Show that if \(d\) is a metric for \(X\), then
\[ d'(x, y) = d(x, y) / (1 + d(x, y)) \]
is a bounded metric that gives the topology of \(X\). [Hint: If \(f(x) = x/(1+x)\) for \(x > 0\), use the mean-value theorem to show that \(f(a+b) - f(b) \leq f(a)\).]
Solution
Let \(f(t)=t/(1+t)=1-1/(1+t)\) for \(t\ge0\), so that \(d'=f\circ d\); \(f\) is increasing, \(f(0)=0\), and \(0\le f<1\).
1. \(f(a+b)\le f(a)+f(b)\) for \(a,b\ge0\).
\[ f(a+b)=\frac a{1+a+b}+\frac b{1+a+b}\le\frac a{1+a}+\frac b{1+b}. \]
2. \(d'\) is a metric bounded by \(1\).
\(d'\ge0\), with \(d'(x,y)=0\) if and only if \(d(x,y)=0\), that is, \(x=y\); \(d'\) is symmetric because \(d\) is; and \(d'<1\) because \(f<1\). Since \(f\) is increasing, step 1 gives \[ d'(x,z)=f(d(x,z))\le f(d(x,y)+d(y,z))\le d'(x,y)+d'(y,z). \]
3. For \(0<r\le\frac12\) and every \(x\), \(B_d(x,r)\subseteq B_{d'}(x,r)\) and \(B_{d'}(x,r)\subseteq B_d(x,2r)\).
\(d'\le d\) gives the first inclusion. If \(d'(x,y)<r\le\frac12\), then \(d(x,y)=d'(x,y)/(1-d'(x,y))<r/(1-r)\le2r\).
4. Q.E.D.
By step 3, every ball of either metric about a point contains a ball of the other metric about that point, so \(d\) and \(d'\) have the same open sets; step 2 shows that \(d'\) is a bounded metric.