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@ -26,7 +26,7 @@ Literature
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\begin{definition}
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\begin{definition}
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Let $A \neq \emptyset$, $B$ be arbitrary sets.
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Let $A \neq \emptyset$, $B$ be arbitrary sets.
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We write $A \le B$ ($A$ is not bigger than $B$ )
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We write $A \le B$ ($A$ is not bigger than $B$)
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iff there is an injection $f\colon A \hookrightarrow B$.
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iff there is an injection $f\colon A \hookrightarrow B$.
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\end{definition}
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\end{definition}
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\begin{lemma}
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\begin{lemma}
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@ -148,12 +148,12 @@ Literature
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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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The \vocab{continuum hypothesis} ($\CH$ )
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The \vocab{continuum hypothesis} ($\CH$)
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says that there is no set $A$ such that
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says that there is no set $A$ such that
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$\N < A < \R$.
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$\N < A < \R$.
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$\CH$ is equivalent to the statement that there is no set $A \subset \R$
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$\CH$ is equivalent to the statement that there is no set $A \subset \R$
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which is uncountable ($\N < A$ )
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which is uncountable ($\N < A$)
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and there is no bijection $A \leftrightarrow \R$.
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and there is no bijection $A \leftrightarrow \R$.
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\end{definition}
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\end{definition}
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@ -9,7 +9,7 @@
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\begin{remark}
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\begin{remark}
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\begin{itemize}
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\begin{itemize}
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\item If $\emptyset \neq O \R$
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\item If $\emptyset \neq O \overset{\text{open}}{\subseteq} \R$
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then $O \sim \R$.
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then $O \sim \R$.
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\item If $O \subseteq \R$
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\item If $O \subseteq \R$
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is open, then $O$ is the union of open intervals
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is open, then $O$ is the union of open intervals
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@ -1,8 +1,7 @@
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\lecture{03}{2023-10–23}{Cantor-Bendixson}
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\lecture{03}{2023-10–23}{Cantor-Bendixson}
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\begin{theorem}[Cantor-Bendixson]
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\begin{theorem}[Cantor-Bendixson]
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\yaref{thm:cantorbendixson}{Cantor-Bendixson}{Cantor-Bendixson}
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\yalabel{Cantor-Bendixson}{Cantor-Bendixson}{thm:cantorbendixson}
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If $A \subseteq \R$ is closed,
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If $A \subseteq \R$ is closed,
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it is either at most countable or else
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it is either at most countable or else
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$A$ contains a perfect set.
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$A$ contains a perfect set.
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@ -128,6 +127,3 @@ all condensation points are accumulation points.
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% is at most countable.
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% is at most countable.
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% Also $A'$ is closed.
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% Also $A'$ is closed.
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% \end{remark}
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% \end{remark}
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