Topologically equivalent and lipschitz equivalent.

Asked Jun 22 '21 at 17:24

Active Jun 22 '21 at 17:29

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I am looking for an example of a metric that is topologically equivalent but not lipschitz equivalent. Can you help me? I would be very happy if you can give such an example. I thought a lot but could not find it, it is difficult to produce an example. I wish you good work.

That is, $d$ and $d'$ will be topologically equivalent, but lipschitz will not meet the equivalence condition.

Lipschitz equivalent metric: Let $(X,d)$,$(X,d')$ be metric spaces.$\forall x,y\in X$ ve $\exists r,s\in \mathbb{R^+}$,

$r d(x,y)\leq d'(x,y)\leq s d(x,y)$

edited Jun 22 '21 at 17:29

Arctic Char

16,007

asked Jun 22 '21 at 17:24

Dilara Yavuz

Hint: Let $\mathbb{D}^2$ be the unit disk. Let $X = \mathbb{D}^2 -(0,0)$ and $X' = \mathbb{R}^2 - (0,0)$. – Elchanan Solomon Jun 22 '21 at 17:27
Why did you buy two different spaces? I do not understand. @ElchananSolomon – Dilara Yavuz Jun 22 '21 at 19:03
The spaces are homeomorphic (topologically equivalent), so it doesn't make a difference. – Elchanan Solomon Jun 22 '21 at 19:17
I am looking for two metrics. @ElchananSolomon – Dilara Yavuz Jun 22 '21 at 19:19

Topologically equivalent and lipschitz equivalent.

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