diff --git a/inputs/lecture_03.tex b/inputs/lecture_03.tex
index 1404146..fb71262 100644
--- a/inputs/lecture_03.tex
+++ b/inputs/lecture_03.tex
@@ -8,7 +8,7 @@
     $A$ contains a perfect set.
 \end{theorem}
 \begin{corollary}
-    If $A se \R$ is closed,
+    If $A \subseteq \R$ is closed,
     then either $A \le \N$ or $A \sim \R$.
 \end{corollary}
 \begin{fact}
@@ -16,12 +16,118 @@
 \end{fact}
 \begin{proof}
     $\supseteq$ is clear.
-    For $\subseteq $, fix $a < x < b$ 
+    For $\subseteq $, fix $a < x < b$
     and let us define $(y_n: n \in  \omega)$
     as well as $((a_n, b_n): n \in \omega)$.
-    
-    % TODO
-    
+    Set $a_0 \coloneqq a$, $b_0 \coloneqq b$.
+    Having defined $(a_n, b_n)$,
+    pick $x \neq y_n \in A \cap (a_n, b_n)$,
+    Then pick $a_n < a_{n+1} < x < b_{n+1} < b_n$
+    such that $y_n \not\in (a_{n+1}, b_{n+1})$.
+    Clearly $y_n \neq y_{n+1}$,
+    hence $\{y_n : n \in \N\}$ is a countable subset of $A \cap (a,b)$.
 \end{proof}
 
+\begin{definition}
+    Let $A \subseteq  \R$.
+    We say that $x \in \R$ 
+    is a \vocab{condensation point} of $A$ 
+    iff for all $a < x < b$, $(a,b) \cap A$ 
+    is uncountable.
+\end{definition}
+By the fact we just proved,
+all condensation points are accumulation points.
+
+\begin{refproof}{thm:cantorbendixson}
+    Fix $A \subseteq \R$ closed.
+    We want to see that $A$ is at most countable
+    or there is some perfect $P \subseteq A$.
+    Let $P \coloneqq  \{x \in \R | x \text{ is a condensation point of $A$}\}$.
+    Since $A $ is closed, $P \subseteq A$.
+
+    \begin{claim}
+        $A \setminus P$ is at most countable.
+    \end{claim}
+    \begin{subproof}
+        For each $x \in A \setminus P$,
+        there is $a_x < x < b_x$ 
+        such that $(a_x, b_x) \cap A$ 
+        is at most countable.
+        Since $\Q$ is dense in $\R$,
+        we may assume that $a_x, b_x \in \Q$.
+
+        Then
+        \begin{IEEEeqnarray*}{rCl}
+            A \setminus P &=& \bigcap_{x \in  A \setminus P} (a_x, b_x) \cap A.
+        \end{IEEEeqnarray*}
+        $\subseteq $ holds by the choice of $a_x$ and $b_x$.
+        For $\supseteq$ let $y$ be an element of the RHS.
+        Then $y \in (a_{x_0}, b_{x_0}) \cap A$ for some $x_0$.
+        As $(a_{x_0}, b_{x_0}) \cap A$ is at most countable,
+        $y \in P$.
+
+        Now we have that $A \setminus P$ is a union
+        of at most countably many sets,
+        each of which is at most countable.
+    \end{subproof}
+
+    \begin{claim}
+        If $P \neq  \emptyset$, the $P$ is perfect.
+    \end{claim}
+    \begin{subproof}
+        $P \neq \emptyset$: $\checkmark$ 
+
+        $P \subseteq  P'$ (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*}
+        
+        % 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 $x \in P'$.
+        Then for $a < x < b$ the set
+        $(a,b) \cap P$
+        always has a member $y$ such that $y \neq x$.
+        Since $y \in P$, we get that $(a,b) \cap A$
+        in uncountable, hence $x \in P$.
+    \end{subproof}
+    But now
+    \[
+        A = \overbrace{P}^{\mathclap{\text{perfect, unless $= \emptyset$}}} \cup \underbrace{(A \setminus P)}_{\mathclap{\text{at most countable}}}.
+    \] 
+\end{refproof}
+
+\todo{Alternative proof of Cantor-Bendixson}
+% \begin{remark}
+%     There is an alternative proof of Cantor-Bendixson, going as follows:
+%     Fix $A \subseteq  \R$ closed.
+%     Define a sequence
+%     \[
+%         A \supseteq A' \supseteq A'' \supseteq \ldots \supseteq \bigcap_{n} A^{(n)}
+%         \supseteq \left( \bigcap_{n} A^{(n)} \right)' \supseteq \ldots
+%     \]
+%     Then $A \setminus A'$ has at most countably many points.
+%     For all $a \in  A \setminus A'$
+%     pick $\Q\ni a_x < x < b_x \in \Q$ 
+%     such that $(a_x, b_x) \cap A = \{x\}$.
+%     Then $A \setminus A' = \bigcup_{x \in A \setminus A'}  [(a_x, b_x) \cap A]$
+%     is at most countable.
+%     Also $A'$ is closed.
+% \end{remark}
+
+
 
diff --git a/logic2.tex b/logic2.tex
index c430f4a..f0624db 100644
--- a/logic2.tex
+++ b/logic2.tex
@@ -26,6 +26,7 @@
 
 \input{inputs/lecture_01}
 \input{inputs/lecture_02}
+\input{inputs/lecture_03}
 
 
 \cleardoublepage