# Metric Space Triangle Inequality

Let $(X, d)$ be a metric space. Define $d_0(x, y) := d(x, y)/(1 + d(x, y))$ for all x, y ∈ X.

i) Prove that $d_0$ is also a metric on $X$.

I assume it suffices to verify that the axioms of Non-negativity, definiteness, symmetry and the triangle inequality.

I have had a bit of trouble proving that $d_0$ satisfies the triangle inequality.

This is what I have so far, I'm pretty new to topology so I'm not sure if it is right:

$2 \geq 1$

$\dfrac{1+d(p,r)}{1+d(p,r)} + \dfrac{1+d(q,r)}{1+d(q,r)}\geq \dfrac{1+d(p,q)}{1+d(p,q)}$

$\dfrac{d(p,r)}{1+d(p,r)} + 1+d(p,r) +\dfrac{d(q,r)}{1+d(q,r)} + 1+d(q,r)\geq \dfrac{d(p,q)}{1+d(p,q)} + 1+d(p,q)$

Since $d$ is a metric on $X$, $d(p,r) + d(q,r) \geq d(p,q)$.

Thus $\dfrac{d(p,r)}{1+d(p,r)}+\dfrac{d(q,r)}{1+d(q,r)}\geq \dfrac{d(p,q)}{1+d(p,q)}$

and $d_0(p,r) + d_0(r,q) \geq d_0(p,q)$ as required.

Cheers.

• You may generalize to the case of $d_0(x,y):=f(d(x,y))$, where $f\colon [0,\infty)\to[0,\infty)$ is concave and $f(0)=0$. In my opinion, the proof for the more general situation is more straightforward. – Hagen von Eitzen Dec 8 '16 at 16:19

Let $f(x)=\frac{x}{1+x}$ which is increasing for $x\geq0$.
Now $d(x,z)+d(z,y)\geq d(x,y)\Rightarrow f(d(x,z)+d(z,y))\geq f(d(x,y))\cdots (\star)$
now $$\frac{d(x,z)}{1+d(x,z)}+\frac{d(y,z)}{1+d(y,z)}\geq \frac{d(x,z)}{1+d(x,z)+d(y,z)}+\frac{d(y,z)}{1+d(x,z)+d(y,z)} \hspace{21cm}= \frac{d(x,z)+d(y,z)}{1+d(x,z)+d(x,z)}\geq \frac{d(x,y)}{1+d(x,y)} \cdots (\star)$$