Exercise 20.11

Proof. First, we show that $d^{\prime }$ is a valid metric. Since $d$ is a metric, we have that

Hence clearly $d^{\prime }(x,y)=d(x,y)∕(1+d(x,y))\ge 0$ as well. If $x=y$ then $d(x,y)=0$ so that

Conversely, if $d^{\prime }(x,y)=0$ then it has to be that $d(x,y)=0$ as well since $1+d(x,y)>0$. It then must be that $x=y$ since $d$ is a metric. This shows part (1) of the definition of a metric.

part (2) is easy to show:

since again $d$ is a metric so that $d(x,y)=d(y,x)$.

For part (3) define the function $f(x)=x∕(1+x)$ over the domain $\left\{x\in ℝ\mid x\ge 0\right\}$ so that clearly $d^{\prime }(x,y)=f(d(x,y))$. We first show that $f$ is monotonically increasing. The easiest way to do this is to show that its derivative is always positive, from which monotonicity follows from elementary calculus. Using the quotient rule, we have

Clearly we have $x\ge 0$ so that $1+x\ge 1>0$, and hence ${(1+x)}^2>0^2=0$ by Lemma 1. It then follows that $f^{\prime }(x)=1∕{(1+x)}^2>0$, and thus $f$ is monotonically increasing.

Now, for any $a,b\ge 0$ we clearly have

since $1+a\ge 0$ and $1+a+b$ and $1+b$ are both non-negative so that their product is as well. It then follows that

by what was just shown before since $a\ge 0$. Hence we have $f(a+b)\le f(a)+f(b)$. Since $d$ is a metric, we then of course have

for any $x,y,z\in X$, which of course shows part (3) of the definition. This completes the proof that $d^{\prime }$ is a valid metric.

It is easy to see that $d^{\prime }$ is bounded by $1$:

so that $d^{\prime }$ is a bounded metric.

Now we must show that the metric topologies induced by $d$ and $d^{\prime }$ are the same. First, we show that the topology induced by $d$ is finer than that induced by $d^{\prime }$ using Lemma 20.2. So consider any $x\in X$ and any $𝜖>0$. If $𝜖\ge 1$ set $\delta =999$, otherwise set $\delta =𝜖∕(1-𝜖)$ noting that in this case $𝜖<1$ so that $0<1-𝜖$ and hence $\delta >0$. Now consider any $y\in B_d(x,\delta )$ so that $d(y,x)<\delta$. If $𝜖\ge 1$ then, of course, $d^{\prime }(y,x)<1\le 𝜖$ since we have previously shown the $d^{\prime }$ is bounded by $1$. On the other hand, if $𝜖<1$, then we have

Thus either was we have $d^{\prime }(y,x)<𝜖$ so that $y\in B_{d^{\prime }}(x,𝜖)$, which of course shows that $B_d(x,\delta )\subset B_{d^{\prime }}(x,𝜖)$ since $y$ was arbitrary. Since $𝜖$ was arbitrary this shows the desired result that the topology induced by $d$ is finer than that induced by $d^{\prime }$.

Now we show the other direction, i.e. that the $d^{\prime }$ topology is also finer than the $d$ topology, again using Lemma 20.2. So consider any $x\in X$ and $𝜖>0$ again. This time set $\delta =𝜖∕(1+𝜖)$ noting that $\delta >0$ follows trivially from the fact that $𝜖>0$. Then, for any $y\in B_{d^{\prime }}(x,\delta )$, we have that

so that $y\in B_d(x,𝜖)$. This of course shows that $B_{d^{\prime }}(x,\delta )\subset B_d(x,𝜖)$, which in turn shows that the $d^{\prime }$ topology is finer than the $d$ topology as desired. Since each topology is finer than the other, of course, they must be the same as desired. □