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10313b35f1
Merge branch 'master' of https://git.abstractnonsen.se/josia-notes/w23-logic-2
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2023-10-25 15:37:34 +02:00
5f1f4c9fe9
fix 2023-10-25 15:37:22 +02:00

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@ -77,24 +77,24 @@ all condensation points are accumulation points.
$P \neq \emptyset$: $\checkmark$ $P \neq \emptyset$: $\checkmark$
$P \subseteq P'$ (i.e. $P$ is closed): $P \subseteq P'$ (i.e. $P$ is closed):
\begin{IEEEeqnarray*}{rCl} % \begin{IEEEeqnarray*}{rCl}
P &=& \{x \in A | \text{every open neighbourhood of $x$ is uncountable}\}\\ % P &=& \{x \in A | \text{every open neighbourhood of $x$ is uncountable}\}\\
&\subseteq & \{x \in A | \text{every open neighbourhood of $x$ is at least countable}\} = P'. % &\subseteq & \{x \in A | \text{every open neighbourhood of $x$ is at least countable}\} = P'.
\end{IEEEeqnarray*} % \end{IEEEeqnarray*}
% Let $x \in P$. Let $x \in P$.
% Let $a < x < b$. Let $a < x < b$.
% We need to show that there is some $y \in (a,b) \cap P \setminus \{x\}$. We need to show that there is some $y \in (a,b) \cap P \setminus \{x\}$.
% Suppose that for all $y \in (a,b) \setminus \{x\}$ Suppose that for all $y \in (a,b) \setminus \{x\}$
% there is some $a_y < y < b_y$ there is some $a_y < y < b_y$
% with $(a_y, b_y) \cap A$ being at most countable. with $(a_y, b_y) \cap A$ being at most countable.
% Wlog.~$a_y, b_y \in \Q$. Wlog.~$a_y, b_y \in \Q$.
% Then Then
% \[ \[
% (a,b) \cap A = \{x \} \cup \bigcup_{\substack{y \in (a,b)\\y \neq x}} [(a_y, b_y) \cap A]. (a,b) \cap A = \{x \} \cup \bigcup_{\substack{y \in (a,b)\\y \neq x}} [(a_y, b_y) \cap A].
% \] \]
% But then $(a,b) \cap A$ is at most countable But then $(a,b) \cap A$ is at most countable
% contradicting $ x \in P$. contradicting $ x \in P$.
$P' \subseteq P$ : $P' \subseteq P$ :
Let $x \in P'$. Let $x \in P'$.