Obviously it has to satisfy the following:
1) For all $x,y\in X$, $0\le d(x,y)$. (positivity)
2) For all $x,y\in X$, $d(x,y)=d(y,x)$. (symmetry)
3) For all $x,y,z\in X$, $d(x,y)\le d(x,z)+d(z,y)$. (triangle-inequality)
This is a homework problem and I'm not sure where to even start. I'm new to the concept of metric spaces and would appreciate any help/direction.
If x=y, then $\rho(x,y)=\rho(x,x)=(x-x)^2=0$. I'm assuming that will suffice for (1).
Best Answer
$d(x,y)=(x-y)^2$
checking for 3rd axiom:
$$(x-y)^2+(y-z)^2 =x^2 -2xy + y^2 +y^2 -2yz +z^2\\=x^2 -2xz +z^2 +2xz -2xy + y^2 +y^2 -2yz\\=(x-z)^2+2x(z-y)-2y(z-y)\\=(x-z)^2+2(z-y)(x-y) $$ It means $$(x-z)^2=(x-y)^2+(y-z)^2-2(z-y)(x-y)$$ Now to satisfy the third axiom $2(z-y)(x-y)\ge0$ which is not possible say for $x<y<z$.