update notations and fix typos

pages/Math4111/Math4111_L9.md

@@ -37,7 +37,7 @@ Let $h>0$. Since $y-h$ is not an upper bound of $E$, $\exists x\in E$ such that
 
 Since $y$ is an upper bound of $E$, $x\leq y$. So $x\in B_n(y)\cap E$, so $B_h(y)\cap E\neq \phi$.
 
-EOP
+QED
 
 #### Remark 2.29
 
@@ -75,7 +75,7 @@ To show $G\cap Y\subset E$.
 
 $G\cap Y=\left(\bigcup_{p\in E}V_p\right)\cap Y=\bigcup_{p\in E}(V_p\cap Y)\subset E$
 
-EOP
+QED
 
 ### Compact sets
 
@@ -101,4 +101,4 @@ as we can build an infinite open cover $\bigcup_{i\in Z} (i,i+2)$ and it does no
 
 Suppose we consider the sub-collection $\{n_i,n_i+2:i=1,..,k\}$, Then $N+3$ is not in the union, where $N=max\{n_1,...,n_k\}$.
 
-EOP
+QED