From bc18850a8fe42d909d26c90090ad91f724f55609 Mon Sep 17 00:00:00 2001
From: Zheyuan Wu <60459821+Trance-0@users.noreply.github.com>
Date: Mon, 22 Sep 2025 11:52:39 -0500
Subject: [PATCH] updates
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content/CSE5519/CSE5519_A2.md | 5 +
content/CSE5519/CSE5519_E2.md | 3 +
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# CSE5519 Advances in Computer Vision (Topic A: 2022: Semantic Segmentation)
+## Masked-Attention Mask Transformer for Universal Image Segmentation
+
+[link to the paper](https://openaccess.thecvf.com/content/CVPR2022/papers/Cheng_Masked-Attention_Mask_Transformer_for_Universal_Image_Segmentation_CVPR_2022_paper.pdf)
+
+### Novelty in Masked-Attention Mask Transformer
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# CSE5519 Advances in Computer Vision (Topic E: 2022: Deep Learning for Geometric Computer Vision)
+## MeshLoc: Mesh-Based Visual Localization
+
+[link to the paper](https://arxiv.org/pdf/2210.05494)
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+# Math4201 Topology I (Lecture 12)
+
+## Metric spaces
+
+### Basic properties and definitions
+
+#### Definition of metric space
+
+A metric space is a set $X$ with a function $d:X\times X\to \mathbb{R}$ that satisfies the following properties:
+
+1. $\forall x,y\in X, d(x,y)\geq 0$ and $d(x,y)=0$ if and only if $x=y$. (positivity)
+2. $\forall x,y\in X, d(x,y)=d(y,x)$. (symmetry)
+3. $\forall x,y,z\in X, d(x,z)\leq d(x,y)+d(y,z)$. (triangle inequality)
+
+
+Example of metric space
+
+Let $X=\mathbb{R}$ and $d(x,y)=|x-y|$.
+
+Check definition of metric space:
+
+1. Positivity: $d(x,y)=|x-y|\geq 0$ and $d(x,y)=0$ if and only if $x=y$.
+2. Symmetry: $d(x,y)=|x-y|=|y-x|=d(y,x)$.
+3. Triangle inequality: $d(x,z)=|x-z|\leq |x-y|+|y-z|=d(x,y)+d(y,z)$ since $|a+b|\leq |a|+|b|$ for all $a,b\in \mathbb{R}$.
+
+---
+
+Let $X$ be arbitrary. The trivial metric is $d(x,y)=\begin{cases}
+0 & \text{if } x=y \\
+1 & \text{if } x\neq y
+\end{cases}$
+
+Check definition of metric space:
+
+1. Positivity: $d(x,y)=\begin{cases}
+0 & \text{if } x=y \\
+1 & \text{if } x\neq y
+\end{cases}\geq 0$ and $d(x,y)=0$ if and only if $x=y$.
+1. Symmetry: $d(x,y)=\begin{cases}
+0 & \text{if } x=y \\
+1 & \text{if } x\neq y
+\end{cases}=d(y,x)$.
+1. Triangle inequality use case by case analysis.
+
+
+
+#### Balls of a metric space forms a basis for a topology
+
+Let $(X,d)$ be a metric space. $x\in X$ and $r>0, r\in \mathbb{R}$. We define the ball of radius $r$ centered at $x$ as $B_r(x)=\{y\in X:d(x,y)0,r\in \mathbb{R}\}\text{ is a basis for a topology on }X
+$$
+
+
+Example of balls of a metric space
+
+Let $X=\mathbb{R}$ and $d(x,y)=\begin{cases}
+0 & \text{if } x=y \\
+1 & \text{if } x\neq y
+\end{cases}$
+
+The balls of this metric space are:
+
+$$
+B_r(x)=\begin{cases}
+\{x\} & \text{if } r<1 \\
+X & \text{if } r\geq 1
+\end{cases}
+$$
+
+> [!NOTE]
+>
+> This basis generate the discrete topology of $X$.
+
+---
+
+Let $X=\mathbb{R}$ and $d(x,y)=|x-y|$.
