format updates

content/Math401/Extending_thesis/Math401_R1.md

@@ -6,9 +6,239 @@
 
 This part will cover the necessary notations and definitions for the remaining parts of the recollection.
 
-### Notations of Hilbert space
+### Notations of Linear algebra
 
-A Hilbert space is a vector space equipped with an inner product.
+#### Definition of vector space
+
+[link to vector space](../../Math429/Math429_L1#definition-1.20)
+
+A vector space over $\mathbb{f}$ is a set $V$ along with two operators $v+w\in V$ for $v,w\in V$, and $\lambda \cdot  v$ for $\lambda\in \mathbb{F}$ and $v\in V$ satisfying the following properties:
+
+* Commutativity: $\forall v, w\in V,v+w=w+v$
+* Associativity: $\forall u,v,w\in V,(u+v)+w=u+(v+w)$
+* Existence of additive identity: $\exists 0\in V$ such that $\forall v\in V, 0+v=v$
+* Existence of additive inverse: $\forall v\in V, \exists w \in V$ such that $v+w=0$
+* Existence of multiplicative identity: $\exists 1 \in \mathbb{F}$ such that $\forall v\in V,1\cdot v=v$
+* Distributive properties: $\forall v, w\in V$ and $\forall a,b\in \mathbb{F}$, $a\cdot(v+w)=a\cdot v+ a\cdot w$ and $(a+b)\cdot v=a\cdot v+b\cdot v$
+
+#### Definition of inner product
+
+[link to inner product](../../Math429/Math429_L25#definition-6.2)
+
+An inner product is a bilinear function $\langle,\rangle:V\times V\to \mathbb{F}$ satisfying the following properties:
+
+* Positivity: $\langle v,v\rangle\geq 0$
+* Definiteness: $\langle v,v\rangle=0\iff v=0$
+* Additivity: $\langle u+v,w\rangle=\langle u,w\rangle+\langle v,w\rangle$
+* Homogeneity: $\langle \lambda u, v\rangle=\lambda\langle u,v\rangle$
+* Conjugate symmetry: $\langle u,v\rangle=\overline{\langle v,u\rangle}$
+
+<details>
+<summary>Examples of inner product</summary>
+
+Let $V=\mathbb{R}^n$.
+
+The dot product is defined by
+
+$$
+\langle u,v\rangle=u_1v_1+u_2v_2+\cdots+u_nv_n
+$$
+
+is an inner product.
+
+---
+
+Let $V=L^2(\mathbb{R}, \lambda)$, where $\lambda$ is the Lebesgue measure. $f,g:\mathbb{R}\to \mathbb{C}$ are complex-valued square integrable functions.
+
+The Hermitian inner product is defined by
+$$
+\langle f,g\rangle=\int_\mathbb{R} \overline{f(x)}g(x) d\lambda(x)
+$$
+
+is an inner product.
+
+---
+
+Let $A,B$ be two linear transformation on $\mathbb{R}^n$.
+
+The Hilbert-Schmidt inner product is defined by
+
+$$
+\langle A,B\rangle=\operatorname{Tr}(A^*B)=\sum_{i=1}^n \sum_{j=1}^n \overline{a_{ij}}b_{ij}
+$$
+
+is an inner product.
+
+</details>
+
+#### Definition of inner product space
+
+A inner product space is a vector space equipped with an inner product.
+
+#### Definition of completeness
+
+[link to completeness](../../Math4111/Math4111_L17#definition-312)
+
+Note that every inner product space is a metric space.
+
+Let $X$ be a metric space. We say $X$ is **complete** if every Cauchy sequence (that is, a sequence such that $\forall \epsilon>0, \exists N$ such that $\forall m,n\geq N, d(p_m,p_n)<\epsilon$) in $X$ converges.
+
+#### Definition of Hilbert space
+
+A Hilbert space is a complete inner product space.
+
+#### Motivation of Tensor product
+
+Recall from the traditional notation of product space of two vector spaces $V$ and $W$, that is, $V\times W$, is the set of all ordered pairs $(v,w)$ where $v\in V$ and $w\in W$.
+
+The space has dimension $\dim V+\dim W$.
+
+We want to define a vector space with notation of multiplication of two vectors from different vector spaces.
+
+That is
+
+$$
+(v_1+v_1)\otimes w=(v_1\otimes w)+(v_2\otimes w)\text{ and } v\otimes (w_1+w_2)=(v\otimes w_1)+(v\otimes w_2)
+$$
+
+and enables scalar multiplication by
+
+$$
+\lambda (v\otimes w)=(\lambda v)\otimes w=v\otimes (\lambda w)
+$$
+
+And we wish to build a way associates the basis of $V$ and $W$ to the basis of $V\otimes W$. That makes the tensor product a vector space with dimension $\dim V\times \dim W$.
+
+#### Definition of linear functional
+
+> [!TIP]
+>
+> Note the difference between a linear functional and a linear map.
+>
+> A generalized linear map is a function $f:V\to W$ satisfying the condition
+>
+> 1. $f(u+v)=f(u)+f(v)$
+> 2. $f(\lambda v)=\lambda f(v)$
+
+A linear functional is a linear map from $V$ to $\mathbb{F}$.
+
+#### Definition of bilinear functional
+
+A bilinear functional is a bilinear function $\beta:V\times W\to \mathbb{F}$ satisfying the condition that $v\to \beta(v,w)$ is a linear functional for all $w\in W$ and $w\to \beta(v,w)$ is a linear functional for all $v\in V$.
+
+The vector space of all bilinear functionals is denoted by $\mathcal{B}(V,W)$.
+
+#### Definition of tensor product
+
+Let $V,W$ be two vector spaces.
+
+Let $V'$ and $W'$ be the dual spaces of $V$ and $W$, respectively, that is $V'=\{\psi:V\to \mathbb{F}\}$ and $W'=\{\phi:W\to \mathbb{F}\}$, $\psi, \phi$ are linear functionals.
