update notations and fix typos

pages/Math4111/Math4111_L15.md

@@ -14,7 +14,7 @@ $$
     Proof:  
     Let $\epsilon>0$ be arbitrary, then $\exists N$ such that $\forall n\geq N,|a_n-a|<\epsilon$.
     Then if $n\geq N$, $|a_n-a|\leq \sqrt{|a_n-a|^2}\leq\sqrt{|a_n-a|^2+|b_n-b|^2}=|x_n-(a,b)|<\epsilon$.  
-    EOP  
+    QED  
     (b) If $x_n\to (a,b)$, then $b_n\to b$.
     This follows from the same argument from (a)
 2. Prove the $\implies$ direction.
@@ -28,7 +28,7 @@ $$
     Since $a_n\to a$, $\exists N_1$ such that $\forall n\geq N_1,|a_n-a|<\frac{\epsilon}{\sqrt{2}}$.  
     Since $b_n\to b$, $\exists N_2$ such that $\forall n\geq N_2,|b_n-b|<\frac{\epsilon}{\sqrt{2}}$.  
     Let $N=\max\{N_1,N_2\}$. Then if $n\geq N$, $|a_n-a|<\epsilon$ and $|b_n-b|<\sqrt{\frac{\epsilon^2}{2}+\frac{\epsilon^2}{2}}=\epsilon$.  
-    EOP  
+    QED  
 
 ## New Materials
 
@@ -82,7 +82,7 @@ $$
 \left|\frac{1}{s_n}-\frac{1}{s}\right|=\frac{|s-s_n|}{|s||s_n|}<\frac{\frac{\epsilon|s|^2}{2}}{|s|^2}=\epsilon
 $$
 
-EOP
+QED
 
 ### Subsequences
 
@@ -108,7 +108,7 @@ $\implies$:
 
 Thought process: show what if the sequence does not converge to $p$, then there exists a subsequence that does not converge to $p$.
 
-EOP
+QED
 
 #### Theorem 3.6