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Let $(M,d)$ be a metric space and $\rho = d/(1+d)$. $A$ is a subset of $M$. Show that:

  1. If $A$ is an open sphere in $(M,d)$ then $A$ is an open sphere on $(M,\rho)$.

My approach: I used the fact that if we have metrics with equivalent topologies then, the open spheres on one would be open on other too. But my instructor said this isn't the correct way of approaching the problem and I should try it a different way.

  1. Except for one particular subset of $M$, if $A$ is an open sphere on $(M,\rho)$ then, $A$ is an open sphere on $(M,d)$.

How else should I go about this and where am I going wrong?

Sepia Glen
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    Do you mean $\rho = d/(1+d)$ ? – Martin R Sep 29 '21 at 12:38
  • Yes. I just edited it. Sorry – Sepia Glen Sep 29 '21 at 13:00
  • I assume you mean "open ball" instead of "open sphere." In that case you're solution is incorrect because all it gives is that $A$ is open in one metric space if and only if it's open in the other. It does not say anything about $A$ being a ball in either metric. For a complete solution you need to show that $A$ is of the form ${x\in M;|;\rho(x,y)<r}$ for some fixed $r>0$ and $y\in M.$ – D. Brogan Sep 29 '21 at 13:13

1 Answers1

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Assume we have $$A:= \{y \in M : d(x,y) < R\}$$ then notice this can be rewritten, dividing by the positive number $1 + d(x,y)$, as $$A:= \left\{y \in M : \frac{d(x,y)}{1 + d(x,y)}< \frac{R}{1 + d(x,y)}\right\}$$

Now what can you do to reach the form of $\{y \in M : \rho(x,y) < R\}$?

Hint: you know that for all $y \in A$ we have $d(x,y) \ge 0$.

Since $y \in A$ we have $d(x,y) \ge 0$ and so we can write $$A:= \left\{y \in M : \frac{d(x,y)}{1 + d(x,y)}< \frac{R}{1 + d(x,y)}< R\right\}$$ hence $$ A=\{y \in M : \rho(x,y) < R\} $$

Wreck it Ralph
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  • For the second part, can we say that since $\rho$ can't be 1 therefore, we can make the radius as large as we require and for that particular subset A wouldn't be an open ball on $(M,d)$ ? – Sepia Glen Sep 29 '21 at 14:54