+
+The tensor product of vectors $v\in V$ and $w\in W$ is the bilinear functional defined by $\forall (\psi,\phi)\in V'\times W'$ given by the notation
+
+$$
+(v\otimes w)(\psi,\phi)\coloneqq\psi(v)\phi(w)
+$$
+
+The tensor product of two vector spaces $V$ and $W$ is the vector space $\mathcal{B}(V',W')$
+
+Notice that the basis of such vector space is the linear combination of the basis of $V'$ and $W'$, that is, if $\{e_i\}$ is the basis of $V'$ and $\{f_j\}$ is the basis of $W'$, then $\{e_i\otimes f_j\}$ is the basis of $\mathcal{B}(V',W')$.
+
+That is, every element of $\mathcal{B}(V',W')$ can be written as a linear combination of the basis.
+
+Since $\{e_i\}$ and $\{f_j\}$ are bases of $V'$ and $W'$, respectively, then we can always find a set of linear functionals $\{\phi_i\}$ and $\{\psi_j\}$ such that $\phi_i(e_j)=\delta_{ij}$ and $\psi_j(f_i)=\delta_{ij}$.
+
+Here $\delta_{ij}=\begin{cases}
+1 & \text{if } i=j \\
+0 & \text{otherwise}
+\end{cases}$ is the Kronecker delta.
+
+$$
+V\otimes W=\left\{\sum_{i=1}^n \sum_{j=1}^m a_{ij} \phi_i(v)\psi_j(w): \phi_i\in V', \psi_j\in W'\right\}
+$$
+
+Note that $\sum_{i=1}^n \sum_{j=1}^m a_{ij} \phi_i(v)\psi_j(w)$ is a bilinear functional that maps $V'\times W'$ to $\mathbb{F}$.
+
+This enables basis free construction of vector spaces with proper multiplication and scalar multiplication.
+
+This vector space is equipped with the unique inner product $\langle v\otimes w, u\otimes x\rangle_{V\otimes W}$ defined by
+
+$$
+\langle v\otimes w, u\otimes x\rangle=\langle v,u\rangle_V\langle w,x\rangle_W
+$$
+
+In practice, we ignore the subscript of the vector space and just write $\langle v\otimes w, u\otimes x\rangle=\langle v,u\rangle\langle w,x\rangle$.
+
+> [!NOTE]
+>
+> All those definitions and proofs can be found in Linear Algebra Done Right by Sheldon Axler.
+
+### Notations in measure theory
+
+#### Definition of Sigma algebra
+
+[link to measure theory](../../Math4121/Math4121_L25#definition-of-sigma-algebra)
+
+A collection of sets $\mathcal{A}$ is called a sigma-algebra if it satisfies the following properties:
+
+1. $\emptyset \in \mathcal{A}$
+2. If $\{A_j\}_{j=1}^\infty \subset \mathcal{A}$, then $\bigcup_{j=1}^\infty A_j \in \mathcal{A}$
+3. If $A \in \mathcal{A}$, then $A^c \in \mathcal{A}$
+
+#### Definition of Measure
+
+A measure is a function $v:\mathcal{A}\to \mathbb{R}$ satisfying the following properties:
+
+1. $v(\emptyset)=0$
+2. If $\{A_j\}_{j=1}^\infty \subset \mathcal{A}$ are pairwise disjoint, then $v(\bigcup_{j=1}^\infty A_j)=\sum_{j=1}^\infty v(A_j)$ (countable additivity)
+3. If $A\in \mathcal{A}$, then $v(A)\geq 0$ (non-negativity)
+
+<details>
+<summary>Examples of measure</summary>
+
+The [Borel measure on $\mathbb{R}$](../../Math4121/Math4121_L25#definition-of-borel-measure) is the collection of all closed, open, and half-open intervals with $m(U)=\ell(U)$ for any open set $U$.
+
+The [Lebesgue measure on $\mathbb{R}$](../../Math4121/Math4121_L27#definition-of-lebesgue-measure) is the collection of all Lebesgue measurable sets with $m_i=\sup_{K\text{ closed},K\subseteq S}m(K)$ and $m_e=\inf_{U\text{ open},S\subseteq U}m(U)$. and $m(S)=m_e(S)=m_i(S)$ for any Lebesgue measurable set $S$.
+
+</details>
+
+#### Definition of Probability measure
+
+Let $\mathscr{F}$ be a sigma-algebra on a set $\Omega$. A probability measure is a function $P:\mathscr{F}\to [0,1]$ satisfying the following properties:
+
+1. $P(\Omega)=1$
+2. $P$ is a measure on $\mathscr{F}$
+
+#### Definition of Measurable space
+
+A measurable space is a pair $(X, \mathscr{B}, v)$, where $X$ is a set and $\mathscr{B}$ is a sigma-algebra on $X$.
+
+In some literatures, $\mathscr{B}$ is ignored and we only denote it as $(X, v)$.
+
+<details>
+<summary>Examples of measurable space</summary>
+
+Let $\Omega$ be arbitrary set.
+
+Let $\mathscr{B}(\mathbb{C})$ be the Borel sigma-algebra on $\mathbb{C}$ generated from rectangles over complex plane with real number axes and $\lambda$ be the Lebesgue measure associated with it.
+
+Let $\mathscr{F}$ be the set of square integrable, that is,
+
+$$
+\int_\Omega |f(x)|^2 d\lambda(x)<\infty
+$$
+
+complex-valued functions on $\Omega$, that is, $f:\Omega\to \mathbb{C}$.
+
+Then the measurable space $(\Omega, \mathscr{B}(\mathbb{C}), \lambda)$ is a measurable space. We usually denote this as $L^2(\Omega, \mathscr{B}(\mathbb{C}), \lambda)$.
+
+If $\Omega=\mathbb{R}$, then we denote such measurable space as $L^2(\mathbb{R}, \lambda)$.
+
+<details>
+
+#### Probability space
+
+A probability space is a triple $(\Omega, \mathscr{F}, P)$, where $\Omega$ is a set, $\mathscr{F}$ is a sigma-algebra on $\Omega$, and $P$ is a probability measure on $\mathscr{F}$.
 
