiom}
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\begin{axiom}
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Every element of a class is a set:
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\[
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\forall X \exists Y.~(X \in Y \to \exists x.~x = X).
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\]
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\forall X .~\exists Y.~(X \in Y \to \exists x.~x = X).
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\]
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\end{axiom}
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\begin{axiom}[Replacement]
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\yalabel{Axiom of Replacement}{(Rep)}{ax:bg:rep}
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If $F$ is a function and inf $a $ is a set,
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then $F"a$ is a set.
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If $F$ is a function and $a$ is a set,
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then $F\,''a$ is a set.
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\end{axiom}
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Here a \vocab[Class function]{(class) function} is a class
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consisting of pairs $(x,y)$,
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such that for every $x$ there is at most one $y$
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with $(x,y) \in F$.
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Furthermore $F"a \coloneqq \{y : \exists x \in a .~(x,y) \in F\}$.
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Furthermore $F\,''a \coloneqq \{y : \exists x \in a .~(x,y) \in F\}$.
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\begin{remark}
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Note that we didn't need to use an axiom schema,
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@ -94,10 +94,9 @@ Furthermore $F"a \coloneqq \{y : \exists x \in a .~(x,y) \in F\}$.
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\begin{axiom}[Comprehension]
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\yalabel{Axiom of Comprehension}{(Comp)}{ax:bg:comp}
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\[
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\forall X_1 \ldots \forall X_k .~\exists Y ( \forall x .~x \in Y \iff \phi(x,X_1,\ldots, X_k))
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\]
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\forall X_1 .~\ldots \forall X_k .~\exists Y.~ ( \forall x .~x \in Y \iff \phi(x,X_1,\ldots, X_k))
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\]
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where $\phi(x, X_1, \ldots, X_k)$
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is a formula which contains exactly $X_1, \ldots, X_k, x$
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as free variables,
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@ -113,6 +113,8 @@
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\DeclareSimpleMathOperator{HOD}
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\DeclareSimpleMathOperator{OD}
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\DeclareSimpleMathOperator{AC}
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\newcommand{\AxC}{\yarefs{ax:c}}
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\newcommand{\AxSep}{\yarefs{ax:sep}} % Separation
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\newcommand{\Choice}{\yarefs{ax:c}}
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% \DeclareSimpleMathOperator{Choice}
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% \DeclareSimpleMathOperator{Fund}
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