Deciding whether two metrics are topologically equivalent Let $(M,d)$ be a metric space. Define $d_1(x,y) = \sqrt{d(x,y)}$ so that $d_1$ is a metric. Is $d_1$ necessarily topologically equivalent to $d$?
I'm inclined to say it it is, but am unsure how to show this. I know to be topologically equivalent any open $d$-open set in M must also be $d_1$-open. Any help would be appreciated. 
 A: Let $U$ be open for the $d$-metric. This says that for any $x \in U$, there exists a positive real $\epsilon$ such that the ball of radius $\epsilon$ is contained in $U$ for the $d$-metric. We would like to show that $U$ is open for the $d_1$-metric. Given $x \in U$, let $\epsilon$ be as above. Then the ball of radius $\epsilon^2$ in the $d_1$-metric is equal to the ball of radius $\epsilon$ in the $d$-metric, so in particular is contained in $U$.
You also need to show the other implication (something open in the $d_1$-metric is open in the $d$-metric), which follows from similar logic (but with one change at the end).
A: More generally there is the following usefull proposition. We say that two metrics are equivalent if the two induced topologies are equal.

Let $d$ and $d'$ be two metrics on a set $M$. Then $d$ and $d'$ are
  equivalent if and only if the following condition is satisfied: for
  every $x \in M$ and every $r > 0$ there exist $r_1,r_2 > 0$ such that
  $B_{r_1}^{(d')}(x) \subseteq B_r^{(d)}(x)$ and $B_{r_2}^{(d)}(x)
 \subseteq B_r^{(d')}(x)$.

A: It is surely, because you can treat it like showing an equivalence of norms. 
ie. $d_1$, $d$ equivalent if there exist $c_1, c_2$ such that $c_1d_1<d<c_2d_1$ which is true if $0<c_1<\sqrt{\frac{1}{d(x,y)}}$ and $\sqrt{\frac{1}{d(x,y)}}<c_2<\inf$. 
Clearly these $c_1$ and $c_2$ exist? Correct me if I'm wrong I'm not ready for the test either dw.
