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In $\mathbb{R}^n$, we have a conclusion $\overline{B_d(a,r)}=B_d[a,r]$, $d$ is usual metric, here $\overline{B(a,r)}$ means closure of $B(a,r)$, $B(a,r)$ is open ball centered at $a$ with radius $r$. $B[a,r]$ means $ \{x|d(x,a)\leq r\}$.

My question is if we have a metric $d'$ which is topologically equivalent to $d$, do we have the conclusion $\overline{B_{d'}(a,r)}=B_{d'}[a,r]$.

noname1014
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1 Answers1

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Not necessarily. Let $d$ be the usual metric on $\Bbb R$, and for $x,y\in\Bbb R$ let $d'(x,y)=\min\{d(x,y),1\}$. Then for each $x\in\Bbb R$ we have $B_{d'}[x,1]=\Bbb R\ne\operatorname{cl}B_{d'}(x,1)$, but $d'$ is topologically equivalent to $d$.

Brian M. Scott
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