update notations and fix typos

pages/Math4111/Math4111_L11.md

@@ -31,7 +31,7 @@ And $(\bigcup_{\alpha\in A} G_\alpha)\cup F^c=X\supset K$ is an open cover of $K
 
 Since $F\subset K$, So $\Phi$ is a cover of $F$ then $\Phi\backslash \{F^c\}$ is a finite subcover of $\{G_\alpha\}_{\alpha \in A}$ of $F$.
 
-EOP
+QED
 
 ##### Corollary 2.35
 
@@ -57,7 +57,7 @@ Let $l=\max\{n_1,...,n_m\}$, then $K_l=\bigcap^{m}_{i=i} K_{n_i}$. So $K_1\subse
 
 which contradicts with $K_l\subset K_1$
 
-EOP
+QED
 
 #### Theorem 2.37
 
@@ -75,7 +75,7 @@ For each $q\in K, q\notin E'$, so $\exists$ neighborhood $\forall q\in B_{r_q}(q
 
 Then $\{V_q\}_{q\in K}$ is an open cover of $K$ , so it has a finite subcover $\{V_{q_i}\}^n_{n=1}$. Then $E\subset K\subset \bigcup_{i=1}^n$, and $\forall i,V_{q_i}\cap E\subset \{q_i\}$, then $E\subset\{q_1,...,q_n\}$
 
-EOP
+QED
 
 #### Theorem 2.38
 
@@ -101,4 +101,4 @@ Since $x$ is an upper bound of $E$ and $a_n\in E$, then $a_n\leq x$.
 
 Since $x$ is the least upper bound of $E$, and $b_n$ is an upper bound of $E$, then $x\leq b_n$. $x\in E,E\neq \phi$
 
-EOP
+QED