Is 'det' a metric space?

Question

suppose d is a metric on space X. I want to know that is det which define as follow a metric space or not? $$det(x,y)=\frac{d(x,y)}{1+(d(x,y))^2}$$ hint: I know that $d^{\prime}(x.y)=\frac{d(x,y)}{1+d(x,y)}$ is a metric space

Answer 1

No, it's not a metric in general, because the triangle inequality doesn't necessarily hold. Consider $\Bbb R$ with $d$ the standard metric, and see what happens when one number is very far from the two others according to $d$ (say the numbers $1,2$ and $1000$, for instance).

Answer 2

The trick is to note that the function $x \mapsto \frac {x}{x + 1}$ is concave over $[0,\infty)$ while $x \mapsto \frac{x}{x^2 + 1}$ is not.

The answer is that det, as you've defined it, will not be a metric (assuming that $X$ is unbounded) since it will fail the triangle inequality.

Interestingly, $x \mapsto \frac{x}{x^2 + 1}$ is concave over $[0,\sqrt{3}]$, which means that for a space $X$ with sufficiently small diameter, the function det will define a metric after all.