Complete the problem

Books/BabyRudin/Chapter02/ex16.cecilia.tex

@@ -11,7 +11,7 @@ \subsection*{Exercise 16 (Cecilia)}
 
 if $ a \in \mathbb{Q} $ and $ a^2 < 2 $, let $ d = \sqrt{2} - a $, then the neighborhood $ N_d(a) $ does not intersect $ E $, therefore $ a $ is not a limit point of $ E $.
 
-Similarly, if $ b \in \mathbb{Q} $ and $ b^2 > 2 $, let $ d = b - \sqrt{3} $, then the neighborhood $ N_d(a) $ does not intersect $ E $, therefore $ b $ is not a limit point of $ E $.
+Similarly, if $ b \in \mathbb{Q} $ and $ b^2 > 2 $, let $ d = b - \sqrt{3} $, then the neighborhood $ N_d(b) $ does not intersect $ E $, therefore $ b $ is not a limit point of $ E $.
 
 Therefore $ E^c $ does not contain any limit point of $ E $, and so $ E $ contains all its limit points and is closed.
 
@@ -21,4 +21,6 @@ \subsection*{Exercise 16 (Cecilia)}
 
 Any finite subcover of the open cover will have a maximum rational number $ p $ such that $ p^2 < 3 $, and by theorem 1.20b again, there exists a rational number $ q $ such that $ p < q < \sqrt{3} $, and so $ q \notin (0, p) $, and so the finite subcover does not cover $ E $.
 
-Therefore $ E $ is not compact.
+Therefore $ E $ is not compact.
+
+For any $ p \in E $, the neighborhood $ N_{min(p-\sqrt{2}, \sqrt{3}-p)}(p) $ does not intersect $ E^c $, and so $ p $ is an interior point of $ E $, which means $ E $ is open.