$\frac{d}{1+d}$ is a metric

Question

Suppose $(X,d)$ is a metric space, then question is to prove that $\rho(x,y)=\frac{d(x,y)}{1+d(x,y)}$ is a metric.

This question is from Munkres Topology book.

Hint : If $f(x)=\frac{x}{1+x}$ for $x>0$ us mean value theorem to show that $f(a+b)-f(b)\leq $ some wierd symbol which i assume to be $f(a)$..

Suppose we prove that $f(a+b)-f(b)\leq f(a)$ then $f(a+b)\leq f(a)+f(b)$. We then have $$\frac{a+b}{1+a+b}\leq \frac{a}{1+a}+\frac{b}{1+b}$$

We have $f'(x)> 0$ so, $f$ is increasing...

We have $d(x,y)\leq d(x,z)+d(z,y)$.. So, we have $f(d(x,y))\leq f(d(x,z)+d(z,y))$ i.e., $$\frac{d(x,y)}{1+d(x,y)}\leq \frac{d(x,z)+d(z,y)}{1+d(x,y)+d(z,y)}$$

With $$\frac{a+b}{1+a+b}\leq \frac{a}{1+a}+\frac{b}{1+b}$$ we have $$\rho(x,y)=\frac{d(x,y)}{1+d(x,y)}\leq \frac{d(x,z)+d(z,y)}{1+d(x,y)+d(z,y)}\leq \frac{d(x,z)}{1+d(x,z)}+\frac{d(z,y)}{1+d(z,y)}=\rho(x,z)+\rho(z,y)$$

So, all that is left to prove is that $f(a+b)-f(b)\leq f(a)$..

Suppose that $a<b$.. By mean value theorem, there exists $c\in (b,a+b)$ such that $$\frac{f(a+b)-f(b)}{(a+b)-b}=f'(c)=\frac{1}{1+c^2}$$ i.e., $$f(a+b)-f(b)=\frac{a}{(1+c)^2}\leq \frac{a}{(1+a)^2}\leq \frac{a}{1+a}=f(a)$$

Last inequality holds because $a<c$

Suppose that $b<a$.. By mean value theorem, there exists $c\in (a,a+b)$ such that $$\frac{f(a+b)-f(a)}{(a+b)-a}=f'(c)=\frac{1}{1+c^2}$$ i.e., $$f(a+b)-f(a)=\frac{b}{(1+c)^2}\leq \frac{b}{(1+b)^2}\leq \frac{b}{1+b}=f(b)$$

Last inequality holds because $b<c$..

So, in any case we have $f(a+b)\leq f(a)+f(b)$ and we are done..

I just want to know if there are any gaps in this solution..

@JimmyR. : I agree that this question has been asked before.. I want to know if what i have done is correct.. — , Feb 19 '16 at 19:10
And so the question Show that db(x,y)=d(x,y)1+d(x,y) is a metric — hamid kamali, Feb 19 '16 at 19:12
@hamidkamali I agree that this question has been asked before.. I want to know if what i have done is correct. — , Feb 19 '16 at 19:13
We have $f(x)=\frac{x}{1+x}$ and you have written $\frac{a}{(1+a)^2}=f(a)$ and $\frac{b}{(1+b)^2}=f(b)$. can you explain what did you mean? — hamid kamali, Feb 19 '16 at 19:22
@hamidkamali : Thanks for pointing out an error.. See the edited version — , Feb 19 '16 at 19:28
It looks okay to me but I didn't go over it with a fine tooth comb. For style points you should point out the you are explicitely only trying to prove the triangle inequality. (All the other conditions are either trivial or very easy.) — fleablood, Feb 19 '16 at 19:31

score 1 · Accepted Answer · answered Feb 19 '16 at 19:19

To be honest I could not follow your proof and hopefully somebody else might help.

Just to let you know that you can generalize your work to n (natural number)

$\rho(x,y)=\frac{d(x,y)}{n+d(x,y)}$ is also metric and proof is simple.

First 3 property of metric space are trivial (non-negativity, zero dist and symmetry) and I think you already did it. Now you need to prove the following:

$\frac{d(x,y)}{n+d(x,y)}\leq \frac{d(x,z)}{n+d(x,z)}+\frac{d(y,z)}{n+d(y,z)}$

let a=d(x,y) ; b=d(x,z) ; c=d(y,z). So accordingly you want to prove:

$0\leq \frac{b}{n+b}+\frac{c}{n+c}-\frac{a}{n+a}$

If you set denominators equal (i don't know the exact term) you will get:

$0\leq \frac{n^2(b+c-a)+2bcn+abc}{(n+b)(n+c)(n+a)}$

Since all a,b,c are non-negative and (b+c-a) is non-negative as well due to triangular inequality in base distance metric, $\rho(x,y)$ is a valid metric as well.

Good luck.

$\frac{d}{1+d}$ is a metric

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