 ### Lipschitz function
 
@@ -28,7 +258,7 @@ That basically means that the function $f$ should not change the distance betwee
 
 Basic definitions
 
-### $SO(n)$
+#### $SO(n)$
 
 The special orthogonal group $SO(n)$ is the set of all **distance preserving** linear transformations on $\mathbb{R}^n$.
 

content/Math401/Extending_thesis/Math401_R2.md

@@ -270,9 +270,26 @@ Not very edible for undergraduates.
 
 #### Definition of m-manifold
 
-An $m$-manifold is a [Hausdorff space](../../Math4201/Math4201_L9#hausdorff-space) $X$ with a countable basis such that each point of $x$ of $X$ has a neighborhood [homeomorphic](../../Math4201/Math4201_L10#definition-of-homeomorphism) to an open subset of $\mathbb{R}^m$.
+An $m$-manifold is a [Hausdorff space](../../Math4201/Math4201_L9#hausdorff-space) $X$ with a **countable basis** (second countable) such that each point of $x$ of $X$ has a neighborhood [homeomorphic](../../Math4201/Math4201_L10#definition-of-homeomorphism) to an open subset of $\mathbb{R}^m$.
 
-Example is trivial that 1-manifold is a curve and 2-manifold is a surface.
+<details>
+<summary>Example of second countable space</summary>
+
+Let $X=\mathbb{R}$ and $\mathcal{B}=\{(a,b)|a,b\in \mathbb{R},a<b\}$ (collection of all open intervals with rational endpoints).
+
+Since the rational numbers are countable, so $\mathcal{B}$ is countable.
+
+So $\mathbb{R}$ is second countable.
+
+Likewise, $\mathbb{R}^n$ is also second countable.
+</details>
+
+<details>
+<summary>Example of manifold</summary>
+
+1-manifold is a curve and 2-manifold is a surface.
+
+</details>
 
 #### Theorem of imbedded space
 
@@ -280,10 +297,51 @@ If $X$ is a compact $m$-manifold, then $X$ can be imbedded in $\mathbb{R}^n$ for
 
 This theorem might save you from imagining abstract structures back to real dimension. Good news, at least you stay in some real numbers.
 
-### Smooth manifold
+### Smooth manifolds and Lie groups
 
 > This section is waiting for the completion of book Introduction to Smooth Manifolds by John M. Lee.
 
