fix #82, thanks @tomvanlier!

Author: jesseli2002

chapters/chapter3/chapter3-2.tex

@@ -32,11 +32,11 @@ \section{Open and Closed Sets}
   \item The set of $B$'s limit points is $[0, 1]$. The set of $A$'s limit points is $\{1, -1\}$.
   \item
     $B$ is not open since every $(a,b) \not\subseteq B$ and $B$ is not closed since we can construct limits to irrational values outside $B$.
-    $A$ is closed since $\{1, -1\} \subseteq A$, but not open as it does not contain any irrationals meaning $(a,b) \not\subseteq A$ for all $a,b \in \mathbf{R}$.
+    $A$ is not closed since $-1 \notin A$, and not open as it does not contain any irrationals meaning $(a,b) \not\subseteq A$ for all $a,b \in \mathbf{R}$.
   \item
-    Every point of $A$ except the limit points $\{1, -1\}$ is isolated, as if it were not isolated it would be a limit point.
+    Every point of $A$ except the limit point $1$ is isolated, as if it were not isolated it would be a limit point.
     $B$ has no isolated points since $B \setminus [0, 1] = \emptyset$, or in other words since $B$ is dense in $[0,1]$ every $b \in B \subseteq [0,1]$ can be reached via a limit.
-  \item $\closure A = A$ as $A$ is already closed, and $\closure B = B \cup [0,1] = [0,1]$.
+  \item $\closure A = A \cup \{-1\}$, and $\closure B = B \cup [0,1] = [0,1]$.
   }
 \end{solution}