If$ (X,d)$ is a metric space, is $(X,d^2)$ a metric space?

Question

If $(X,d)$ is a metric space, is $(X,d^2)$ a metric space or not? Thanks a lot.

Answer 1

Take $X=\mathbb{R}$ with $d(x, y)=|x-y|$. Observe that $d^{2}(1, 3)=4$, $d^2(1, 2)=1$ and $d^2(2,3)=1$. Conclude that the $(X, d^2)$ is not a metric space.

Answer 2

I will assume that $d^2(x, y)$ denotes the square of the original distance between $x$ and $y$.

Going down the list of requirements a metric space must satisfy, the only one that is not immediately obvious is the triangle inequality.

We know $d(x, z) \leq d(x, y) + d(y, z)$. Now square both sides. Since both sides are positive, the direction of the inequality is preserved:

$$d^2(x, z) \leq \Big(d(x, y) + d(y, z)\Big)^2 = d^2(x, y) + 2d(x,y)d(y,z) + d^2(y,z)$$

Herein lies a potential problem: to get this into the form we desire, we must subtract $2d(x,y)d(y,z)$ from the right side of the inequality, which might affect the direction of the inequality. With some cleverness, you should be able to think of a metric where this would be problematic.