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1 changed files with 17 additions and 17 deletions
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@ -78,24 +78,24 @@ all condensation points are accumulation points.
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$P \neq \emptyset$: $\checkmark$
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$P \subseteq P'$ (i.e. $P$ is closed):
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\begin{IEEEeqnarray*}{rCl}
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P &=& \{x \in A | \text{every open neighbourhood of $x$ is uncountable}\}\\
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&\subseteq & \{x \in A | \text{every open neighbourhood of $x$ is at least countable}\} = P'.
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\end{IEEEeqnarray*}
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% \begin{IEEEeqnarray*}{rCl}
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% P &=& \{x \in A | \text{every open neighbourhood of $x$ is uncountable}\}\\
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% &\subseteq & \{x \in A | \text{every open neighbourhood of $x$ is at least countable}\} = P'.
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% \end{IEEEeqnarray*}
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% Let $x \in P$.
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% Let $a < x < b$.
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% We need to show that there is some $y \in (a,b) \cap P \setminus \{x\}$.
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% Suppose that for all $y \in (a,b) \setminus \{x\}$
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% there is some $a_y < y < b_y$
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% with $(a_y, b_y) \cap A$ being at most countable.
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% Wlog.~$a_y, b_y \in \Q$.
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% Then
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% \[
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% (a,b) \cap A = \{x \} \cup \bigcup_{\substack{y \in (a,b)\\y \neq x}} [(a_y, b_y) \cap A].
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% \]
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% But then $(a,b) \cap A$ is at most countable
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% contradicting $ x \in P$.
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Let $x \in P$.
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Let $a < x < b$.
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We need to show that there is some $y \in (a,b) \cap P \setminus \{x\}$.
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Suppose that for all $y \in (a,b) \setminus \{x\}$
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there is some $a_y < y < b_y$
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with $(a_y, b_y) \cap A$ being at most countable.
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Wlog.~$a_y, b_y \in \Q$.
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Then
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\[
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(a,b) \cap A = \{x \} \cup \bigcup_{\substack{y \in (a,b)\\y \neq x}} [(a_y, b_y) \cap A].
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\]
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But then $(a,b) \cap A$ is at most countable
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contradicting $ x \in P$.
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$P' \subseteq P$ :
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Let $x \in P'$.
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