+
+The balls of this metric space are:
+
+$$
+B_r(x)=\{(x-r,x+r)\}
+$$
+
+This basis is the set of all open sets in $\mathbb{R}$, which generates the standard topology of $\mathbb{R}$.
+
+
+
+
+Proof
+
+Let's check the two properties of basis:
+
+1. $\forall x\in X$, $\exists B_r(x)\in \{B_r(x)|x\in X,r>0,r\in \mathbb{R}\}$ such that $x\in B_r(x)$. (Trivial by definition of non-zero radius ball)
+2. $\forall B_r(x),B_r(y)\in \{B_r(x)|x\in X,r>0,r\in \mathbb{R}\}$, $\forall z\in B_r(x)\cap B_r(y)$, $\exists B_r(z)\in \{B_r(x)|x\in X,r>0,r\in \mathbb{R}\}$ such that $z\in B_r(z)\subseteq B_r(x)\cap B_r(y)$.
+
+Observe that for any $z\in B_r(x)$, then there exists $\delta>0$ such that $B_\delta(z)\subseteq B_r(x)$.
+
+Let $\delta=r-d(x,z)$, then $B_\delta(z)\subseteq B_r(x)$ (by triangle inequality)
+
+Similarly, there exists $\delta'>0$ such that $B_\delta'(z)\subseteq B_r(y)$.
+
+Take $\lambda=min\{\delta,\delta'\}$, then $B_\lambda(z)\subseteq B_r(x)\cap B_r(y)$.
+
+
+
+#### Definition of Metric topology
+
+For any metric space $(X,d)$, the topology generated by the balls of the metric space is called metric topology.
+
+#### Definition of metrizable
+
+A topological space $(X,\mathcal{T})$ is metrizable if it is the metric topology for some metric $d$ on $X$.
+
+> Q: When is a topological space metrizable?
+
+#### Lemma: Every metric topology is Hausdorff
+
+If a topology isn't Hausdorff, then it isn't metrizable.
+
+
+Example of non-metrizable space
+
+Trivial topology **with at least two points** is not Hausdorff, so it isn't metrizable.
+
+---
+
+Finite complement topology on infinite set is not Hausdorff.
+
+Suppose there exists $x,y\in X$ such that $x\neq y$ and $x\in U\subseteq X$ and $y\in V\subseteq X$ such that $X-U$ and $X-V$ are finite.
+
+Since $U\cap V=\emptyset$, we have $V\subseteq X-U$, which is finite. So $X-V$ is infinite. (contradiction that $X-V$ is finite)
+
+So $X$ with finite complement topology is not Hausdorff, so it isn't metrizable.
+
+
+
+
+Proof
+
+Let $x,y\in (X,d)$ and $x\neq y$. To show that $X$ is Hausdorff, it is suffices to show that there exists $r,r'>0$ such that $B_r(x)\cap B_r'(y)=\emptyset$.
+
+Take $r=r'=\frac{1}{2}d(x,y)$, then $B_r(x)\cap B_r'(y)=\emptyset$. (by triangle inequality)
+
+We prove this by contradiction.
+
+Suppose $\exists z\in B_r(x)\cap B_r'(y)$, then $d(x,z)
+
+### Other metrics on $\mathbb{R}^n$
+
+Let $\mathbb{R}^n$ be the set of all $n$-tuples of real numbers with standard topology.
+
+Let $d: \mathbb{R}^n\times \mathbb{R}^n\to \mathbb{R}$ be defined by (the Euclidean distance)
+
+$$
+d(u,v)=\sqrt{\sum_{i=1}^n (u_i-v_i)^2}
+$$
+
+In $\mathbb{R}^2$ the ball is a circle.
+
+Let $\rho(u,v)=\max_{i=1}^n |u_i-v_i|$. (Square metric)
+
+In $\mathbb{R}^2$ the ball is a square.
+
+Let $m(u,v)=\sum_{i=1}^n |u_i-v_i|$. (Manhattan metric)
+
+In $\mathbb{R}^2$ the ball is a diamond.
+
+#### Lemma: Square metric, Manhattan metric, and Euclidean metric are well defined metrics on $\mathbb{R}^n$
+
+Proof ignored. Hard part is to show the triangle inequality. May use Cauchy-Schwarz inequality.