Euclidean, Taxicab, and Sup Metrics Give the Same Topology

Three Distances on the Same Coordinate Space

A point in \(\mathbb{R}^n\) can be measured in several natural ways. The Euclidean distance is the familiar straight-line distance; the taxicab distance adds coordinate differences; and the sup distance looks only at the largest coordinate difference. Their numerical values and their balls can differ. The key fact is that, for each fixed finite \(n\), inequalities between the distances let us find small balls for one metric inside balls for another. Consequently, all three metrics give the same open sets.

Throughout, let \(n\in\mathbb{N}\), and give \(\mathbb{R}^n\) its usual coordinate structure. For \(x=(x_1,\ldots,x_n)\) and \(y=(y_1,\ldots,y_n)\), define the following distances:

These formulas do define metrics. Each distance is nonnegative and symmetric, and is zero exactly when every coordinate difference is zero, which is equivalent to \(x=y\). For the triangle inequality, let \(x,y,z\in\mathbb{R}^n\). The coordinate inequality \(|x_i-z_i|\le |x_i-y_i|+|y_i-z_i|\), summed over \(i\), gives \(d_1(x,z)\le d_1(x,y)+d_1(y,z)\). Taking the maximum over \(i\) gives the same inequality for \(d_\infty\). The Euclidean triangle inequality follows from the inner product inequality \(|u\mathbin{\cdot}v|\le \|u\|\|v\|\): expanding \(\|u+v\|^2\) gives \(\|u+v\|^2\le(\|u\|+\|v\|)^2\), and taking nonnegative square roots gives \(\|u+v\|\le\|u\|+\|v\|\). Apply this with \(u=x-y\), \(v=y-z\). The inner product inequality is the standard Cauchy–Schwarz inequality from linear algebra.

Coordinate Inequalities

For the comparison, fix \(x,y\), and write \(a_i=|x_i-y_i|\). Each \(a_i\) is nonnegative. The largest coordinate difference cannot exceed the square root of the sum of all squared differences; and that square root cannot exceed the sum of the differences. The final bound below follows from Cauchy–Schwarz applied to \((a_1,\ldots,a_n)\) and \((1,\ldots,1)\).

Proof. For each \(i\), \(a_i^2\le\sum_{j=1}^n a_j^2\), so \(a_i\le\sqrt{\sum_{j=1}^n a_j^2}\). Taking the maximum over \(i\) proves \(d_\infty(x,y)\le d_2(x,y)\). Also, \(\sum_i a_i^2\le(\sum_i a_i)^2\), because the right side equals \(\sum_i a_i^2+2\sum_{i<j}a_i a_j\), and each additional term is nonnegative. Taking square roots proves \(d_2(x,y)\le d_1(x,y)\). Cauchy–Schwarz gives \(\sum_i a_i\le\sqrt{n}\sqrt{\sum_i a_i^2}\), proving \(d_1(x,y)\le\sqrt{n}\,d_2(x,y)\). Finally, \(d_2(x,y)\le\sqrt{n}\max_i a_i=\sqrt{n}\,d_\infty(x,y)\), since each \(a_i^2\le d_\infty(x,y)^2\). Combining this with the preceding bound gives \(d_1(x,y)\le n\,d_\infty(x,y)\). \(\square\)

The comparison is useful in both directions. For example, \(d_\infty\le d_2\) means a point close in Euclidean distance is also close in sup distance. Conversely, \(d_2\le\sqrt{n}\,d_\infty\) says that being sufficiently close in sup distance forces Euclidean closeness. The factors involving \(n\) may change the radii, but they do not prevent us from choosing a positive radius.

Ball Inclusions and Equal Topologies

The distance estimates give explicit ball inclusions. Each inclusion uses the same center; the radii are chosen so that a point in the smaller ball satisfies the required strict distance bound.

Proof. The inequalities of Theorem 1 imply the following inclusions for every \(x\in\mathbb{R}^n\) and every \(r>0\): \(B_{d_1}(x,r)\subseteq B_{d_2}(x,r)\subseteq B_{d_\infty}(x,r)\). For example, if \(y\in B_{d_1}(x,r)\), then \(d_2(x,y)\le d_1(x,y)<r\), so \(y\in B_{d_2}(x,r)\). The second inclusion follows because \(d_\infty(x,y)\le d_2(x,y)\). The other direction of comparison is supplied by \(B_{d_\infty}(x,r/n)\subseteq B_{d_1}(x,r)\), \(B_{d_2}(x,r/\sqrt{n})\subseteq B_{d_1}(x,r)\), and \(B_{d_\infty}(x,r/\sqrt{n})\subseteq B_{d_2}(x,r)\). Indeed, if \(d_\infty(x,y)<r/n\), then \(d_1(x,y)\le n d_\infty(x,y)<r\). If \(d_2(x,y)<r/\sqrt{n}\), then \(d_1(x,y)\le\sqrt{n}d_2(x,y)<r\). If \(d_\infty(x,y)<r/\sqrt{n}\), then \(d_2(x,y)\le\sqrt{n}d_\infty(x,y)<r\).

Now let \(U\) be open in the topology induced by one of the three metrics, and fix \(x\in U\). By the definition of metric openness, there is \(r>0\) such that the ball for that metric, centered at \(x\) with radius \(r\), is contained in \(U\). The displayed inclusions show that there is a ball centered at \(x\) for either of the other metrics, with a positive radius among \(r\), \(r/n\), and \(r/\sqrt{n}\), contained in that same ball and hence in \(U\). Thus \(U\) is open for each of the other metrics. Applying this argument to each choice of the original metric proves equality of all three topologies. \(\square\)

The proof establishes equality of open sets, not equality of distances or of balls. It also uses that \(n\) is finite: the bounds by \(n\) and \(\sqrt{n}\) are finite positive constants for each fixed \(n\). No assertion about an infinite-dimensional coordinate space follows from this argument.

Common Pitfall and Why the Result Matters

Common pitfall. A tempting but false calculation is to identify the Euclidean and taxicab distances. For the points \(x=(0,0)\) and \(y=(1,1)\), the coordinate differences are \(1\) and \(1\). The taxicab distance is \(1+1=2\), whereas the Euclidean distance is \(\sqrt{1^2+1^2}=\sqrt{2}\). They are not equal. The correction is to use the inequalities: in \(\mathbb{R}^2\), \(\sqrt{2}\le2\le\sqrt{2}\sqrt{2}=2\), as required. Different distances can generate the same topology without assigning the same value to each pair.

The conclusion lets us choose the distance formula that makes a calculation convenient. Coordinatewise bounds can be easier with \(d_\infty\); sums of coordinate errors can be natural with \(d_1\); Euclidean geometry uses \(d_2\). For questions about open sets in fixed finite-dimensional real space, changing among these metrics changes the ball shapes and radii but not which subsets are open.