Topological Equivalence of Norms

Question

Proposition: Prove that $d,d'$ generate the same topology on M if and only if $\forall x \in M$ and $\forall r>0$ $\exists$ $r_1,r_2$ such that $B^{d'}(x,r_1)$ $\subseteq$ $B^{d}(x,r)$ and $B^{d}(x,r_2)$ $\subseteq$ $B^{d'}(x,r)$.

I was able to prove the above proposition. The problem I want to solve is the following:

Define a metric $d'$ on $\mathbb{R}^n$ by $d'(x,y)=$ $max_{i=1,2...,n}$ $\{$ $|x_i-y_i |$ $\}$. Show that the euclidean metric and $d'$ generate the same topology.

I know that for any $x,y \in \mathbb{R}^n$, $|x_i-y_i|$ $\leq$ $||x-y||$, consequently $max|x_i-y_i|$ $\leq$ ||x-y||,.

Now, I'm trying to find an $r_1>0$ so that $B^{d'}(x,r)$ $\subseteq$ $B^{d}(x,r)$ where $r>0$ is arbitrary.

May I have some hints which is something other than the equivalence of the metrics, as I know (and was able to prove) the equivalence of the metrics, please?

Hint: $|x-y|{max} \leq |x-y|{eucl} \leq \sqrt{n}|x-y|_{max}$. — Aphelli, Sep 24 '19 at 18:52
Relevant IMHO. For both parts. $d_\infty$ is your $d'$ etc. and the Euclidean metric is $d_2$ there. — Henno Brandsma, Sep 24 '19 at 21:13

score 1 · Answer 1 · answered Sep 24 '19 at 18:53

1

Hint: Use the inequality: $$|x_j|\le\sqrt{\sum_{1\le j\le n}^nx_j ^2}\le\sqrt{n}\max_{1\le j\le n}|x_j|.$$ May be try to prove it first and, then apply for you problem appropriately.

answered Sep 24 '19 at 18:53

Bumblebee

18,220
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score 0 · Answer 2 · answered Sep 24 '19 at 18:54

0

$ΗΙΝΤ$

In $\Bbb{R}^d,$ the open ball centered at $x$, with respect to the maximum metric is an $n-$ dimensional cube centered at $x$ and with edge width $2r$

answered Sep 24 '19 at 18:54

Marios Gretsas

21,906

Topological Equivalence of Norms

2 Answers2