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is the following true: if (X,d) is a complete metric space.let $A\subset$X .If (A,d) is also complete,then (A,$d_1$) is complete iff $d_1$ is equivalent to d or is the above claim true in only the direction that (A,$d_1$) is complete if $d_1$ is equivalent to d.Any help would be appericiated..

Abhishek Shrivastava
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1 Answers1

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It's not true. Consider $X=A=\Bbb R$, $d(x,y)=\lvert x-y\rvert$ and $d_1(x,y)=\begin{cases}1&\text{if }x\ne y\\0&\text{if }x= y\end{cases}$

$(\Bbb R,d)$ and $(\Bbb R,d_1)$ are non-homeomorphic complete metric spaces.