updates

content/CSE5519/CSE5519_C1.md

@@ -58,3 +58,6 @@ using inverse transform sampling, evaluate our "fine" network at the union of th
 >
 > 1. This paper reminds me of Gaussian Splatting. In this paper setting, we can treat the scene as a function of 5D coordinates. (all the cameras are focusing on the world origin) However, in general settings, we have 6D coordinates (3D position and 3D direction). Is there any way to use Gaussian Splatting to reconstruct the scene?
 > 2. In the positional encoding, the function $\gamma(p)$ reminds me of the Fourier transform. Is there any connection between the two?
+### Volume Rendering
+
+Output of color and density.

content/Math4201/Math4201_L4.md

@@ -0,0 +1,151 @@
+# Math4201 Lecture 4
+
+## Recall from last lecture
+
+Assignment due next Thursday. 10PM
+
+Let $\mathcal{B}$ be a basis for a topology. Then the topology ($\mathcal{T}_{\mathcal{B}}$) **generated** by $\mathcal{B}$ is $\{U\in \mathcal{T}_{\mathcal{B}} \mid \forall x\in U, \exists B\in \mathcal{B} \text{ such that } x\in B\subseteq U\}$.
+
+## New materials
+
+### Topology basis
+
+Given a topology on a set $X$, When is a given collection of subsets of $X$ a basis for a topology?
+
+Suppose $U\in\mathcal{T}$ is an open set in $X$. If an arbitrary set $\mathcal{C}$ is a basis for $\mathcal{T}$, then by the definition of a topology generated by a basis, we should have the following:
+
+$$
+\exists C\in \mathcal{C} \text{ such that } x\in C\subseteq U
+$$
+
+#### Theorem of basis of topology
+
+> [!CAUTION]
+>
+> In this course, we use lowercase letters to denote element of a set, and uppercase letters to denote sets. We use $\mathcal{X}$ to denote set of subsets of $X$.
+
+Let $(X,\mathcal{T})$ be a topological space. Let $\mathcal{C}\subseteq \mathcal{T}$ be a collection of subsets of $X$ satisfying the following property:
+
+$$
+\forall U\in \mathcal{T}, \exists C\in \mathcal{C} \text{ such that } U\subseteq C
+$$
+
+Then $\mathcal{C}$ is a basis and the topology generated by $\mathcal{C}$ is $\mathcal{T}$.
+
+<details>
+<summary>Proof</summary>
+
+We want to show that $\mathcal{C}$ is a basis.
+
+> Recall the definition of a basis:
+>
+> 1. $\forall x\in X$, there is $B\in \mathcal{B}$ such that $x\in B$
+> 2. $\forall B_1,B_2\in \mathcal{B}$, $\forall x\in B_1\cap B_2$, there is $B_3\in \mathcal{B}$ such that $x\in B_3\subseteq B_1\cap B_2$
+
+First, we want to show that $\mathcal{C}$ satisfies the first property.
+
+Take $x\in X$. Since $X\in \mathcal{T}$, we can apply the given condition ($
+\forall U\in \mathcal{T}, \exists C\in \mathcal{C} \text{ such that } U\subseteq C
+$) to get $C\in \mathcal{C}$ such that $x\in C\subseteq X$.
+
+Next, we want to show that $\mathcal{C}$ satisfies the second property.
+
+Let $C_1,C_2\in \mathcal{C}$ and $x\in C_1\cap C_2$. Since $C_1,C_2\in \mathcal{T}$, by the definition of $\mathcal{T}$, we have $U=C_1\cap C_2\in \mathcal{T}$. 
+
+We can apply the given condition to get $C_3\in \mathcal{C}$ such that $x\in C_3\subseteq U=C_1\cap C_2$.
+
+---
+
+Then we want to show that the topology generated by $\mathcal{C}$ is $\mathcal{T}$.
+
+> Recall the definition of the topology generated by a basis:
+>
+> To prove this, we need to show that $\forall U\in \mathcal{T}\implies U\in \mathcal{T}_{\mathcal{C}}$ and $\forall U\in \mathcal{T}_{\mathcal{C}}\implies U\in \mathcal{T}$.
+>
+> Moreover, from last lecture, we have $U\in \mathcal{T}_{\mathcal{B}}\iff U=\bigcup_{\alpha \in I} B_\alpha$ for some $\{B_\alpha\}_{\alpha \in I}\subseteq \mathcal{B}$.
+
