fix typo and formatting errors

content/Math4201/Exam_reviews/Math4201_E1.md

@@ -1 +1,93 @@
-# Math 4201 Exam 1 review
+# Math 4201 Exam 1 review
+
+> [!NOTE]
+>
+> This is a review for definitions we covered in the classes. It may serve as a cheat sheet for the exam if you are allowed to use it.
+
+The exam will have 5 problems, roughly covering the following types of questions:
+
+- Define concepts from class (e.g. what is the definition of the interior of a set?)
+- Give an example of a space/map which satisfies/does not satisfy a certain property (e.g. give an example of a map that is not continuous.)
+- Proofs from the lectures
+- Homework problems
+- A new problem at the same level of difficulty as homework problems
+
+## Topological space
+
+### Basic definitions
+
+#### Definition for topological space
+
+A topological space is a pair of set $X$ and a collection of subsets of $X$, denoted by $\mathcal{T}$ (imitates the set of "open sets" in $X$), satisfying the following axioms:
+
+1. $\emptyset \in \mathcal{T}$ and $X \in \mathcal{T}$
+2. $\mathcal{T}$ is closed with respect to arbitrary unions. This means, for any collection of open sets $\{U_\alpha\}_{\alpha \in I}$, we have $\bigcup_{\alpha \in I} U_\alpha \in \mathcal{T}$
+3. $\mathcal{T}$ is closed with respect to finite intersections. This means, for any finite collection of open sets $\{U_1, U_2, \ldots, U_n\}$, we have $\bigcap_{i=1}^n U_i \in \mathcal{T}$
+
+#### Definition of open set
+
+$U\subseteq X$ is an open set if $U\in \mathcal{T}$
+
+#### Definition of closed set
+
+$Z\subseteq X$ is a closed set if $X\setminus Z\in \mathcal{T}$
+
+### Basis
+
+#### Definition of topological basis
+
+For a set $X$, a topology basis, denoted by $\mathcal{B}$, is a collection of subsets of $X$, such that the following properties are satisfied:
+
+1. For any $x \in X$, there exists a $B \in \mathcal{B}$ such that $x \in B$ (basis covers the whole space)
+2. If $B_1, B_2 \in \mathcal{B}$ and $x \in B_1 \cap B_2$, then there exists a $B_3 \in \mathcal{B}$ such that $x \in B_3 \subseteq B_1 \cap B_2$ (every non-empty intersection of basis elements are also covered by a basis element)
+
+#### Definition of topology generated by basis
+
+Let $\mathcal{B}$ be a basis for a topology on a set $X$. Then the topology generated by $\mathcal{B}$ is defined by the set as follows:
+
+$$
+\mathcal{T}_{\mathcal{B}} \coloneqq \{ U \subseteq X \mid \forall x\in U, \exists B\in \mathcal{B} \text{ such that } x\in B\subseteq U \}
+$$
+
+> This is basically a closure of $\mathcal{B}$ under arbitrary unions and finite intersections
+
+#### Lemma of topology generated by basis
+
+$U\in \mathcal{T}_{\mathcal{B}}\iff \exists \{B_\alpha\}_{\alpha \in I}\subseteq \mathcal{B}$ such that $U=\bigcup_{\alpha \in I} B_\alpha$
+
+#### Definition of basis generated from a topology
+
+Let $(X, \mathcal{T})$ be a topological space. Then the basis generated from a topology is $\mathcal{C}\subseteq \mathcal{B}$ such that $\forall U\in \mathcal{T}$, $\forall x\in U$, $\exists B\in \mathcal{C}$ such that $x\in B\subseteq U$.
+
+#### Definition of subbasis of topology
+
+A subbasis of a topology is a collection $\mathcal{S}\subseteq \mathcal{T}$ such that $\bigcup_{U\in \mathcal{S}} U=X$.
+
+#### Definition of topology generated by subbasis
+
+Let $\mathcal{S}\subseteq \mathcal{T}$ be a subbasis of a topology on $X$, then the basis generated by such subbasis is the closure of finite intersection of $\mathcal{S}$
+
+$$
+\mathcal{B}_{\mathcal{S}} \coloneqq \{B\mid B\text{ is the intersection of a finite number of elements of }\mathcal{S}\}
+$$
+
+Then the topology generated by $\mathcal{B}_{\mathcal{S}}$ is the subbasis topology denoted by $\mathcal{T}_{\mathcal{S}}$.
+
+Note that all open set with respect to $\mathcal{T}_{\mathcal{S}}$ can be written as a union of finitely intersections of elements of $\mathcal{S}$
+
+### Comparing topologies
+
+#### Definition of finer and coarser topology
+
+Let $(X,\mathcal{T})$ and $(X,\mathcal{T}')$ be topological spaces. Then $\mathcal{T}$ is finer than $\mathcal{T}'$ if $\mathcal{T}'\subseteq \mathcal{T}$. $\mathcal{T}$ is coarser than $\mathcal{T}'$ if $\mathcal{T}\subseteq \mathcal{T}'$.
+
+#### Lemma of comparing basis
+
+Let $(X,\mathcal{T})$ and $(X,\mathcal{T}')$ be topological spaces with basis $\mathcal{B}$ and $\mathcal{B}'$. Then $\mathcal{T}$ is finer than $\mathcal{T}'$ if and only if for any $x\in X$, $x\in B'$, $B'\in \mathcal{B}'$, there exists $B\in \mathcal{B}$, such that $x\in B$ and $x\in B\subseteq B'$.
+
+### Product space
+
+#### Definition of cartesian product
+
+Let $X,Y$ be sets. The cartesian product of $X$ and $Y$ is the set of all ordered pairs $(x,y)$ where $x\in X$ and $y\in Y$, denoted by $X\times Y$.
+

