Let $X$ be a set and $d,d^\prime$ two metrics on $X$ generating the same topology. Is it true that there exists positive constants $\alpha,\beta$ such that $$ \alpha d^\prime(x,y) \le d(x,y) \le \beta d^\prime(x,y) $$ for all $x,y \in X$?
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Let $X = \mathbb{R}$ and $d$ be the euclidean metric. Define another metric on $X$ by $d'(x,y) = \min\{1,d(x,y)\}$. You should prove that $d'$ is in fact a metric, and that it generates the same topology for $X$ as $d$. Suppose there existed a $\beta$ such that $d(x,y) \leqslant \beta d'(x,y)$ for all $x,y \in X$. Then $d(x,y) \leqslant \beta \cdot 1 = \beta$ for all $x,y \in X$ which is certainly not true.
joeb
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For what its worth, $d'$ is called the standard bounded metric corresponding to $d$. Here is a thread which discusses why it is a metric and induces the same topology as $d$: http://math.stackexchange.com/questions/275841/standard-bounded-metric-induces-same-topology – joeb Mar 23 '17 at 18:43
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I know that there are different notions of "equivalent metric spaces" (e.g., homeomorphic spaces, existence of a uniformly continuous bijection, or bilipschitz bijection), which coincide in case the metric is induced by a norm. However, the aim of this question is that I was searching a sufficient condition which ensures the above inequality, for some $\alpha,\beta>0$. – Paolo Leonetti Mar 23 '17 at 18:47