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