+First, we want to show that $\forall U\in \mathcal{T}_{\mathcal{C}}\implies U\in \mathcal{T}$.
+
+Let $U=\bigcup_{\alpha \in I} C_\alpha$ for some $\{C_\alpha\}_{\alpha \in I}\subseteq \mathcal{C}$. Then since $C_\alpha\in \mathcal{T}$, by the definition of $\mathcal{T}$, we have $U\in \mathcal{T}$.
+
+Next, we want to show that $\forall U\in \mathcal{T}\implies U\in \mathcal{T}_{\mathcal{C}}$.
+
+Let $U\in \mathcal{T}$. Then $\forall x\in U$ by the given condition, we have $C\in \mathcal{C}$ such that $x\in C\subseteq U$.
+
+So, $U=\bigcup_{\alpha \in I} C_\alpha\in \mathcal{T}_{\mathcal{C}}$. (using the [same trick last time](https://notenextra.trance-0.com/Math4201/Math4201_L3#lemma))
+
+</details>
+
+Let $\mathcal{T}$ be the topology on $X$. Then $\mathcal{T}$ itself satisfies the basis condition.
+
+#### Definition of subbasis of topology
+
+A subbasis of a topology on a set $X$ is a collection $\mathcal{S}\subseteq \mathcal{T}$ of subsets of $X$ such that their union is $X$.
+
+$$
+\mathcal{S}=\{S_{\alpha}\mid S_\alpha\subseteq X\}_{\alpha \in I}\text{  and  }\bigcup_{\alpha \in I} S_\alpha=X
+$$
+
+#### Definition of topology generated by a subbasis
+
+If we consider the basis generated by the subbasis $\mathcal{S}$ by the following:
+
+$$
+\mathcal{B}=\{B\mid B\text{ is the intersection of a finite number of elements of }\mathcal{S}\}
+$$
+
+Then $\mathcal{B}$ is a basis.
+
+<details>
+<summary>Proof</summary>
+
+First, $\forall x\in X$, there is $S_\alpha\in \mathcal{S}$ such that $x\in S_\alpha$. In particular, $x\in \mathcal{B}$.
+
+Second, let $B_1,B_2\in \mathcal{B}$. Since $B_1$ is the intersection of a finite number of elements of $\mathcal{S}$, we have $B_1=\bigcap_{i=1}^n S_{i_1}, B_2=\bigcap_{i=1}^n S_{i_2}$ for some $S_{i_1},S_{i_2}\in \mathcal{S}$.
+
+So $B_1\cap B_2$ is the intersection of finitely many elements of $\mathcal{S}$.
+
+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 $\mathcal{T}_{\mathcal{S}}$.
+
+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 $\mathcal{S}$.
+
+<details>
+
+<summary>Example (standard topology on $\mathbb{R}$)</summary>
+
+Let $X=\mathbb{R}$. Take $\mathcal{S}=\{(-\infty, a)|a\in \mathbb{R}\}\cup \{(a,+\infty)|a\in \mathbb{R}\}$.
+
+We claim this is a subbasis of the standard topology on $\mathbb{R}$.
+
+The basis $\mathcal{B}$ associated with $\mathcal{S}$ is the collection of all open intervals.
+
+$$
+\mathcal{B}=\{(a,b)=(-\infty, b)\cap (a,+\infty)\}
+$$
+
+So, $\mathcal{B}=\mathcal{B}_{st}$ (the standard basis).
+
+This topology on $\mathbb{R}$ is the same as the standard topology on $\mathbb{R}$.
+
+</details>
+
+<details>
+<summary>Example (finite complement topology)</summary>
+
+Let $X$ be an arbitrary set. Let $\mathcal{S}$ defined as follows:
+
+$$
+\mathcal{S}=\{S\subseteq X\mid S=X\setminus \{x\} \text{ for some } x\in X\}
+$$
+
+Let $x,y\in X$ and $x\neq y$. Then $S_x=X\setminus \{x\}$ and $S_y=X\setminus \{y\}$ are two elements of $\mathcal{S}$. Since $x\neq y$, we have $S_x\cup S_y=X\setminus \{x\}\cup X\setminus \{y\}=X$. So $\mathcal{S}$ is a subbasis of $X$.
+
+So, the basis associated with $\mathcal{S}$, $\mathcal{B}$, is the collection of subsets of $X$ with finite complement.
+
+This is in fact a topology, which is the **finite complement topology** on $X$.
+
+</details>

content/Math4201/_meta.js

@@ -6,4 +6,5 @@ export default {
     Math4201_L1: "Topology I (Lecture 1)",
     Math4201_L2: "Topology I (Lecture 2)",
     Math4201_L3: "Topology I (Lecture 3)",
+    Math4201_L4: "Topology I (Lecture 4)",
 }