+#### Partial derivatives
+
+Let $U\subseteq \mathbb{R}^n$ and $f:U\to \mathbb{R}^n$ be a map.
+
+For any $a=(a_1,\cdots,a_n)\in U$, $j\in \{1,\cdots,n\}$, the $j$-th partial derivative of $F$ at $a$ is defined as
+
+$$
+\begin{aligned}
+\frac{\partial f}{\partial x_j}(a)&=\lim_{h\to 0}\frac{f(a_1,\cdots,a_j+h,\cdots,a_n)-f(a_1,\cdots,a_j,\cdots,a_n)}{h} \\
+&=\lim_{h\to 0}\frac{f(a+he_j)-f(a)}{h}
+\end{aligned}
+$$
+
+#### Continuously differentiable maps
+
+Let $U\subseteq \mathbb{R}^n$ and $f:U\to \mathbb{R}^n$ be a map.
+
+If for any $j\in \{1,\cdots,n\}$, the $j$-th partial derivative of $f$ is continuous at $a$, then $f$ is continuously differentiable at $a$.
+
+If $\forall a\in U$, $\frac{\partial f}{\partial x_j}$ exists and is continuous at $a$, then $f$ is continuously differentiable on $U$. or $C^1$ map. (Note that $C^0$ map is just a continuous map.)
+
+#### Smooth maps
+
+A function $f:U\to \mathbb{R}^n$ is smooth if it is of class $C^k$ for every $k\geq 0$ on $U$. Such function is called a diffeomorphism if it is also a **bijection** and its **inverse is also smooth**.
+
+#### Charts
+
+Let $M$ be a smooth manifold. A **chart** is a pair $(U,\phi)$ where $U\subseteq M$ is an open subset and $\phi:U\to \hat{U}\subseteq \mathbb{R}^n$ is a homeomorphism (a continuous bijection map and its inverse is also continuous).
+
+If $p\in U$ and $\phi(p)=0$, then we say that $p$ is the origin of the chart $(U,\phi)$.
+
+#### Atlas
+
+Let $M$ be a smooth manifold. An **atlas** is a collection of charts $\mathcal{A}=\{(U_\alpha,\phi_\alpha)\}_{\alpha\in I}$ such that $M=\bigcup_{\alpha\in I} U_\alpha$.
+
+An atlas is said to be **smooth** if the transition maps $\phi_\alpha\circ \phi_\beta^{-1}:\phi_\beta(U_\alpha\cap U_\beta)\to \phi_\alpha(U_\alpha\cap U_\beta)$ are smooth for all $\alpha, \beta\in I$.
+
+#### Smooth manifold
+
+A smooth manifold is a pair $(M,\mathcal{A})$ where $M$ is a topological manifold and $\mathcal{A}$ is a smooth atlas.
+
 ### Riemannian manifolds
 
 A Riemannian manifold is a smooth manifold equipped with a **Riemannian metric**, which is a smooth assignment of an inner product to each tangent space $T_pM$ of the manifold.

content/Math4121/Math4121_L17.md

@@ -24,7 +24,8 @@ Given a set $S$, the power set of $S$, denoted $\mathscr{P}(S)$ or $2^S$, is the
 
 Cardinality of $2^S$ is not equal to the cardinality of $S$.
 
-Proof:
+<details>
+<summary>Proof of Cantor's Theorem</summary>
 
 Assume they have the same cardinality, then $\exists \psi: S \to 2^X$ which is one-to-one and onto. (this function returns a subset of $S$)
 
@@ -38,7 +39,7 @@ If $b\in T$, then by definition of $T$, $b \notin \psi(b)$, but $\psi(b) = T$, w
 
 If $b \notin T$, then $b \in \psi(b)$, which is also a contradiction since $b\in T$. Therefore, $2^S$ cannot have the same cardinality as $S$.
 
-QED
+</details>
 
 ### Back to Hankel's Conjecture
 

content/Math4121/Math4121_L18.md

@@ -12,13 +12,16 @@ By modifying this example, we can find similar with any outer content between 0
 
 $S\subseteq[0,1]$ is perfect if $S=S'$.
 
-Example:
+<details>
+<summary>Examples of perfect set</summary>
 
 - $[0,1]$ is perfect
   - perfect sets are closed
 - Finite collection of points is not perfect because they do not have limit points.
   - perfect sets are uncountable (no countable sets can be perfect)
 
+</details>
+
 #### Middle third Cantor set
 
 We construct the set by removing the middle third of the interval.
@@ -49,7 +52,8 @@ $$
 
 $C$ is perfect and nowhere dense, and outer content is 0.
 
-Proof:
+<details>
+<summary>Proof</summary>
 
 (i) $c_e(C)=0$
 
@@ -70,3 +74,4 @@ It is sufficient to show $C$ contains no intervals.
 Any open intervals has a real number with 1 in it's base 3 decimal expansion (proof in homework)
 
 _take some interval in $(a,b)$ we can change the digits that is small enough and keep the element still in the set_
+</details>

content/Math4121/Math4121_L19.md

@@ -44,11 +44,12 @@ The outer content of $SVC(n)$ is $\frac{n-3}{n-2}$.
 
