Does $d(x,y) = \lvert N(x) - N(y)\rvert$ satisfy the triangular inequality?

Question

Let $N(x)$ be the norm of the vector $X$ and efine

$$d(x,y) = |N(x) - N(y)|$$

I want to prove that $d(x,y)$ satisfies the triangular inequality. Here is my attempt:

$$|N(x) - N(y)| \leq |N(x)| + |N(y)| = $$ $$=|\left[N(x)-N(z)\right] + N(z)| + |\left[N(y)-N(z)\right] + N(z)| \leq $$ $$ \leq |N(x)-N(z)| + |N(z)| + |N(y)-N(z)| + |N(z)| \leq $$ $$ \leq d(x,z) + d(y,z) + 2|N(z)| \leq d(x,z) + d(y,z).$$

Is this right?

Additions Ok, the last line is wrong. Then, can I prove that $d$ satisfies the triangular inequality???

Remark I don't want to prove that $d$ is a metric!

Answer 1

I imagine that you want to show that:

If $N :X\to\mathbb R$ is $1-1$, then $d(x,y)=\lvert N(x)-N(y)\rvert$ is a metric.

Clearly, the only property which needs checking is the triangular inequality.

We have that $$ d(x,z)=\lvert N(x)-N(z)\rvert\le\lvert N(x)-N(y)\rvert+\lvert N(y)-N(z)\rvert=d(x,y)+d(y,z). $$

Note. The fact that $N$ is 1-1 guarantees that $d(x,y)>0$, if $x\ne 0$.