inputs/lecture_03.tex

@@ -77,24 +77,24 @@ 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 $x \in P'$.