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.


  • $\begingroup$ 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. $\endgroup$ – 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 $$\frac{d(x,z)}{1+d(x,z)}+\frac{d(y,z)}{1+d(y,z)}\geq \frac{d(x,z)}{1+d(x,z)+d(y,z)}+\frac{d(y,z)}{1+d(x,z)+d(y,z)} \hspace{21cm}= \frac{d(x,z)+d(y,z)}{1+d(x,z)+d(x,z)}\geq \frac{d(x,y)}{1+d(x,y)} \cdots (\star)$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.