NoteNextra-origin/content/Math4201/Math4201_L13.md

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)

Proof

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.

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.