First, we show that $d'$ is a bounded metric on $X$ . Using the hint, on $\overline{\mathbb{R}}_{+}$ , $f(x)=1-\tfrac{1}{1+x}:\overline{\mathbb{R}}_{+}\rightarrow[0,1)$ is well-defined, nonnegative, zero at $x=0$ only, strictly increasing and bounded. Hence, $d'=f\circ d:X\times X\rightarrow[0,1)$ is well-defined, nonnegative, zero iff $d$ is zero iff $x=y$ , symmetric and bounded. So, we only need to show the triangle inequality. For $a,b\ge0$ , $f(a+b)-f(b)=$ $\tfrac{a}{(1+b)(1+a+b)}\le$ $\tfrac{a}{1+a}=f(a)$ . Using this, and the facts that $d'=f\circ d$ and $f$ is increasing, $d'(x,y)+d'(y,z)\ge$ $f(d(x,y)+d(y,z))\ge d'(x,z)$ . So, $d'$ is a metric.

Now, we show that $d'$ induces the same topology as $d$ . Since $f$ and $f^{-1}(y)=\tfrac{y}{1-y}:[0,1)\rightarrow\overline{\mathbb{R}}_{+}$ are continuous, $d'=f\circ d$ and $d=f^{-1}\circ d'$ , $d'$ is continuous in the $d$ -topology (using Exercise 3(a)), and $d$ is continuous in the $d'$ -topology, implying that the topologies are the same (Exercise 3(b)).