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.
Best Answer
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).