Topology-Preserving Transformations of a Metric

Changing Distances Without Changing Open Sets

A metric records distances, but the topology it induces records which sets are open. Different distance formulas can therefore give the same topology. Two useful transformations of a metric are truncating every distance at one and replacing each distance \(t\) by \(t/(1+t)\). Both change numerical distances, and both produce bounded metrics. We will verify the metric axioms and then compare balls to show that neither transformation changes the open sets.

Let \(d\) be a metric on a set \(X\). Thus \(d(x,y)\) is a nonnegative real number for \(x,y\in X\); \(d(x,y)=0\) exactly when \(x=y\); \(d(x,y)=d(y,x)\); and \(d(x,z)\le d(x,y)+d(y,z)\). For a metric \(d\), the open ball of radius \(r>0\) about \(x\) is \(B_d(x,r)=\{y\in X:d(x,y)<r\}\). A set \(U\subseteq X\) is open in the metric topology if, for every \(x\in U\), some ball \(B_d(x,r)\) is contained in \(U\).

Verifying the Metric Axioms

Proof. For either transformation, the value at \((x,y)\) is nonnegative, is zero when \(x=y\), and is symmetric in \(x,y\), because these properties hold for \(d\). If \(x\ne y\), then \(d(x,y)>0\). Consequently \(\min(d(x,y),1)>0\), and \(d(x,y)/(1+d(x,y))>0\). It remains to prove the triangle inequality for each transformed function.

Write \(a=d(x,y)\) and \(b=d(y,z)\), so \(d(x,z)\le a+b\). For the truncated function, if \(a+b<1\), then \(a<1\) and \(b<1\), and \(\min(d(x,z),1)\le d(x,z)\le a+b=\min(a,1)+\min(b,1)\). If \(a+b\ge1\), then \(\min(a,1)+\min(b,1)\ge1\): if either \(a\) or \(b\) is at least one, its truncated value is one; otherwise their truncated values sum to \(a+b\). Since \(\min(d(x,z),1)\le1\), the triangle inequality follows in this case as well.

For the fractional function, put \(f(t)=t/(1+t)\) for \(t\ge0\). If \(a,b\ge0\), direct subtraction gives \(\displaystyle f(a)+f(b)-f(a+b)=\frac{ab(2+a+b)}{(1+a)(1+b)(1+a+b)}\ge0.\) The denominator is positive, and the numerator is nonnegative. Also, if \(0\le s\le t\), then \(f(t)-f(s)=\frac{t-s}{(1+t)(1+s)}\ge0\), so \(f\) is increasing. Therefore \(f(d(x,z))\le f(a+b)\le f(a)+f(b)\). This is the triangle inequality for \(d_{\mathrm{frac}}\). Both transformed functions satisfy all the metric axioms. \(\square\)

The algebra in the fractional case matters: being increasing alone would not prove the triangle inequality. The additional inequality \(f(a+b)\le f(a)+f(b)\) is what allows the original triangle inequality to pass through this transformation.

Comparing Balls and Open Sets

The ball comparisons below are exact, with the strict inequality in the definition of an open ball retained. For \(d_{\min}\), any radius at most one gives the same ball as \(d\); larger radii give the whole space.

Proof. If \(r\le1\), then \(\min(d(x,y),1)<r\) holds exactly when \(d(x,y)<r\). Indeed, \(d(x,y)<r\) implies the minimum is less than \(r\); conversely, if the minimum is less than \(r\le1\), it cannot equal one, so it must be \(d(x,y)<r\). If \(r>1\), then \(d_{\min}(x,y)\le1<r\) for every \(y\in X\), giving the whole space.

Now suppose \(U\) is open for \(d\), and \(x\in U\). There is an \(\varepsilon>0\) such that \(B_d(x,\varepsilon)\subseteq U\). Set \(\delta=\min(\varepsilon,1)\), which is positive and at most one. The ball comparison gives \(B_{d_{\min}}(x,\delta)=B_d(x,\delta)\subseteq B_d(x,\varepsilon)\subseteq U\). Thus \(U\) is open for \(d_{\min}\). Conversely, suppose \(U\) is open for \(d_{\min}\), and \(x\in U\). Some \(B_{d_{\min}}(x,r)\) is contained in \(U\). If \(r\le1\), this is the \(d\)-ball \(B_d(x,r)\). If \(r>1\), it is \(X\), so \(U=X\), which is open for \(d\). Hence the two metrics induce the same open sets. \(\square\)

For the fractional transformation, the ball-radius conversion is different. A transformed radius below one corresponds to a finite original radius, while a transformed radius at least one gives the whole space.

Proof. For \(0<s<1\), all quantities in the following rearrangement have positive denominators: \(\frac{d(x,y)}{1+d(x,y)}<s\) if and only if \(d(x,y)<s(1+d(x,y))\), if and only if \(d(x,y)(1-s)<s\), if and only if \(d(x,y)<s/(1-s)\). This proves the first ball identity. If \(s\ge1\), then \(d(x,y)/(1+d(x,y))<1\le s\) for every \(y\), so the transformed ball is \(X\). Taking \(s=r/(1+r)\), which lies strictly between zero and one, gives \(s/(1-s)=r\), proving the second identity.

If \(U\) is open for \(d\), then for each \(x\in U\) there is an \(r>0\) with \(B_d(x,r)\subseteq U\). The second identity expresses this same ball as \(B_{d_{\mathrm{frac}}}(x,r/(1+r))\), so \(U\) is open for \(d_{\mathrm{frac}}\). Conversely, suppose \(U\) is open for \(d_{\mathrm{frac}}\), and choose a ball \(B_{d_{\mathrm{frac}}}(x,s)\subseteq U\) around each \(x\in U\). If \(s<1\), this ball is \(B_d(x,s/(1-s))\). If \(s\ge1\), the ball is \(X\), so \(U=X\). In either case, \(U\) is open for \(d\). The open sets are equal. \(\square\)

Worked Ball Comparisons

What the Transformations Preserve—and a Common Mistake

Each transformation changes the distances but preserves the open sets, and therefore preserves the topology induced by the metric. In both cases the new distances between any two points are at most one; for \(d_{\mathrm{frac}}\), each distance is strictly less than one. This does not mean the original and transformed metrics assign the same numerical distances, or that balls of an arbitrary fixed radius agree. The radius conversions are part of the comparison.

Common pitfall. For \(d_{\min}\), it is wrong to say that a ball of radius one is the whole space because every transformed distance is at most one. A ball uses a strict inequality. For example, if \(d(x,y)=1\), then \(d_{\min}(x,y)=\min(1,1)=1\), and \(y\notin B_{d_{\min}}(x,1)\), because \(1\) is not less than \(1\). The correct identity is \(B_{d_{\min}}(x,1)=B_d(x,1)\). Only when the radius is strictly greater than one is the truncated-metric ball necessarily all of \(X\).

The practical method is to translate ball inequalities rather than compare distances informally. For the truncated metric, radii at most one give exact original balls. For the fractional metric, a radius \(s<1\) translates to \(s/(1-s)\), and an original radius \(r\) translates to \(r/(1+r)\). These formulas account for the strict endpoints and prove that the transformations preserve the topology.