content/Math4201/Math4201_L16.md

@@ -119,7 +119,7 @@ Then $Z$ is a closed subset of $(\operatorname{Map}(X,Y),\rho)$.
 
 We need to show that $\overline{Z}=Z$.
 
-Since $\operatorname{Map}(X,Y)$ is a metric space, this is equivalent to showing that: $f_n:X\to Y\in Z$ continuous,
+Since $\operatorname{Map}(X,Y)$ is a metric space, this is equivalent to showing that: Let $f_n:X\to Y\in Z$ be a sequence of continuous maps,
 
 Which is to prove the uniform convergence,
 
@@ -127,28 +127,33 @@ $$
 f_n \to f \in \operatorname{Map}(X,Y)
 $$
 
-Then we want to show that $f$ is continuous.
+Then we want to show that $f$ is also continuous.
 
-Let $B_r(y)$ be an arbitrary ball in $Y$, it suffices to show that $f^{-1}(B_r(y))$ is open in $X$.
+It is to show that for any open subspace $V$ of $Y$, $f^{-1}(V)$ is open in $X$.
 
-Take $N$ to be large enough such that for $n\geq N$, we have
+Take $x_0\in f^{-1}(V)$, we'd like to show that there is an open neighborhood $U$ of $x_0$ such that $U\subseteq f^{-1}(V)$.
 
-$$
-\rho(f_n(x), f(x)) < \frac{r}{3}
-$$
+Since $x_0\in f^{-1}(V)$, then $f(x_0)\in V$. By metric definition, there is $r>0$ such that $B_r(f(x_0))\subseteq V$.
 
-In particular, this holds for $n=N$. So we have
+Take $N$ to be large enough such $\rho(f_N(x), f(x)) < \frac{r}{3}$
 
-$$
-d(f_N(x), f(x)) < \frac{r}{3},\forall x\in X
-$$
+So $\forall x\in X$, $d(f(x),f_N(x))<\frac{r}{3}$
 
-Take $x_0\in f^{-1}(B_r(y))$, we'd like to show that there is an open ball around $x_0$ in $f^{-1}(B_r(y))$.
+Since $f_N$ is continuous, $f_N^{-1}(B_{r/3}(f(x_0)))$ is an open set $U\subseteq X$ containing $x_0$.
 
-Since $f_N$ is continuous, $f^{-1}_N(B_{\frac{r}{3}}(y))$ is open in $X$.
+Take $x\in U$, $d(f(x),f(x_0))<d(f(x),f_N(x_0))+d(f_N(x),f_N(x_0))+d(f_N(x_0),f(x_0))$ using triangle inequality.
 
-$d(f(x_0), f(x_0))<\frac{r}{3}$
+Note that,
 
-continue the proof in bonus video
+$d(f(x),f_N(x))<\frac{r}{3}$ (using $N$ large enough),
 
+$d(f_N(x),f_N(x_0))<\frac{r}{3}$ (using $x\in U$, then $f_N(x)\in B_{r/3}(f_N(x_0))$, so $d(f_N(x),f_N(x_0))<\frac{r}{3}$),
+
+$d(f_N(x_0),f(x_0))<\frac{r}{3}$ (using $N$ large enough),
+
+So $d(f(x),f(x_0))<\frac{r}{3}+\frac{r}{3}+\frac{r}{3}=r$.
+
+So $f(x)\in B_r(f(x_0))\implies x\in f^{-1}(B_r(f(x_0)))\implies x\in f^{-1}(V)\implies U\subseteq f^{-1}(V)$.
+
+So $f^{-1}(V)$ is open in $X$.
 </details>

content/Math4201/Math4201_L18.md

@@ -1,4 +1,4 @@
-# Math 4201 Topology (Lecture 18)
+# Math4201 Topology I (Lecture 18)
 
 ## Quotient topology
 

content/Math4201/Math4201_L2.md

@@ -20,7 +20,9 @@ The elements of $\mathcal{T}$ are called **open sets**.
 
