update notations and fix typos

pages/Math4111/Math4111_L18.md

@@ -79,7 +79,7 @@ Case 2: $(s_n)$ is bounded above.
 
 Let $M$ be an upper bound for $(s_n)$. Then $\{n\in\mathbb{N}:s_n\in[M,x]\}$ is infinite, by **Theorem 3.6 (b)** ($\exists$ subsequence $(s_{n_k})$ in $[x,M)$ and $\exists t\in[x,M]$ such that $s_{n_k}\to t$. This implies $t\in E$, so $x\leq t\leq S^*$).
 
-EOP
+QED
 
 #### Theorem 3.19 ("one-sided squeeze theorem")
 
@@ -104,7 +104,7 @@ By **Theorem 3.17**, $\{n\in\mathbb{N}:t_n\geq x\}$ is finite $\implies$ $\{n\in
 
 Thus $\limsup_{n\to\infty} s_n\leq \limsup_{n\to\infty} t_n$.
 
-EOP
+QED
 
 > Normal squeeze theorem: If $s_n\leq t_n\leq u_n$ for all $n\in\mathbb{N}$, and $\lim_{n\to\infty} s_n=\lim_{n\to\infty} u_n=L$, then $\lim_{n\to\infty} t_n=L$.
 >