some fixes
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This commit is contained in:
Josia Pietsch 2023-11-13 19:42:48 +01:00
parent 883ee88516
commit 0771440511
Signed by: josia
GPG key ID: E70B571D66986A2D
4 changed files with 28 additions and 26 deletions

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@ -41,7 +41,8 @@ all condensation points are accumulation points.
Fix $A \subseteq \R$ closed. Fix $A \subseteq \R$ closed.
We want to see that $A$ is at most countable We want to see that $A$ is at most countable
or there is some perfect $P \subseteq A$. or there is some perfect $P \subseteq A$.
Let $P \coloneqq \{x \in \R | x \text{ is a condensation point of $A$}\}$. Let
\[P \coloneqq \{x \in \R | x \text{ is a condensation point of $A$}\}.\]
Since $A $ is closed, $P \subseteq A$. Since $A $ is closed, $P \subseteq A$.
\begin{claim} \begin{claim}

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@ -9,7 +9,7 @@
$\ZFC$ stands for $\ZFC$ stands for
\begin{itemize} \begin{itemize}
\item \textsc{Zermelo}s axioms (1905), % crises around 19000 \item \textsc{Zermelo}s axioms (1905), % crises around 19000
\item \vocab{Fraenkel}'s axioms, \item \textsc{Fraenkel}'s axioms,
\item the axiom of choice. \item the axiom of choice.
\end{itemize} \end{itemize}
\begin{notation} \begin{notation}