 If $S\subseteq T$, then $c_e(S)\leq c_e(T)$.
 
-Proof: 
+<details>
+<summary>Proof of Monotonicity of outer content</summary>
 
 If $C$ is cover of $T$, then $S\subseteq T\subseteq C$, so $C$ is a cover of $S$. Since $c_e(s)$ takes the inf over a larger set that $c_e(T)$, $c_e(S) \leq c_e(T)$.
 
-QED
+</details>
 
 #### Theorem Osgood's Lemma
 

content/Math4121/Math4121_L20.md

@@ -30,7 +30,8 @@ If $S=\bigcup_{n=1}^{\infty} I_n$, $T=\bigcup_{n=1}^{\infty} J_n$, where $I_n$ a
 
 Let $S$ be a closed, bounded set in $\mathbb{R}$, and $S_1\subseteq S_2\subseteq \ldots$, and $S=\bigcup_{n=1}^{\infty} S_n$. Then $\lim_{k\to\infty} c_e(S_k)=c_e(S)$.
 
-Proof:
+<details>
+<summary>Proof of Osgood's Lemma</summary>
 
 Trivial that $c_e(S_k)\leq c_e(S)$.
 
@@ -70,7 +71,7 @@ c_e(S)&\leq c_e(U)\\
 \end{aligned}
 $$
 
-QED
+</details>
 
 ### Convergence Theorems for sequences of functions
 
@@ -96,7 +97,8 @@ $$
 \lim_{n\to\infty}\int_a^b f_n(x)\ dx=\int_a^b f(x)\ dx
 $$
 
-Proof:
+<details>
+<summary>Proof of Arzela-Osgood Theorem (incomplete)</summary>
 
 Define $\Gamma_{\alpha}=\{x:\forall m\in \mathbb{N} \textup{ and }\forall \delta>0, \exists n\geq m \textup{ s.t. } |y-x|<\delta \textup{ and } |f_n(y)-f_m(y)|>\alpha\}$.
 
@@ -105,3 +107,4 @@ _$\Gamma_{\alpha}$ is the negation of $(\alpha,\delta)$ definition of limit._
 $\Gamma_{\alpha}$ is closed and nowhere dense.
 
 Continue on next lecture.
+</details>

content/Math4121/Math4121_L21.md

@@ -20,7 +20,8 @@ $$
 
 Fact: $\Gamma_{\alpha}$ is closed and nowhere dense.
 
-Proof:
+<details>
+<summary>Proof</summary>
 
 Without loss of generality, we can assume $f=0$. Given any $\alpha > 0$, $\exists N$ such that
 
@@ -70,5 +71,5 @@ This implies $\ell(P_2)\leq \frac{\alpha}{4B}$.
 
 Continue on Friday.
 
-QED
+</details>
 

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>

content/Math4121/Math4121_L23.md

@@ -110,7 +110,8 @@ $$
 
 So $S$ is Jordan measurable if and only if $c_e(\partial S)=0$.
 
-Proof:
+<details>
+<summary>Proof</summary>
 
 Let $\epsilon > 0$, and $\{R_j\}_{j=1}^N$ be an open cover of $\partial S$. such that $\sum_{j=1}^N \text{vol}(R_j) < c_e(\partial S)+\frac{\epsilon}{2}$.
 
@@ -136,4 +137,4 @@ If $\eta$ is small enough (depends on $\delta$), then $\mathcal{C}_\eta=\{Q\in K
 
 Suppose $\exists x\in S$ but not in $\mathcal{C}_\eta$. Then $x$ is closed to $\partial S$ so in some $Q_j$. (This proof is not rigorous, but you get the idea. Also not clear in book actually.)
 
-EOP
+</details>

content/Math4121/Math4121_L24.md

@@ -14,7 +14,8 @@ $$
 
 where $\partial S$ is the boundary of $S$ and $c_e(\partial S)=0$.
 
-Example:
+<details>
+<summary>Examples for Jordan measurable</summary>
 
 1. $S=\mathbb{Q}\cap [0,1]$ is not Jordan measurable.
 
@@ -56,6 +57,8 @@ So $c_e(SVC(4))=\frac{1}{2}$.
 
 > General formula for $c_e(SVC(n))=\frac{n-3}{n-2}$, and since $SVC(n)$ is nowhere dense, $c_i(SVC(n))=0$.
 
