update notations and fix typos

pages/Math4111/Math4111_L24.md

@@ -33,7 +33,7 @@ $\impliedby$: Suppose for every open set $V\subset Y$, $f^{-1}(V)$ is open in $X
 
 Since $p\in f^{-1}(B_\epsilon(f(p)))$ and $f^{-1}(B_\epsilon(f(p)))$ is open, $\exists \delta > 0$ such that $B_\delta(p)\subset f^{-1}(B_\epsilon(f(p)))$. Therefore, $f(B_\delta(p))\subset B_\epsilon(f(p))$. This shows that $f$ is continuous.
 
-EOP
+QED
 
 #### Corollary 4.8
 
@@ -82,7 +82,7 @@ Proof:
 
 Let $p\in \mathbb{R}$ and $\epsilon > 0$. Let $\delta = \epsilon$. Then, $\forall x\in \mathbb{R}$, if $|x-p|<\delta$, then $|f(x)-f(p)| = |x-p| < \delta = \epsilon$.
 
-EOP
+QED
 
 Therefore, by **Theorem 4.9**, $f(x) = x^2$ is continuous. $f(x) = x^3$ is continuous... So all polynomials are continuous.
 
@@ -131,7 +131,7 @@ By Theorem 2.41, $f(X)$ is closed and bounded.
 
 By Theorem 2.28, $\sup f(X)$ and $\inf f(X)$ exist and are in $f(X)$. Let $p_0\in X$ such that $f(p_0) = \sup f(X)$. Let $q_0\in X$ such that $f(q_0) = \inf f(X)$.
 
-EOP
+QED
 
 ---
 
@@ -151,7 +151,7 @@ Proof:
 
 See the textbook.
 
-EOP
+QED
 
 ---
 
@@ -191,7 +191,7 @@ Since $A$ and $B$ are separated, $\overline{A}\cap B = \phi$ and $\overline{B}\c
 
 Therefore, $\overline{G}\cap H = \phi$ and $\overline{H}\cap G = \phi$.
 
-EOP
+QED
 
 #### Theorem 4.23 (Intermediate Value Theorem)
 
@@ -207,4 +207,4 @@ $f(a)$ and $f(b)$ are real numbers in $f([a,b])$, and $c$ is a real number betwe
 
 By **Theorem 2.47**, $c\in f([a,b])$.
 
-EOP
+QED