Metric space in which $d(x, y)\neq d(x - y, 0)$?

Question

When working in familiar metric spaces like $\mathbb{R}^n$ (with the Pythagorean distance function), it seems that $d(x, y) = d(x - y, 0)$ for all $x, y\in\mathbb{R}^n$.

Does this always hold in any metric space?

Answer 1

No. There are metric spaces that haven't a "$0$" element. Consider this example:

Let $A$ be an arbitrary set (of whatever you want). Define in $A$ the function $d:A\times A \to [0,\infty[$ by $$d(a,b)=\left\{\begin{array}{c} 0 & \text{if }a=b\\ 1 & \text{otherwise} \end{array}\right.$$ We can show that $(A,d)$ is a metric space.

Choose, for example, $A=\Bbb{N}$ or $A=C^{\infty}([0,1])$ or whatever and have fun.

Answer 2

Counter-example:

Consider the British Rail metric on $\mathbb R$, given by $ d(x,y)=\lVert x\rVert +\lVert y\rVert $ for distinct points $x$ and $y$, and $d(x,x) = 0$. Then,

$$ d(2,1)=\lVert 2\rVert +\lVert 1\rVert = 3 \neq 1 = \lVert 2- 1\rVert +\lVert 0\rVert= d(2-1,0).$$