From 5f1f4c9fe978be0daceda70ab1a64caef415fb77 Mon Sep 17 00:00:00 2001
From: Josia Pietsch <git@jrpie.de>
Date: Wed, 25 Oct 2023 15:37:22 +0200
Subject: [PATCH] fix

---
 inputs/lecture_03.tex | 34 +++++++++++++++++-----------------
 1 file changed, 17 insertions(+), 17 deletions(-)

diff --git a/inputs/lecture_03.tex b/inputs/lecture_03.tex
index fb71262..8229f8b 100644
--- a/inputs/lecture_03.tex
+++ b/inputs/lecture_03.tex
@@ -78,24 +78,24 @@ all condensation points are accumulation points.
         $P \neq \emptyset$: $\checkmark$ 
 
         $P \subseteq  P'$ (i.e. $P$ is closed):
-        \begin{IEEEeqnarray*}{rCl}
-            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'.
-        \end{IEEEeqnarray*}
+        % \begin{IEEEeqnarray*}{rCl}
+        %     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'.
+        % \end{IEEEeqnarray*}
         
-        % Let $x \in P$.
-        % Let $a < x < b$.
-        % We need to show that there is some $y \in  (a,b) \cap P \setminus \{x\}$.
-        % Suppose that for all $y \in  (a,b) \setminus \{x\}$
-        % there is some $a_y < y < b_y$
-        % with $(a_y, b_y) \cap A$ being at most countable.
-        % Wlog.~$a_y, b_y \in \Q$.
-        % Then
-        % \[
-        %     (a,b) \cap A = \{x \} \cup \bigcup_{\substack{y \in (a,b)\\y \neq x}} [(a_y, b_y) \cap A].
-        % \] 
-        % But then $(a,b) \cap A$ is at most countable
-        % contradicting $ x \in P$.
+        Let $x \in P$.
+        Let $a < x < b$.
+        We need to show that there is some $y \in  (a,b) \cap P \setminus \{x\}$.
+        Suppose that for all $y \in  (a,b) \setminus \{x\}$
+        there is some $a_y < y < b_y$
+        with $(a_y, b_y) \cap A$ being at most countable.
+        Wlog.~$a_y, b_y \in \Q$.
+        Then
+        \[
+            (a,b) \cap A = \{x \} \cup \bigcup_{\substack{y \in (a,b)\\y \neq x}} [(a_y, b_y) \cap A].
+        \] 
+        But then $(a,b) \cap A$ is at most countable
+        contradicting $ x \in P$.
 
         $P' \subseteq P$ :
         Let $x \in P'$.