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.
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).
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 $B_{r_2}^{(d)}(x) \subseteq B_r^{(d')}(x)$.
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 Clearly these $c_1$ and $c_2$ exist? Correct me if I'm wrong I'm not ready for the test either dw.