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@ -18,74 +18,74 @@ has the following axioms:
\begin{axiom}[Extensionality] \begin{axiom}[Extensionality]
\yalabel{Axiom of Extensionality}{(Ext)}{ax:bg:ext} \yalabel{Axiom of Extensionality}{(Ext)}{ax:bg:ext}
\[ \[
\forall x, y. \left( x = y \iff \left( \forall z.~(z \in x \iff z \in y \right) \right). \forall x.~\forall y.~ \left( x = y \iff \left( \forall z.~(z \in x \iff z \in y \right) \right).
\] \]
\end{axiom} \end{axiom}
\begin{axiom}[Foundation] \begin{axiom}[Foundation]
\yalabel{Axiom of Foundation}{(Fund)}{ax:bg:fund} \yalabel{Axiom of Foundation}{(Fund)}{ax:bg:fund}
\[ \[
\forall x .(x \neq \emptyset \implies \exists y \in x . y \cap x = \emptyset). \forall x .~(x \neq \emptyset \implies \exists y \in x .~ y \cap x = \emptyset).
\] \]
\end{axiom} \end{axiom}
\begin{axiom}[Pairing] \begin{axiom}[Pairing]
\yalabel{Axiom of Pairing}{(Pair)}{ax:bg:pair} \yalabel{Axiom of Pairing}{(Pair)}{ax:bg:pair}
\[ \[
\forall x \forall y \exists z . z = \{x,y\}. \forall x.~\forall y . ~\exists z .~ z = \{x,y\}.
\] \]
\end{axiom} \end{axiom}
\begin{axiom}[Union] \begin{axiom}[Union]
\yalabel{Axiom of Union}{(Union)}{ax:bg:union} \yalabel{Axiom of Union}{(Union)}{ax:bg:union}
\[ \[
\forall x \exists y .~ y = \bigcup x. \forall x .~\exists y .~ y = \bigcup x.
\] \]
\end{axiom} \end{axiom}
\begin{axiom}[Power] \begin{axiom}[Power]
\yalabel{Powerset Axiom}{(Pow)}{ax:bg:pow} \yalabel{Powerset Axiom}{(Pow)}{ax:bg:pow}
\[ \[
\forall x \exists y .~ y = \cP(x). \forall x .~\exists y .~ y = \cP(x).
\] \]
\end{axiom} \end{axiom}
\begin{axiom}[Infinity] \begin{axiom}[Infinity]
\yalabel{Axiom of Infinity}{(Infinity)}{ax:bg:inf} \yalabel{Axiom of Infinity}{(Infinity)}{ax:bg:inf}
\[ \[
\exists x . ~(\emptyset \in x \land \left( \forall y \in x .~y \cup \{y\} \in x \right)). \exists x .~(\emptyset \in x \land \left( \forall y \in x .~y \cup \{y\} \in x \right)).
\] \]
\end{axiom} \end{axiom}
Together with the following axioms for classes: Together with the following axioms for classes:
\begin{axiom}[Extensionality for classes] \begin{axiom}[Extensionality for classes]
\[ \[
\forall X \forall Y \left( \forall x(x \in X \iff y \in X) \implies X = Y). \forall X .~\forall Y.~ \left( \forall x.~(x \in X \iff x \in Y) \implies X = Y\right).
\] \]
\end{axiom} \end{axiom}
\begin{axiom} \begin{axiom}
Every set is a class: Every set is a class:
\[ \[
\forall x \exists X. x = X. \forall x.~ \exists X.~ x = X.
\] \]
\end{axiom} \end{axiom}
\begin{axiom} \begin{axiom}
Every element of a class is a set: Every element of a class is a set:
\[ \[
\forall X \exists Y.~(X \in Y \to \exists x.~x = X). \forall X .~\exists Y.~(X \in Y \to \exists x.~x = X).
\] \]
\end{axiom} \end{axiom}
\begin{axiom}[Replacement] \begin{axiom}[Replacement]
\yalabel{Axiom of Replacement}{(Rep)}{ax:bg:rep} \yalabel{Axiom of Replacement}{(Rep)}{ax:bg:rep}
If $F$ is a function and inf $a $ is a set, If $F$ is a function and $a$ is a set,
then $F"a$ is a set. then $F\,''a$ is a set.
\end{axiom} \end{axiom}
Here a \vocab[Class function]{(class) function} is a class Here a \vocab[Class function]{(class) function} is a class
consisting of pairs $(x,y)$, consisting of pairs $(x,y)$,
such that for every $x$ there is at most one $y$ such that for every $x$ there is at most one $y$
with $(x,y) \in F$. with $(x,y) \in F$.
Furthermore $F"a \coloneqq \{y : \exists x \in a .~(x,y) \in F\}$. Furthermore $F\,''a \coloneqq \{y : \exists x \in a .~(x,y) \in F\}$.
\begin{remark} \begin{remark}
Note that we didn't need to use an axiom schema, Note that we didn't need to use an axiom schema,
@ -94,10 +94,9 @@ Furthermore $F"a \coloneqq \{y : \exists x \in a .~(x,y) \in F\}$.
\begin{axiom}[Comprehension] \begin{axiom}[Comprehension]
\yalabel{Axiom of Comprehension}{(Comp)}{ax:bg:comp} \yalabel{Axiom of Comprehension}{(Comp)}{ax:bg:comp}
\[ \[
\forall X_1 \ldots \forall X_k .~\exists Y ( \forall x .~x \in Y \iff \phi(x,X_1,\ldots, X_k)) \forall X_1 .~\ldots \forall X_k .~\exists Y.~ ( \forall x .~x \in Y \iff \phi(x,X_1,\ldots, X_k))
\] \]
where $\phi(x, X_1, \ldots, X_k)$ where $\phi(x, X_1, \ldots, X_k)$
is a formula which contains exactly $X_1, \ldots, X_k, x$ is a formula which contains exactly $X_1, \ldots, X_k, x$
as free variables, as free variables,

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@ -113,6 +113,8 @@
\DeclareSimpleMathOperator{HOD} \DeclareSimpleMathOperator{HOD}
\DeclareSimpleMathOperator{OD} \DeclareSimpleMathOperator{OD}
\DeclareSimpleMathOperator{AC} \DeclareSimpleMathOperator{AC}
\newcommand{\AxC}{\yarefs{ax:c}}
\newcommand{\AxSep}{\yarefs{ax:sep}} % Separation
\newcommand{\Choice}{\yarefs{ax:c}} \newcommand{\Choice}{\yarefs{ax:c}}
% \DeclareSimpleMathOperator{Choice} % \DeclareSimpleMathOperator{Choice}
% \DeclareSimpleMathOperator{Fund} % \DeclareSimpleMathOperator{Fund}