 The topological space is denoted by $(X, \mathcal{T})$.
 
-#### Examples
+<details>
+
+<summary>Examples of topological spaces</summary>
 
 Trivial topology: Let $X$ be arbitrary. The trivial topology is $\mathcal{T}_0 = \{\emptyset, X\}$
 
@@ -40,12 +42,18 @@ $\mathcal{T}_2 = \{\emptyset, \{a\}, \{a,b\}\}$
 
 $\mathcal{T}_3 = \{\emptyset, \{b\}, \{a,b\}\}$
 
-Non-examples:
+</details>
+
+<details>
+
+<summary>Non-example of topological space</summary>
 
 Let $X=\{a,b,c\}$
 
 The set $\mathcal{T}_1=\{\emptyset, \{a\}, \{b\}, \{a,b,c\}\}$ is not a topology because it is not closed under union $\{a\} \cup \{b\} = \{a,b\} \notin \mathcal{T}_1$
 
+</details>
+
 #### Definition of Complement finite topology
 
 Let $X$ be arbitrary. The complement finite topology is $\mathcal{T}\coloneqq \{U\subseteq X|X\setminus U \text{ is finite}\}\cup \{\emptyset\}$

content/Math4201/Math4201_L3.md

@@ -12,7 +12,7 @@ Let $X$ be a set. A basis for a topology on $X$ is a collection $\mathcal{B}$ (e
 2. $\forall B_1,B_2\in \mathcal{B}$, $\forall x\in B_1\cap B_2$, $\exists B_3\in \mathcal{B}$ such that $x\in B_3\subseteq B_1\cap B_2$
 
 <details>
-<summary>Example 1</summary>
+<summary>Example of standard basis in real numbers</summary>
 
 Let $X=\mathbb{R}$ and $\mathcal{B}=\{(a,b)|a,b\in \mathbb{R},a<b\}$ (collection of all open intervals).
 
@@ -27,7 +27,7 @@ let $B_1=(a,b)$ and $B_2=(c,d)$ be two basis elements, and $x\in B_1\cap B_2=(\m
 </details>
 
 <details>
-<summary>Example 2</summary>
+<summary>Example of lower limit basis in real numbers</summary>
 
 Let $X=\mathbb{R}$ and $\mathcal{B}_{LL}=\{[a,b)|a,b\in \mathbb{R},a<b\}$ (collection of all open intervals).
 
@@ -48,7 +48,7 @@ Extend this to $\mathbb{R}^2$.
 Let $X$ and $Y$ be sets. The cartesian product of $X$ and $Y$ is the set $X\times Y=\{(x,y)|x\in X,y\in Y\}$.
 
 <details>
-<summary>Example 3</summary>
+<summary>Example of open rectangles basis for real plane</summary>
 
 Let $X=\mathbb{R}^2$ and $\mathcal{B}$ be the collection of rectangle of the form $(a,b)\times (c,d)$ where $a,b,c,d\in \mathbb{R}$ and $a<b,c<d$. (boundary is not included)
 
@@ -63,7 +63,7 @@ let $B_1=(a,b)\times (c,d)$ and $B_2=(e,f)\times (g,h)$ be two basis elements, a
 </details>
 
 <details>
-<summary>Example 4</summary>
+<summary>Example of open disks basis for real plane</summary>
 
 Let $X=\mathbb{R}^2$ and $\mathcal{B}$ be the collection of open disks.
 

content/Math4201/Math4201_L4.md

@@ -109,7 +109,7 @@ So $B_1\cap B_2\in \mathcal{B}$.
 
 </details>
 
-We call $\mathcal{B}$ the topology generated by the subbasis $\mathcal{S}$. Denote it by $\mathcal{T}_{\mathcal{S}}$.
+We call the topology generated by $\mathcal{B}$ the topology generated by the subbasis $\mathcal{S}$. Denote it by $\mathcal{T}_{\mathcal{S}}$.
 
 An open set with respect to $\mathcal{T}_{\mathcal{S}}$ is a subset of $X$ such that it can be written as a union of finitely intersections of elements of $\mathcal{S}$.
 

content/Math4201/Math4201_L5.md

@@ -51,7 +51,7 @@ So the discrete topology is finer than the trivial topology.
 Let $\mathcal{T}$ and $\mathcal{T}'$ be topologies on $X$ associated with bases $\mathcal{B}$ and $\mathcal{B}'$. Then
 
 $$
-\mathcal{T}\text{ is finer than } \mathcal{T}'\iff \text{ for any } B'\in \mathcal{B}', \exists B\in \mathcal{B} \text{ such that } B'\subseteq B
+\mathcal{T}\text{ is finer than } \mathcal{T}'\iff \text{ for any } x\in X, x\in B'\in \mathcal{B}', \exists B\in \mathcal{B} \text{ such that } x\in B\subseteq B'
 $$
 
 <details>