$D(x, y) = d (x, y) + \mid \frac{1}{d(x, A^c)} - \frac{1}{d(y, A^c)}\mid$. Then show that $D$ and $d$ are equivalent on $A$.

Question

Let $(X,d)$ is a metric space. $A \subset X$ and $A \neq \phi$. Let $D$ be a metric defined by $D(x, y) = d (x, y) + \mid \frac{1}{d(x, A^c)} - \frac{1}{d(y, A^c)}\mid$. Then show that $D$ and $d$ are equivalent on $A$.

TO show equivalence of two metrics we have to show there exists constants $c,d$ such that $c\ d(x,y) \leq D(x,y) \leq d\ d(x,y)$. One side is trivial that $c = 1$ but finding the constant $d$ is where I am stuck.

Please Help.

Answer 1

Some hints: Suppose $A$ open in $(X,d)$. Show that $D$ defines a metric on $A$ (triangular inequality etc...).

Second you need to show that for any $a\in A$ an open ball for one of the metrics (centered at $a$) contains an open ball (centered at $a$) for the other metric.

The difficult part is when $B_D(a,R)\subset A$. For this it may be helpful to first show that $|d(x,A^c)-d(y,A^c)|\leq d(x,y)$ for $x,y\in A$. Also show that if $d(x,A^c)<s$ and $d(y,A^c)<s$ then you have: $$ D(x,y) \leq d(x,y) \left( 1 + \frac{1}{s^2} \right) $$