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:

$\forall B_1\in \mathcal{T}, \exists B_2\in \mathcal{T}'$ such that $\forall x\in B_1, \exists x\in B_2\subseteq B_1$ .
$\forall B_2\in \mathcal{T}', \exists B_1\in \mathcal{T}$ 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.

$\forall x\in X, \forall \delta>0, \exists \epsilon>0$ such that $B_\delta(x)\subseteq B_\epsilon(x)$
$\forall x\in X, \forall \epsilon>0, \exists \delta>0$ 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.