Defining a metric space I'm studying for actuarial exams, but I always pick up mathematics books because I like to challenge myself and try to learn new branches.  Recently I've bought Topology by D. Kahn and am finding it difficult.  Here is a problem that I think I'm am answering sufficiently but any help would be great if I am off.

If $d$ is a metric on a set $S$, show that 
$$d_1(x,y)=\frac{d(x,y)}{1+d(x,y)}$$ is a metric on $S$.

The conditions for being a metric are $d(X,Y)\ge{0}, d(X,Y)=0$ iff $X=Y$, $d(X,Y)=d(Y,X)$, and $d(X,Y)\le{d(X,Z)+d(Z,Y)}$.  Thus, we simply go axiom by axiom.

1)  Since both $d(x,y)\ge{0}$ and $1+d(x,y)\ge{0},$ it is clear that $d_1(x,y)\ge{0}$.  (Is this a sufficient analysis?)
2) $d_1(x,x)=\frac{d(x,x)}{1+d(x,x)}=\frac{0}{1+0}=0$.
3) $d_1(x,y)=\frac{d(x,y)}{1+d(x,y)}=\frac{d(y,x)}{1+d(y,x)}=d_1(y,x).$
4) $d_1(x,y)=\frac{d(x,y)}{1+d(x,y)}\le{\frac{d(x,z)+d(z,y)}{1+d(x,z)+d(z,y)}}=\frac{d(x,z)}{1+d(x,z)+d(z,y)}+\frac{d(z,y)}{1+d(x,z)+d(z,y)}\lt\frac{d(x,z)}{1+d(x,z)}+\frac{d(z,y)}{1+d(z,y)}=d_1(x,z)+d_1(z,y).$

However, #4 is strictly less, not less than or equal to, according to my analysis, so where did I go wrong?
There is someing wrong in 4), just as Brian comments. Here I offered a proof for you:


  Proof: Notice that $f(x)=\frac{x} {1+x}$ is increasing on $\mathbb R^+$: to see this, let $g(x)=\frac{1}{x+1}$. It is easily to see that $g(x)$ is descreasing on $\mathbb R^+$. And note that $f(x)+g(x)=1$. Therefore, $f(x)$ is a increasing function on $\mathbb R^+$.


Since $d(x,y) \le d(x,z)+d(z,y)$, we have $d_1(x,y)=\frac{d(x,y)}{1+d(x,y)}\le{\frac{d(x,z)+d(z,y)}{1+d(x,z)+d(z,y)}}=\frac{d(x,z)}{1+d(x,z)+d(z,y)}+\frac{d(z,y)}{1+d(x,z)+d(z,y)}\lt\frac{d(x,z)}{1+d(x,z)}+\frac{d(z,y)}{1+d(z,y)}=d_1(x,z)+d_1(z,y).$

Hope this be helpful for you.



ADDed: $d(z,y)$ and $d(x,z)$ could be zero.
Yeah, you forgot that z=y. Remember, no where is it stated that each of these points are different. This will give the in equality that you are looking for.