Make sure negative points are considered as well

Books/BabyRudin/Chapter02/ex16.cecilia.tex

@@ -1,26 +1,26 @@
 \subsection*{Exercise 16 (Cecilia)}
 For all point $ p $ in $ E $
-\begin{eqnarray*}
-  p^2 < 3 \\
-  \implies p^2 < 4 \\
-  \implies p < 2 \\
-  \implies |0 - p| < 2 \\
-  \implies d(0, p) < 2
-\end{eqnarray*}
+\begin{flalign*}
+  p^2 &< 3 \\
+  p^2 &< 4 \\
+  |p| &< 2 \\
+  |0 - p| &< 2 \\
+  d(0, p) &< 2
+\end{flalign*}
 Therefore $ E $ is bounded.
 
-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 $.
+if $ a \in \mathbb{Q} $ and $ a^2 < 2 $, let $ d = \min(|a - \sqrt{2}|, |a + \sqrt{2}|) $, 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(b) $ does not intersect $ E $, therefore $ b $ is not a limit point of $ E $.
+Similarly, if $ b \in \mathbb{Q} $ and $ b^2 > 3 $, let $ d = \min(|b - \sqrt{3}|,|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.
 
-Consider the open cover $ (0, p) $ for all rational $ p $ such that $ 0 < p^2 < 3 $. 
+Consider the open cover $ (-p, p) $ for all rational $ p $ such that $ 0 < p < \sqrt{3} $. 
 
-The open cover does cover $ E $ because for all $ x \in E $, there exists a rational number $ p $ such that $ x < p < \sqrt{3} $ by theorem 1.20b, and so $ x \in (0, p) $.
+The open cover does cover $ E $ because for all $ x \in E $, there exists a rational number $ p $ such that $ |x| < p < \sqrt{3} $ by theorem 1.20b, and so $ x \in (-p, p) $.
 
-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 $.
+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 (-p, p) $, and so the finite subcover does not cover $ E $.
 
 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.
+For any $ p \in E $, the neighborhood $ N_{\min(|p-\sqrt{2}|, |p+\sqrt{2}|, |p-\sqrt{3}|, |p+\sqrt{3}|}(p) $ does not intersect $ E^c $, and so $ p $ is an interior point of $ E $, which means $ E $ is open.