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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?

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    $\begingroup$ You meant to ask "show that... also is a metric distance ", right? $\endgroup$ – DonAntonio Apr 9 '13 at 2:31
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    $\begingroup$ If you prove that $a<b$, then you’ve certainly proved that $a\le b$. However, your calculation in (4) is incorrect: how do you justify the claim that $$\frac{d(x,y)}{1+d(x,y)}\le\frac{d(x,z)+d(z,y)}{1+d(x,z)+d(z,y)}\;?$$ The numerator on the right is at least as big as the numerator on the left, but the denominators aren’t the same. $\endgroup$ – Brian M. Scott Apr 9 '13 at 2:32
  • $\begingroup$ It looks correct: $\frac{a}{1+a} \leq \frac{b}{1+b}$ whenever $0 \leq a \leq b$. $\endgroup$ – Lord Soth Apr 9 '13 at 2:34
  • $\begingroup$ @LordSoth: Yes, but it still ought to be justified; the justification isn’t hard, but it isn’t one of the obvious inequalities that everyone recognizes right off the bat. $\endgroup$ – Brian M. Scott Apr 9 '13 at 2:35
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    $\begingroup$ There is a little slip in 2), the conditon says $d(x,y)=0$ iff (if and only if) $x=y$. You verified only one direction. $\endgroup$ – André Nicolas Apr 9 '13 at 2:43
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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.

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  • $\begingroup$ how does that show $\frac{d(x,y)}{1+d(x,y)}\le{\frac{d(x,z)+d(z,y)}{1+d(x,z)+d(y,z)}}$? It looked trivial when I did it originally, so i offered no justification? Is it because if $a<b$ then $\frac{1}{1+a}>\frac{1}{1+b}$? Which, i assume by your proof, implies that $\frac{a}{1+a}<\frac{b}{1+b}$? $\endgroup$ – Eleven-Eleven Apr 9 '13 at 3:51
  • $\begingroup$ @CHristopher: Since $f(x)$ is a increasing function, just as i pointed out above, then $f(d(x,y))\le f(d(x,z)+d(z,y))$. $\endgroup$ – Paul Apr 9 '13 at 3:55
  • $\begingroup$ @ChristopherErnst: You are welcome. $\endgroup$ – Paul Apr 9 '13 at 4:02
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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.

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