Two metrics \(d\) and \(d'\) on a set \(X\) are said to be metrically equivalent if the identity map \(i: (X, d) \to (X, d')\) and its inverse are both uniformly continuous.
Show that \(d\) is metrically equivalent to the standard bounded metric \(\bar{d}\) derived from \(d\).
Show that if \(d\) and \(d'\) are metrically equivalent, then \(X\) is complete under \(d\) if and only if it is complete under \(d'\).
Solution
- Recall that the standard bounded metric is \[ \bar d(x,y)=\min\{d(x,y),1\}. \] The identity \[ i:(X,d)\to(X,\bar d) \] is uniformly continuous because \(\bar d\le d\): given \(\varepsilon>0\), take \(\delta=\varepsilon\).
For the inverse identity, given \(\varepsilon>0\), choose \[ \delta=\tfrac12\min\{\varepsilon,1\}. \] If \(\bar d(x,y)<\delta<1\), then necessarily \(d(x,y)<1\), so \(\bar d(x,y)=d(x,y)<\delta\le\varepsilon\). Thus the inverse identity is uniformly continuous. Hence \(d\) and \(\bar d\) are metrically equivalent.
- Suppose \(d\) and \(d'\) are metrically equivalent. A uniformly continuous map sends Cauchy sequences to Cauchy sequences. Since both identity maps \[ (X,d)\rightleftarrows(X,d') \] are uniformly continuous, a sequence is \(d\)-Cauchy if and only if it is \(d'\)-Cauchy.
Assume \((X,d)\) is complete and let \((x_n)\) be \(d'\)-Cauchy. Then it is \(d\)-Cauchy, hence \(x_n\to x\) in \(d\) for some \(x\in X\). Continuity of the identity \((X,d)\to(X,d')\) gives \(x_n\to x\) in \(d'\). Thus \((X,d')\) is complete. The converse is symmetric.