Let d be a metric on X. Is $d^{2}$ then a metric on X?

Question

I am supposed to solve the following problem.

Let d be a metric on X. Is $d^{2}$ then a metric on X?

I verify the three conditions determining the metric space:

$ \forall x,y,z \in X: d(x,y )\geq 0\Rightarrow \left ( d \left ( x,y \right ) \right )^{2}\geq 0 $
$ \forall x,y,z \in X: d(x,y )=d(y,x )\Rightarrow \left ( d \left ( x,y \right ) \right )^{2}\Rightarrow \left ( d \left ( y,x \right ) \right )^{2}$

Is that correct? How should I verify the triangle inequality?

For the triangular equality, write $a=d(x,y), b=d(y,z)$ and $c=d(x,z)$. You have that $a+b\geqslant c$, does this mean $a^2+b^2\geqslant c^2$? That is what you need to show either way. — Alec Barns-Graham, Feb 16 '20 at 14:16
On the real line, with the usual metric, $d(0,2)=2\le d(0,1)+d(1,2)=1+1$, but $d(0,2)^2=4>d(0,1)^2+d(1,2)^2=1+1=2$. — kimchi lover, Feb 16 '20 at 14:18

score 3 · Accepted Answer · answered Feb 16 '20 at 14:18

3

Consider $\mathbb R$ with the usual metric.

$d(0,1)^2+d(1,3)^2<d(0,3)^2$,

so the triangle inequality does not hold for $d^2$.

answered Feb 16 '20 at 14:18

J. W. Tanner

1 Answers1