From ffcf2d90ebb3951df1090a73f2e70670c9514f47 Mon Sep 17 00:00:00 2001
From: Zheyuan Wu <60459821+Trance-0@users.noreply.github.com>
Date: Wed, 24 Sep 2025 11:52:40 -0500
Subject: [PATCH] updates

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+# Math4201 Topology I (Lecture 13)
+
+## Metic spaces
+
+### Three different metrics on $\mathbb{R}^n$
+
+Euclidean metric:
+
+$$
+d(x,y)=\sqrt{\sum_{i=1}^n (x_i-y_i)^2}
+$$
+
+Square metric:
+
+$$
+d(x,y)=\max_{i=1}^n |x_i-y_i|
+$$
+
+Manhattan metric:
+
+$$
+d(x,y)=\sum_{i=1}^n |x_i-y_i|
+$$
+
+So to prove our proposition, we need to show that any pair of metrics $d$ and $d'$ with basis generated by balls defined
+
+$$
+\mathcal{B}=\{B_r^{(d)}(x)|x\in X,r>0,r\in \mathbb{R}\}
+$$
+
+and
+
+$$
+\mathcal{B}'=\{B_r^{(d')}(x)|x\in X,r>0,r\in \mathbb{R}\}
+$$
+
+are equivalent.
+
+#### Proposition: The metrics induce the same topology on $\mathbb{R}^n$
+
+The three metrics induce the same topology on $\mathbb{R}^n$, and it's the standard topology.
+
+#### Lemma of equivalent topologies
+
+If $\mathcal{T}$ and $\mathcal{T}'$ are two topologies on $X$, we say $\mathcal{T}$ and $\mathcal{T}'$ are equivalent to each other if and only if the following two conditions are satisfied:
+
+1. $\forall B_1\in \mathcal{T}, \exists B_2\in \mathcal{T}'$ such that $\forall x\in B_1, \exists x\in B_2\subseteq B_1$.
+2. $\forall B_2\in \mathcal{T}', \exists B_1\in \mathcal{T}$ such that $\forall x\in B_2, \exists x\in B_1\subseteq B_2$.
+
+#### Lemma of equivalent metrics
+
+Let $d$ and $d'$ be two metrics on $X$. If the following holds, then the metric topology associated to $d$ and $d'$ are equivalent.
+
+1. $\forall x\in X, \forall \delta>0, \exists \epsilon>0$ such that $B_\delta(x)\subseteq B_\epsilon(x)$
+2. $\forall x\in X, \forall \epsilon>0, \exists \delta>0$ such that $B_\epsilon(x)\subseteq B_\delta(x)$
+
+<details>
+<summary>Proof</summary>
+
+To apply the lemma, we try to compute the three metrics on $\mathbb{R}^n$.
+
+$u=(u_1,u_2,\dots,u_n), v=(v_1,v_2,\dots,v_n)\in \mathbb{R}^n$
+
+For Euclidean metric:
+
+$$
+d(u,v)=\sqrt{\sum_{i=1}^n (u_i-v_i)^2}
+$$
+
+For square metric:
+
+$$
+\rho(u,v)=\max_{i=1}^n |u_i-v_i|
+$$
+
+For Manhattan metric:
+
+$$
+m(u,v)=\sum_{i=1}^n |u_i-v_i|
+$$
+
+**First** we will show that $d$ and $\rho$ are equivalent.
+
+Note that
+
+$$
+\max_{i=1}^n |u_i-v_i|\leq \sqrt{\sum_{i=1}^n (u_i-v_i)^2}
+$$
+
+So $\forall u\in B_r^{(d)}(x), d(u,v)<r\implies \rho(u,v)<r$
+
+So $u\in B_r^{(\rho)}(x)$.
+
+$$
+B_r^{(d)}(u)\subseteq B_r^{(\rho)}(u)
+$$
+
+Note that
+
+$$
+\sqrt{\sum_{i=1}^n (u_i-v_i)^2}\leq \sqrt{\sum_{i=1}^n max\{|u_i-v_i|\}^2}=\sqrt{n}\times max\{|u_i-v_i|\}
+$$
+
+So $\forall u\in B_{r/\sqrt{n}}^{(\rho)}(x), \rho(u,v)<r/\sqrt{n}\implies d(u,v)<r$
+
+So $u\in B_r^{(d)}(x)$.
+
+So $B_{r/\sqrt{n}}^{(\rho)}(x)\subseteq B_r^{(d)}(x)$.
+
+> imagine two square capped circle inside
+
+**Then we will show** that $\rho$ and $m$ are equivalent.
+
+Observing that
+
+$$
+\max_{i=1}^n |u_i-v_i|\leq \sum_{i=1}^n |u_i-v_i|\leq n\times \max_{i=1}^n |u_i-v_i|
+$$
+
+Then we have
+
+$$
+B_{r/n}^{(\rho)}(x)\subseteq B_r^{(m)}(x)\subseteq B_n^{(\rho)}(x)
+$$
+
+> imagine two square capped a diamonds inside
+
+**Finally**, we will show that the topology generated by the square metric is the same as the product topology on $\mathbb{R}^n$.
+
+Recall the basis for the product topology on $\mathbb{R}^n$ with standard topology.
+
+$$
+\mathcal{B}=\{(a_1,b_1)\times (a_2,b_2)\times \cdots \times (a_n,b_n)|a_i,b_i\in \mathbb{R},a_i<b_i\}
+$$
+
+Let $x=(x_1,x_2,\dots,x_n)\in (a_1,b_1)\times (a_2,b_2)\times \cdots \times (a_n,b_n)$.
+
+Let $\delta=\min_{i=1}^n \{|x_i-a_i|,|x_i-b_i|\}$.
+
+Then $x\in B_\delta^{(\rho)}(x)\subseteq (a_1,b_1)\times (a_2,b_2)\times \cdots \times (a_n,b_n)$.
+
+In the other direction, let $x\in B_r^{(\rho)}(x)$, $x=(x_1,x_2,\dots,x_n)$.
+
+Then $B_r^{(\rho)}(x)\subseteq (x_1-r,x_1+r)\times (x_2-r,x_2+r)\times \cdots \times (x_n-r,x_n+r)$.
+
+This is an element of $\mathcal{B}$, by combining the two directions, we have the standard topology is the same as the topology generated by the square metric.
+
+</details>
+
+#### Proposition of metric induced product topology
+
+Let $(X,d),(Y,d')$ be two metric spaces with metric topology $\mathcal{T},\mathcal{T}'$. On $X\times Y$, we can define a metric $\rho$ by $\rho((x,y),(x',y'))\coloneqq \max\{d(x,x'),d'(y,y')\}$, $(x,y),(x',y')\in X\times Y$.
+
+Then this metric topology on $X\times Y$ is the same as the product topology on $X\times Y$.
+
+> [!NOTE]
+>
+> Product of metrizable topological spaces is metrizable.
+
diff --git a/content/Math4201/_meta.js b/content/Math4201/_meta.js
index baf4154..d236f84 100644
--- a/content/Math4201/_meta.js
+++ b/content/Math4201/_meta.js
@@ -15,4 +15,5 @@ export default {
     Math4201_L10: "Topology I (Lecture 10)",
     Math4201_L11: "Topology I (Lecture 11)",
     Math4201_L12: "Topology I (Lecture 12)",
+    Math4201_L13: "Topology I (Lecture 13)",
 }