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Let $(X,d)$ be a complete metric space and $U$ be an open subset , $A:=X \setminus U$ , define a metric on $U$ as

$$D(x,y)=d(x,y)+ \left|\frac1{\operatorname{dist}(x,A)}-\frac 1{\operatorname{dist}(y,A)}\right| , \forall x,y \in U$$ note that the distances are not zero as

$x \in U$ so $x \notin A=\bar A$ ; then is $(U,D)$ complete as a metric space ? Is $(U,D)$ equivalent with the

induced $d$ metric $(A,d)$ ?

  • Please notice my edits including the proper use of \setminus. ${}\qquad{}$ – Michael Hardy Mar 01 '15 at 15:48
  • What equivalence relation of metrics are you asking about? For example, are you asking whether $d$ and $D$ generate the same topology? – Lee Mosher Mar 01 '15 at 16:02
  • @LeeMosher : Yes , I am asking about topological equivalence i.e. whether the identity map is homeomorphic or not ... –  Mar 02 '15 at 11:00

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the map $U \to \mathbb{R}$, $x\mapsto dist(x,A)$ is continuous and non vanishing over $U$ ...so $1/dist(x,A)$ is also continuous on $U$ ... given any $r>0$ and fix $x \in U$ there exists $\delta >0$ s.t whenever $d(x-u)$ <$\delta$/2 then |$1/dist(x,A) - 1/dist(u,A)$|<$r/2$... consider $k= min ${$\delta,r$} ...then $D(x,y) < r$ if $d(x,y)< k$ (followed directly from the hypothysis)...on the other hand if $D(x,y)< r$ then $d(x,y)<r$ ...since $x$ was fixed but an arbitrary element of $U$ so $D$ and $d$ are equivalent metric on $U$.

Anubhav Mukherjee
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