How to show that a norm space is a metric space?

Question

Given $X$ a set, and $\mid\mid\cdot\mid\mid:X\rightarrow \mathbb{R}$ a function, we define $\mid\mid\cdot\mid\mid$ to be a norm on $X$ following the normed space axioms.

My textbook states that if we let $d(x,y)=\mid\mid x-y \mid\mid$, we can show that this distance is indeed a metric space following the usual metric space axioms.

This is fairly straightforward, but got me thinking - surely I could use $d(x,y)=\mid\mid x+y \mid\mid$ and still be able to prove it is a metric using the usual metric space axioms?

Is there a reason why the textbook suggests to use $d(x,y)=\mid\mid x-y \mid\mid$ and would both be valid metric spaces?

Answer 1

The mapping $$d: (x,y)\mapsto \|x+y\|$$

is not a metric, because it does not satisfy the axiom

$$\forall x\in X:d(x,x)=0$$

(except if $X\neq \{0\}$, but then you only have one metrix on $X$ anyway).

Answer 2

Other answers are pointing to the formal reasons why $d(x,y)=\|x+y\|$ is not a metric. However, I would also suggest that you build your geometrical intuition here.

In geometry, when you have points $A$ and $B$ and the origin $O$ in, say, a plane, you would calculate the vector $\vec{AB}=\vec{OB}-\vec{OA}$ and then you would use the ordinary vector norm to calculate the distance of $A$ to $B$:

$$d(A, B)=\|\vec{AB}\|=\|\vec{OB}-\vec{OA}\|=\|\vec{B}-\vec{A}\|$$

(if we agree to not write $O$ and just identify any point $X$ with the vector $\vec{X}=\vec{OX}$).

That is where the minus sign comes from. It would not make much sense to use the plus sign ($\|\vec{OB}+\vec{OA}\|$). You can calculate with it, but you certainly don't have a clear geometrical intuition behind it.