181 lines
5.2 KiB
Markdown
181 lines
5.2 KiB
Markdown
# Math4201 Topology I (Lecture 12)
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## Metric spaces
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### Basic properties and definitions
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#### Definition of metric space
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A metric space is a set $X$ with a function $d:X\times X\to \mathbb{R}$ that satisfies the following properties:
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1. $\forall x,y\in X, d(x,y)\geq 0$ and $d(x,y)=0$ if and only if $x=y$. (positivity)
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2. $\forall x,y\in X, d(x,y)=d(y,x)$. (symmetry)
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3. $\forall x,y,z\in X, d(x,z)\leq d(x,y)+d(y,z)$. (triangle inequality)
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<details>
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<summary>Example of metric space</summary>
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Let $X=\mathbb{R}$ and $d(x,y)=|x-y|$.
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Check definition of metric space:
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1. Positivity: $d(x,y)=|x-y|\geq 0$ and $d(x,y)=0$ if and only if $x=y$.
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2. Symmetry: $d(x,y)=|x-y|=|y-x|=d(y,x)$.
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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 $a,b\in \mathbb{R}$.
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---
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Let $X$ be arbitrary. The trivial metric is $d(x,y)=\begin{cases}
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0 & \text{if } x=y \\
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1 & \text{if } x\neq y
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\end{cases}$
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Check definition of metric space:
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1. Positivity: $d(x,y)=\begin{cases}
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0 & \text{if } x=y \\
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1 & \text{if } x\neq y
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\end{cases}\geq 0$ and $d(x,y)=0$ if and only if $x=y$.
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1. Symmetry: $d(x,y)=\begin{cases}
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0 & \text{if } x=y \\
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1 & \text{if } x\neq y
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\end{cases}=d(y,x)$.
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1. Triangle inequality use case by case analysis.
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</details>
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#### Balls of a metric space forms a basis for a topology
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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 $B_r(x)=\{y\in X:d(x,y)<r\}$.
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Goal: Show that the balls of a metric space forms a basis for a topology.
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$$
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\{B_r(x)|x\in X,r>0,r\in \mathbb{R}\}\text{ is a basis for a topology on }X
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$$
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<details>
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<summary>Example of balls of a metric space</summary>
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Let $X=\mathbb{R}$ and $d(x,y)=\begin{cases}
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0 & \text{if } x=y \\
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1 & \text{if } x\neq y
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\end{cases}$
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The balls of this metric space are:
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$$
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B_r(x)=\begin{cases}
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\{x\} & \text{if } r<1 \\
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X & \text{if } r\geq 1
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\end{cases}
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$$
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> [!NOTE]
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>
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> This basis generate the discrete topology of $X$.
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---
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Let $X=\mathbb{R}$ and $d(x,y)=|x-y|$.
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The balls of this metric space are:
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$$
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B_r(x)=\{(x-r,x+r)\}
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$$
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This basis is the set of all open sets in $\mathbb{R}$, which generates the standard topology of $\mathbb{R}$.
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</details>
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<details>
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<summary>Proof</summary>
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Let's check the two properties of basis:
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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)
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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 $z\in B_r(z)\subseteq B_r(x)\cap B_r(y)$.
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Observe that for any $z\in B_r(x)$, then there exists $\delta>0$ such that $B_\delta(z)\subseteq B_r(x)$.
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Let $\delta=r-d(x,z)$, then $B_\delta(z)\subseteq B_r(x)$ (by triangle inequality)
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Similarly, there exists $\delta'>0$ such that $B_\delta'(z)\subseteq B_r(y)$.
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Take $\lambda=min\{\delta,\delta'\}$, then $B_\lambda(z)\subseteq B_r(x)\cap B_r(y)$.
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</details>
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#### Definition of Metric topology
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For any metric space $(X,d)$, the topology generated by the balls of the metric space is called metric topology.
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#### Definition of metrizable
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A topological space $(X,\mathcal{T})$ is metrizable if it is the metric topology for some metric $d$ on $X$.
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> Q: When is a topological space metrizable?
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#### Lemma: Every metric topology is Hausdorff
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If a topology isn't Hausdorff, then it isn't metrizable.
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<details>
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<summary>Example of non-metrizable space</summary>
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Trivial topology **with at least two points** is not Hausdorff, so it isn't metrizable.
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---
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Finite complement topology on infinite set is not Hausdorff.
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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.
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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)
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So $X$ with finite complement topology is not Hausdorff, so it isn't metrizable.
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</details>
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<details>
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<summary>Proof</summary>
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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 $B_r(x)\cap B_r'(y)=\emptyset$.
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Take $r=r'=\frac{1}{2}d(x,y)$, then $B_r(x)\cap B_r'(y)=\emptyset$. (by triangle inequality)
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We prove this by contradiction.
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Suppose $\exists z\in B_r(x)\cap B_r'(y)$, then $d(x,z)<r$ and $d(y,z)<r'$.
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Then $d(x,y)\leq d(x,z)+d(z,y)<r+r'=\frac{1}{2}d(x,y)+\frac{1}{2}d(x,y)=d(x,y)$. (contradiction $d(x,y)<d(x,y)$)
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Therefore, $X$ is Hausdorff.
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</details>
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### Other metrics on $\mathbb{R}^n$
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Let $\mathbb{R}^n$ be the set of all $n$-tuples of real numbers with standard topology.
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Let $d: \mathbb{R}^n\times \mathbb{R}^n\to \mathbb{R}$ be defined by (the Euclidean distance)
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$$
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d(u,v)=\sqrt{\sum_{i=1}^n (u_i-v_i)^2}
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$$
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In $\mathbb{R}^2$ the ball is a circle.
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Let $\rho(u,v)=\max_{i=1}^n |u_i-v_i|$. (Square metric)
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In $\mathbb{R}^2$ the ball is a square.
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Let $m(u,v)=\sum_{i=1}^n |u_i-v_i|$. (Manhattan metric)
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In $\mathbb{R}^2$ the ball is a diamond.
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#### Lemma: Square metric, Manhattan metric, and Euclidean metric are well defined metrics on $\mathbb{R}^n$
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Proof ignored. Hard part is to show the triangle inequality. May use Cauchy-Schwarz inequality.
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