+</details>
+
 ### Additivity of Content
 
 Recall that outer content is sub-additive. Let $S,T\subseteq \mathbb{R}^n$ be disjoint.
@@ -80,7 +83,8 @@ $$
 c(\bigcup_{i=1}^N S_i)=\sum_{i=1}^N c(S_i)
 $$
 
-Proof:
+<details>
+<summary>Proof</summary>
 
 $$
 \begin{aligned}
@@ -96,7 +100,7 @@ $$
 c(\bigcup_{i=1}^N S_i)=\sum_{i=1}^N c(S_i)
 $$
 
-QED
+</details>
 
 ##### Failure for countable additivity for Jordan content
 

content/Math4121/Math4121_L25.md

@@ -48,7 +48,8 @@ The Borel sets are Borel measurable.
 
 (proof in the following lectures)
 
-Examples:
+<details>
+<summary>Examples for Borel measurable</summary>
 
 1. Let $S=\{x\in [0,1]: x\in \mathbb{Q}\}$
 
@@ -62,6 +63,8 @@ Since $c_e(SVC(4))=\frac{1}{2}$ and $c_i(SVC(4))=0$, it is not Jordan measurable
 
 $S$ is Borel measurable with $m(S)=\frac{1}{2}$. (use setminus and union to show)
 
+</details>
+
 #### Proposition 5.3
 
 Let $\mathcal{B}$ be the Borel sets in $\mathbb{R}$. Then the cardinality of $\mathcal{B}$ is $2^{\aleph_0}=\mathfrak{c}$. But the cardinality of the set of Jordan measurable sets is $2^{\mathfrak{c}}$.

content/Math4121/Math4121_L27.md

@@ -14,7 +14,7 @@ where $I_j$ is an open interval
 
 1. $m_e(I)=\ell(I)$
 2. Countably sub-additive: $m_e\left(\bigcup_{n=1}^\infty S_n\right)\leq \sum_{n=1}^\infty m_e(S_n)$ (Prove today)
-3. does not repect complementation (Build in to Borel measure)
+3. does not respect complementation (Build in to Borel measure)
 
 Why does Jordan content respect complementation?
 
@@ -64,7 +64,8 @@ $$
 m_e\left(\bigcup_{n=1}^\infty S_n\right)\leq\sum_{n=1}^\infty m(S_n)
 $$
 
-Proof:
+<details>
+<summary>Proof</summary>
 
 Let $\epsilon>0$ and for each $j$, let $\{I_{i,j}\}_{i=1}^\infty$ be a cover of $S_j$ s.t.
 
@@ -84,21 +85,22 @@ $$
 m_e\left(\bigcup_{j=1}^\infty S_j\right)\leq\sum_{j=1}^\infty m_e(S_j)=\sum_{j=1}^\infty m(S_j)
 $$
 
-QED
+</details>
 
-#### Corollary
+#### Corollary: inner measure is always less than or equal to outer measure
 
 $$
 m_i(S)\leq m_e(S)
 $$
 
-Proof:
+<details>
+<summary>Proof</summary>
 
 $$
 m_i(S)=m(I)-m_e(I\setminus S)\leq m(I)-m_i(I\setminus S)=m_e(S)
 $$
 
-QED
+</details>
 
 ### Caratheodory's Criterion
 
@@ -110,7 +112,8 @@ $$
 m_e\left(S\cap \left(\bigcup_{j=1}^\infty I_j\right)\right)=m_e\left(\bigcup_{j=1}^\infty (S\cap I_j)\right)=\sum_{j=1}^\infty m_e(S\cap I_j)
 $$
 
-Proof:
+<details>
+<summary>Proof</summary>
 
 For each $j$, let $\{J_i\}_{i=1}^\infty$ be a cover of $S\cap \left(\bigcup_{j=1}^\infty I_j\right)$ such that $\sum_{i=1}^\infty \ell(J_i)<c_e(S\cap \left(\bigcup_{j=1}^\infty I_j\right))+\epsilon$. Since $\{I_j\}_{j=1}^\infty$ are pairwise disjoint, so is $\{J_i\cap I_j\}_{j=1}^\infty$ for each $i$.
 
@@ -132,7 +135,7 @@ $$
 m_e\left(S\cap \left(\bigcup_{j=1}^\infty I_j\right)\right)\leq \sum_{j=1}^\infty m_e(S\cap I_j)
 $$
 
-QED
+</details>
 
 #### Theorem 5.6 (Caratheodory's Criterion)
 

content/Math4121/Math4121_L28.md

@@ -31,7 +31,8 @@ $$
 > $$m_e\left(S\cap \bigcup_{j=1}^{\infty} I_j\right) = \sum_{j=1}^{\infty} m_e(S\cap I_j)$$
 > Proved on Friday
 
-Proof:
+<details>
+<summary>Proof</summary>
 
 $\implies$ If Lebesgue criterion holds for $S$, then for any $X$ of finite outer measure,
 
@@ -72,7 +73,7 @@ m_e(X)&\leq m_e(X\cap S)+m_e(S^c\cap X)\\
 \end{aligned}
 $$
 
-QED
+</details>
 
 ### Revisit Borel's criterion
 
@@ -88,7 +89,8 @@ $$
 m_e(S)=\sum_{j=1}^{\infty} m_e(S_j)
 $$
 
-Proof:
+<details>
+<summary>Proof</summary>
 
 First we prove $m_e(\bigcup_{j=1}^{\infty} S_j)=\sum_{j=1}^{\infty} m(S_j)$ by induction.
 
