Nice Metrics Induce a Norm Theorem

Theorem

1. With $V$ as a vector space, and $d$ a Nice Metric
   1. This means $\lambda \in \mathbb{F}⟹d(\lambda x,\lambda y)=|\lambda |d(x,y)$
   2. This means $x\in V⟹d(x+z,y+z)\le d(x,y)$
2. Then, $||x||=d(x,0)$ is a Norm for $V$

Proof

Showing $||x||\ge 0$

1. Let $x\in V$
2. Then, $||x||=d(x,0)\ge 0$

Showing $||x||=0⟺x=0$

1. Let $x\in V$
2. Then, $||x||=0⟺d(x,0)=0⟺\lambda =0$

Showing $||\lambda x||+||\lambda ||\ast ||x||$

1. Let $\lambda \in \mathbb{F}$
2. Let $x\in V$
3. Then, $||\lambda x||=d(\lambda x,0)$
4. $=d(\lambda x,\lambda 0)=|\lambda |d(x,0)$ by nice metric property
5. $=|\lambda |\ast ||x||$

Showing $||x+y||\le ||x||+||y||$

1. Let $x,y\in V$
2. $||x+y||=d(x+y,0)$
3. $=d(x+y-y,-y)$ by adding $-y$ to both terms, assuming this is allowed
4. $\le d(x,0)+d(0,-y)$
5. $\le d(x,0)+d(y,y-y)$
6. $\le d(x,0)+d(y,0)$ (Note we could also use $|\lambda ||y|$ symmetry)
7. $\le d(x,0)+d(y,0)$
8. $\le ||x||+||y||$