format updates

content/Math4121/Math4121_L22.md

@@ -2,7 +2,8 @@
 
 ## Continue on Arzela-Osgood Theorem
 
-Proof:
+<details>
+<summary>Proof continuation of Arzela-Osgood Theorem</summary>
 
 Part 2: Control the integral on $\mathcal{U}$
 
@@ -40,7 +41,7 @@ $$
 
 $\forall N\geq K$.
 
-QED
+</details>
 
 ### Baire Category Theorem
 
@@ -50,7 +51,8 @@ Nowhere dense sets can be large, but they canot cover an open (or closed) interv
 
 An open interval cannot be covered by a countable union of nowhere dense sets.
 
-Proof:
+<details>
+<summary>Proof</summary>
 
 Suppose $(0,1)\subset \bigcup_{n=1}^\infty S_n$ where each $S_n$ is nowhere dense. In particular, $\exists I_1$ closed interval such that $I_1\subset (0,1)$ and $I_1\cap S_1=\emptyset$.
 
@@ -62,7 +64,7 @@ Then $x\in (0,1)$ and $x\notin \bigcup_{n=1}^\infty S_n$.
 
 Contradiction with the assumption that $(0,1)\subset \bigcup_{n=1}^\infty S_n$.
 
-QED
+</details>
 
 #### Definition First Category
 
@@ -72,13 +74,14 @@ A countable union of nowhere dense sets is called a set of **first category**.
 
 Complement of a set of first category in $\mathbb{R}$ is dense in $\mathbb{R}$.
 
-Proof:
+<details>
+<summary>Proof</summary>
 
 We need to show that for every interval $I$, $\exists x\in I\cap S^c$. ($\exists x\in I$ and $x\notin S$)
 
 This is equivalent to the Baire Category Theorem.
 
-QED
+</details>
 
 Recall a function is pointwise discontinuous if $\mathcal{C}=\{c\in [a,b]: f\text{ is continuous at } c\}$ is dense in $[a,b]$.
 
@@ -88,7 +91,8 @@ $\mathcal{D}=[a,b]\setminus \mathcal{C}$ is called the set of points of disconti
 
 $f$ is pointwise discontinuous if and only if $\mathcal{D}$ is of first category.
 
-Proof:
+<details>
+<summary>Proof</summary>
 
 Part 1: If $\mathcal{D}$ is of first category, then $f$ is pointwise discontinuous.
 
@@ -104,13 +108,14 @@ Let $I\subseteq [a,b]$ so $\exists c\in \mathcal{C}\cap I$. So by definition of
 
 Thus, $P_k$ is nowhere dense.
 
-QED
+</details>
 
 #### Corollary 4.10
 
 Let $\{f_n\}$ be a sequence of pointwise discontinuous functions. The set of points at which all $f_n$ are simultaneously continuous is dense (it's also uncountable).
 
-Proof:
+<details>
+<summary>Proof</summary>
 
 $$
 \bigcap_{n=1}^\infty \mathcal{C}_n=\left(\bigcup_{n=1}^\infty \mathcal{D}_n\right)^c
@@ -118,4 +123,4 @@ $$
 
 The complement of a set of first category is dense.
 
-QED
+</details>