@@ -116,16 +118,17 @@ Therefore, $\sum_{j=1}^{\infty} m(S_j)\leq m_e(S)\leq \sum_{j=1}^{\infty} m(S_j)
 
 So $S$ is measurable.
 
-QED
+</details>
 
 #### Proposition 5.9 (Preview)
 
 Any finite union (and intersection) of measurable sets is measurable.
 
-Proof:
+<details>
+<summary>Proof</summary>
 
 Let $S_1, S_2$ be measurable sets.
 
 We prove by verifying the Caratheodory's criteria for $S_1\cup S_2$.
 
-QED
+</details>

content/Math4121/Math4121_L29.md

@@ -34,7 +34,8 @@ Towards proving $\mathfrak{M}$ is closed under countable unions:
 
 Any finite union/intersection of Lebesgue measurable sets is Lebesgue measurable.
 
-Proof:
+<details>
+<summary>Proof</summary>
 
 Suppose $S_1, S_2$ is a measurable, and we need to show that $S_1\cup S_2$ is measurable. Given $X$, need to show that
 
@@ -61,13 +62,14 @@ $$
 
 by measurability of $S_1$ again.
 
-QED
+</details>
 
 #### Theorem 5.10 (Countable union/intersection of Lebesgue measurable sets is Lebesgue measurable)
 
 Any countable union/intersection of Lebesgue measurable sets is Lebesgue measurable.
 
-Proof:
+<details>
+<summary>Proof</summary>
 
 Let $\{S_j\}_{j=1}^{\infty}\subset\mathfrak{M}$. Definte $T_j=\bigcup_{k=1}^{j}S_k$ such that $T_{j-1}\subset T_j$ for all $j$.
 
@@ -109,7 +111,7 @@ Therefore, $m_e(X\cap S)=m_e(X)$.
 
 Therefore, $S$ is measurable.
 
-QED
+</details>
 
 #### Corollary from the proof
 

content/Math4121/Math4121_L30.md

@@ -20,7 +20,8 @@ $$
 m_i(S)=\sup_{K\text{ closed},K\subseteq S}m(K)
 $$
 
-Proof:
+<details>
+<summary>Proof</summary>
 
 Inner regularity:
 
@@ -34,7 +35,7 @@ $$
 
 So $m_i(S)<m(K)+\epsilon$. Since $\epsilon$ is arbitrary, $m_i(S)\leq m_e(S)$.
 
-QED
+</details>
 
 We can approximate $m(S)$ from outside by open sets. If we are just concerned with "approximating" $m(S)$, we can use finite union of intervals.
 
@@ -58,7 +59,8 @@ $$
 
 where $U=\bigcup_{j =1}^n I_j$.
 
-Proof:
+<details>
+<summary>Proof</summary>
 
 Let $\epsilon>0$ and $m(V)<m(S)+\frac{\epsilon}{2}$. Let $K\subseteq S$ be closed set such that $m(S)-\frac{\epsilon}{2}<m(K)$. $V$ is an open cover of closed and bounded set $K$. By Heine-Borel theorem, $K$ has a finite subcover. Let $I_1,I_2,\cdots,I_n$ be the open intervals in the subcover.
 
@@ -68,7 +70,7 @@ $$
 m(S\Delta U)=m(S\setminus U)+m(U\setminus S)\leq m(S\setminus K)+m(U\setminus S)<\epsilon
 $$
 
-QED
+</details>
 
 Recall $\{T_j\}_{j=1}^\infty$ are disjoint measurable sets. Then $T=\bigcup_{j=1}^\infty T_j$ is measurable and
 

content/Math429/Math429_L25.md

@@ -23,17 +23,17 @@ Some properties
 
 #### Definition 6.2
 
-An inner product $<,>:V\times V\to \mathbb{F}$
+An inner product $\langle,\rangle:V\times V\to \mathbb{F}$
 
-Positivity: $<v,v>\geq 0$
+Positivity: $\langle v,v\rangle\geq 0$
 
-Definiteness: $<v,v>=0\iff v=0$
+Definiteness: $\langle v,v\rangle=0\iff v=0$
 
-Additivity: $<u+v,w>=<u,w>+<v,w>$
+Additivity: $\langle u+v,w\rangle=\langle u,w\rangle+\langle v,w\rangle$
 
-Homogeneity: $<\lambda u, v>=\lambda<u,v>$
+Homogeneity: $\langle \lambda u, v\rangle=\lambda\langle u,v\rangle$
 
-Conjugate symmetry: $<u,v>=\overline{<v,u>}$
+Conjugate symmetry: $\langle u,v\rangle=\overline{\langle v,u\rangle}$
 
 Note: the dot product on $\mathbb{R}^n$ satisfies these properties
 
@@ -43,39 +43,39 @@ $V=C^0([-1,-])$
 
 $L_2$ - inner product.
 
