updates

content/Math4111/Math4111_L1.md

@@ -24,7 +24,8 @@ $\equiv\cancel{\exist} p\in \mathbb{Q}, p^2=2$
 
 $\equiv p\in \mathbb{Q},p^2\neq 2$
 
-#### Proof
+<details>
+<summary>Proof</summary>
 
 Suppose for contradiction, $\exist p\in \mathbb{Q}$ such that $p^2=\mathbb{Q}$.
 
@@ -36,7 +37,7 @@ So $m^2$ is divisible by 4, $2n^2$ is divisible by 4.
 
 So $n^2$ is even. but they are not both even.
 
-QED
+</details>
 
 ### Theorem (No closest rational for a irrational number)
 

content/Math4111/Math4111_L2.md

@@ -47,19 +47,18 @@ Let $S$ be an ordered set and $E\subset S$. We say $\alpha\in S$ is the LUB of $
 1. $\alpha$ is the UB of $E$. ($\forall x\in E,x\leq \alpha$)
 2. if $\gamma<\alpha$, then $\gamma$ is not UB of $E$. ($\forall \gamma <\alpha, \exist x\in E$ such that $x>\gamma$ )
 
-#### Lemma 
-
-Uniqueness of upper bounds.
+#### Lemma (Uniqueness of upper bounds)
 
 If $\alpha$ and $\beta$ are LUBs of $E$, then $\alpha=\beta$.
 
-Proof:
+<details>
+<summary>Proof</summary>
 
 Suppose for contradiction $\alpha$ and $\beta$ are both LUB of $E$, then $\alpha\neq\beta$
 
 WLOG $\alpha>\beta$ and $\beta>\alpha$.
 
-QED
+</details>
 
 We write $\sup E$ to denote the LUB of $E$.
 

content/Math4111/Math4111_L3.md

@@ -26,7 +26,8 @@ Let $S=\mathbb{Z}$.
 
 Proof that $LUBP\implies GLBP$.
 
-Proof:
+<details>
+<summary>Proof</summary>
 
 Let $S$ be an ordered set with LUBP. Let $B<S$ be non-empty and bounded below.
 
@@ -57,7 +58,7 @@ Let's say $\alpha=sup\ L$. We claim that $\alpha=inf\ B$. We need to show $2$ th
 
 Thus $\alpha=inf\ B$
 
-QED
+</details>
 
 ### Field
 

content/Math4111/Math4111_L4.md

@@ -27,7 +27,8 @@
 
 (Archimedean property) If $x,y\in \mathbb{R}$ and $x>0$, then $\exists n\in \mathbb{N}$ such that $nx>y$.
 
-Proof
+<details>
+<summary>Proof</summary>
 
 Suppose the property is false, then $\exist x,y\in \mathbb{R}$ with $x>0$ such that $\forall v\in \mathbb{N}$, nx\leq y$
 
@@ -39,7 +40,7 @@ This implies $(m+1)x>\alpha$
 
 Since $(m+1)x\in \alpha$, this contradicts the fact that $\alpha$ is an upper bound of $A$.
 
-QED
+</details>
 
 ### $\mathbb{Q}$ is dense in $\mathbb{R}$
 
@@ -51,7 +52,8 @@ $$
 x<\frac{m}{n}<y\iff nx<m<ny
 $$
 
-Proof:
+<details>
+<summary>Proof</summary>
 
 Let $x,y\in\mathbb{R}$, with $x<y$. We'll find $n\in \mathbb{N},\mathbb{m}\in \mathbb{Z}$ such that $nx<m<ny$.
 
@@ -59,7 +61,7 @@ By Archimedean property, $\exist n\in \mathbb{N}$ such that $n(y-x)>1$, and $\ex
 
 So $-m_2<nx<m_1$. Thus $\exist m\in \mathbb{Z}$ such that $m-1\leq nx<m$ (Here we use a property of $\mathbb{Z}$) We have $ny>1+nx\geq 1+(m-1)=m$
 
-QED
+</details>
 
 ### $\sqrt{2}\in \mathbb{R}$, $(\sqrt[n]{x}\in\mathbb{R})$
 

content/Math4111/Math4111_L5.md

@@ -30,7 +30,8 @@ $\forall x\in \mathbb{R}_{>0},\forall n\in \mathbb{N},\exist$ unique $y\in \math
 
 (Because of this Theorem we can define $x^{1/x}=y$ and $\sqrt{x}=y$)
 
-Proof:
+<details>
+<summary>Proof</summary>
 
 We cna assume $n\geq 2$ (For $n=1,y=x$)
 
@@ -94,7 +95,7 @@ So want $k\leq \frac{y^n-x}{ny^{n-1}}$
 
 [For actual proof, see the text.]
 
-QED
+</details>
 
 ### Complex numbers
 
@@ -151,7 +152,8 @@ $$
 (\sum a_j b_j)^2=(\sum a_j^2)(\sum b_j^2)
 $$
 
-Proof:
+<details>
+<summary>Proof</summary>
 
 For real numbers:
 
@@ -169,7 +171,7 @@ let $t=C/B$ to get $0\leq A-2(C/B)C+(C/B)^2B=A-\frac{C^2}{B}$
 
 to generalize this to $\mathbb{C}$, $A=\sum |a_j|^2,B=\sum |b_j|^2,C=\sum |a_j \bar{b_j}|$.
 
-QED
+</details>
 
 ### Euclidean spaces
 

content/Math4111/Math4111_L6.md

@@ -126,7 +126,8 @@ $A$ is countable, $n\in \mathbb{N}$,
 
 $\implies A^n=\{(a_{1},...,a_{n}):a_1\in A, a_n\in A\}$, is countable.
 
-Proof:
+<details>
+<summary>Proof</summary>
 
 Induct on $n$,
 
@@ -138,7 +139,7 @@ Induction step: suppose $A^{n-1}$ is countable. Note $A^n=\{(b,a):b\in A^{n-1},a
 
 Since $b$ is fixed, so this is in 1-1 correspondence with $A$, so it's countable by Theorem 2.12.
 
-QED
+</details>
 
 #### Theorem 2.14
 

content/Math4111/Math4111_L7.md

@@ -80,7 +80,8 @@ Let $(X,d)$ be  a metric space, $\forall p\in X,\forall r>0$, $B_r(p)$ is an ope
 
 *every ball is an open set*
 
-Proof:
+<details>
+<summary>Proof</summary>
 
 Let $q\in B_r(p)$.
 
@@ -88,7 +89,7 @@ Let $h=r-d(p,q)$.
 
 Since $q\in B_r(p),h>0$. We claim that $B_h(q)$. Then $d(q,s)<h$, so $d(p,s)\leq d(p,q)+d(q,s)<d(p,q)+h=r$. (using triangle inequality) So $S\in B_r(p)$.
 
-QED
+</details>
 
 ### Closed sets