Difference between metric space and norms space?

Question

I just heard the conclusion that a normed space must be a metric space, but not vise versa.

Is this right? I just consider that the norm of x can be described as d(x,0), and vice versa, so when we have a metric, we have a norm. I really want a proof! Any suggestions shall be appreciated!

Have you tried to verify that $d(x,0)$ actually satisfies the definition of a norm? — Lee Mosher, Mar 24 '20 at 12:22

score 2 · Accepted Answer · answered Mar 24 '20 at 10:56

Take the discrete metric on $\mathbb{R}$, i.e.

$$d(x,y)=\left\{\begin{array} {cc} 0 &\text{if }x=y\\ 1&\text{otherwise}\end{array}\right..$$

Then, if you set $||x||=d(x,0)$, you have to check for example the homogeneity, namely do we have $||\lambda x||=|\lambda|\cdot||x||$ for all $\lambda\in\mathbb{R}$? The answer is no, since taking $x=1$ and $\lambda=2$ leads to

$$||\lambda x||=d(\lambda x,0)=d(2,0)=1\neq 2=|2|\cdot d(1,0)=|\lambda|\cdot||x||.$$

score 1 · Answer 2 · answered Mar 24 '20 at 13:09

In general, there is following statement: a metric $\rho(x,y)$ defined on a normed linear space $E$ is generated by a norm $\|x-y\|$ if and only if $\forall x,y,z\in E$ $\rho(x+z,y+z)=\rho(x,y)$ and $\forall x\in E$ $\forall\lambda\in \mathbb{C}$ $\rho(\lambda x,0)=|\lambda|\rho(x,0)$.

Difference between metric space and norms space?

2 Answers2