-$<f,g>=\int^1_{-1} f\cdot g$
+$\langle f,g\rangle=\int^1_{-1} f\cdot g$
 
-$<f,f>=\int ^1_{-1}f^2\geq 0$
+$\langle f,f\rangle=\int ^1_{-1}f^2\geq 0$
 
-$<f+g,h>=<f,h>+<g,h>$
+$\langle f+g,h\rangle=\langle f,h\rangle+\langle g,h\rangle$
 
-$<\lambda f,g>=\lambda<f,g>$
+$\langle \lambda f,g\rangle=\lambda\langle f,g\rangle$
 
-$<f,g>=\int^1_{-1} f\cdot g=\int^1_{-1} g\cdot f=<g,f>$
+$\langle f,g\rangle=\int^1_{-1} f\cdot g=\int^1_{-1} g\cdot f=\langle g,f\rangle$
 
 The result is in real vector space so no conjugate...
 
 #### Theorem 6.6
 
-For $<,>$ an inner product
+For $\langle,\rangle$ an inner product
 
-(a) Fix $V$, then the map given by $u\mapsto <u,v>$ is a linear map (Warning: if $\mathbb{F}=\mathbb{C}$, then $u\mapsto<u,v>$ is not linear).
+(a) Fix $V$, then the map given by $u\mapsto \langle u,v\rangle$ is a linear map (Warning: if $\mathbb{F}=\mathbb{C}$, then $u\mapsto\langle u,v\rangle$ is not linear).
 
-(b,c) $<0,v>=<v,0>=0$
+(b,c) $\langle 0,v\rangle=\langle v,0\rangle=0$
 
-(d) $<u,v+w>=<u,v>+<u,w>$ (second terms are additive.)
+(d) $\langle u,v+w\rangle=\langle u,v\rangle+\langle u,w\rangle$ (second terms are additive.)
 
-(e) $<u,\lambda v>=\bar{\lambda}<u,v>$
+(e) $\langle u,\lambda v\rangle=\bar{\lambda}\langle u,v\rangle$
 
 #### Definition 6.4
 
-An **inner product space** is a pair of vector space and inner product on it. $(v,<,>)$. In practice, we will say "$V$ is an inner product space" and treat $V$ as the vector space.
+An **inner product space** is a pair of vector space and inner product on it. $(v,\langle,\rangle)$. In practice, we will say "$V$ is an inner product space" and treat $V$ as the vector space.
 
 For the remainder of the chapter. $V,W$ are inner product vector spaces...
 
 #### Definition 6.7
 
-For $v\in V$ the **norm of $V$** is given by $||v||:=\sqrt{<v,v>}$
+For $v\in V$ the **norm of $V$** is given by $||v||:=\sqrt{\langle v,v\rangle}$
 
 #### Theorem 6.9
 
@@ -86,13 +86,13 @@ Suppose $v\in V$.
 
 Proof:
 
-$||\lambda v||^2=<\lambda v,\lambda v> =\lambda<v,\lambda v>=\lambda\bar{\lambda}<v,v>$
+$||\lambda v||^2=\langle \lambda v,\lambda v\rangle =\lambda\langle v,\lambda v\rangle=\lambda\bar{\lambda}\langle v,v\rangle$
 
-So $|\lambda|^2 <v,v>=|\lambda|^2||v||^2$, $||\lambda v||=|\lambda|\ ||v||$
+So $|\lambda|^2 \langle v,v\rangle=|\lambda|^2||v||^2$, $||\lambda v||=|\lambda|\ ||v||$
 
 #### Definition 6.10
 
-$v,u\in V$ are **orthogonal** if $<v,u>=0$.
+$v,u\in V$ are **orthogonal** if $\langle v,u\rangle=0$.
 
 #### Theorem 6.12 (Pythagorean Theorem)
 
@@ -102,9 +102,9 @@ Proof:
 
 $$
 \begin{aligned}
-    ||u+v||^2&=<u+v,u+v>\\
+    ||u+v||^2&=\langle u+v,u+v\rangle\\
-    &=<u,u+v>+<v,u+v>\\
+    &=\langle u,u+v\rangle+\langle v,u+v\rangle\\
-    &=<u,u>+<u,v>+<v,u>+<v,v>\\
+    &=\langle u,u\rangle+\langle u,v\rangle+\langle v,u\rangle+\langle v,v\rangle\\
     &=||u||^2+||v||^2
 \end{aligned}
 $$