update notations and fix typos

pages/Math4111/Math4111_L16.md

@@ -17,7 +17,7 @@ Let $n\geq N$. Since $(s_n)$ is monotonic increasing, $s_n\geq s_N>t-\epsilon$.
 
 So $(s_n)$ converges to $t$.
 
-EOP
+QED
 
 ## New materials
 
@@ -47,7 +47,7 @@ Step 3: By induction, we can get a sequence $n_1,n_2,\cdots$ such that $\forall
 
 Then $(p_{n_i})$ is a subsequence of $(p_n)$ and $p_{n_i}\to q$.
 
-EOP
+QED
 
 ### Cauchy Sequences
 
@@ -93,7 +93,7 @@ If $m,n\geq N$, then $d(p_m,p_n)\leq d(p_m,p)+d(p,p_n)<\epsilon+\epsilon=2\epsil
 
 *You can also use $\frac{\epsilon}{2}$ instead of $\epsilon$ in the above proof, just for fun.*
 
-EOP
+QED
 
 #### Lemma 3.11 (b)
 
@@ -107,7 +107,7 @@ Let $r=max\{d(p_i,p_j);1\leq i,j\leq N\}+1$.
 
 Then $\forall n\in \mathbb{N}$, $p_n\in B_r(p_N)$.
 
-EOP
+QED
 
 > Note: This proof is nearly identical to the proof of convergent sequences implies bounded.
 
@@ -148,5 +148,5 @@ By definition of supremum, the claim implies that $\forall \epsilon>0$, $diam(\o
 
 (b) By **Theorem 2.36**, $\bigcap_{n=1}^{\infty}K_n\neq \phi$. Suppose for contradiction that there are at least two distinct points $p,q\in \bigcap_{n=1}^{\infty}K_n$. Then for all $n\in \mathbb{N}$, $x,y\in K_n$ so $diam K_n\geq d(p,q)>0$. Then diameter of $K_n$ does not converge to 0.
 
-EOP
+QED