Euclidean Space R^n

From Real Coordinates to Euclidean Space

The fixed-point results just developed apply to complete metric spaces, but many problems in multivariable analysis take place in coordinate spaces. The notation \(\mathbb{R}^n\) means the set of all ordered \(n\)-tuples of real numbers. To use the tools of analysis there, we need a distance that measures separation across all coordinates at once. The usual choice is Euclidean distance, which extends the familiar distance on the real line.

Throughout, let \(n\) be a positive integer and write points of \(\mathbb{R}^n\) as \(x=(x_1,\ldots,x_n)\) and \(y=(y_1,\ldots,y_n)\). Define the coordinate dot product and Euclidean length by

The Euclidean distance between two points is the length of their coordinate difference:

In one dimension, this is just \(d_2(x,y)=|x_1-y_1|\). In higher dimensions, the sum accounts for discrepancies in every coordinate. We first verify the inequality that makes this length behave as expected.

Proof. If \(y=0\), then \(x\cdot y=0\), so the inequality holds. Suppose \(y\ne0\). For every real number \(t\), the sum of squares \(\|x-ty\|_2^2\) is nonnegative. Expanding it gives

This quadratic in \(t\) has positive leading coefficient \(\|y\|_2^2\). A quadratic that is nonnegative for every real \(t\) cannot have a positive discriminant. Therefore

Dividing by \(4\) and taking nonnegative square roots yields \(|x\cdot y|\leq\|x\|_2\|y\|_2\), as required. \(\square\)

Euclidean Distance Is a Metric

The Cauchy–Schwarz Inequality gives the triangle inequality for Euclidean length. Indeed, expanding the square of the length of a sum gives

Both sides whose squares are compared are nonnegative, so \(\|x+y\|_2\leq\|x\|_2+\|y\|_2\). This also proves the triangle inequality for distances, by applying the length inequality to \(x-z=(x-y)+(y-z)\).

Proof. Each square in the defining sum is nonnegative, so \(d_2(x,y)\geq0\). If \(d_2(x,y)=0\), then the sum of the nonnegative numbers \((x_i-y_i)^2\) is zero. Every term must be zero, hence \(x_i=y_i\) for every \(i\), and \(x=y\). Conversely, if \(x=y\), every term is zero, so \(d_2(x,y)=0\). Symmetry holds because \((x_i-y_i)^2=(y_i-x_i)^2\) for each coordinate. Finally, the triangle inequality for the length, applied to \(x-z=(x-y)+(y-z)\), gives

Thus \(d_2\) satisfies nonnegativity, separation of distinct points, symmetry, and the triangle inequality. It is a metric. \(\square\)

Euclidean Balls and Coordinate Boxes

A metric determines its open balls: for \(x\in\mathbb{R}^n\) and \(r>0\), the Euclidean open ball of radius \(r\) is

For a second way to measure coordinate separation, define the maximum metric by \(d_\infty(x,y)=\max_{1\leq i\leq n}|x_i-y_i|\). Comparing the largest coordinate difference with the sum of all squared differences gives

For the first inequality, each \(|x_i-y_i|\) is at most the square root of the sum of all the squared differences. For the second, each squared difference is at most \(d_\infty(x,y)^2\), so their sum is at most \(n\,d_\infty(x,y)^2\).

Proof. The maximum metric induces the product topology on a finite product of metric spaces, as established in the earlier tutorial on Product Topology. It remains to show that \(d_2\) and \(d_\infty\) induce the same topology. Fix \(x\in\mathbb{R}^n\) and \(r>0\). The inequalities above imply

For the first inclusion, if \(d_\infty(x,y)<r/\sqrt n\), then \(d_2(x,y)\leq\sqrt n\,d_\infty(x,y)<r\). For the second, if \(d_2(x,y)<r\), then \(d_\infty(x,y)\leq d_2(x,y)<r\). Thus every Euclidean ball contains a maximum-metric ball about the same center, and every maximum-metric ball contains a Euclidean ball about that center. These inclusions show that each metric has the same open sets as the other. Since the maximum metric gives the product topology, so does \(d_2\). \(\square\)

Completeness of Euclidean Space

A metric space is complete if every Cauchy sequence in it converges to a point of that space. Euclidean space is complete because its finitely many real coordinates can be handled separately. This fact is fundamental when applying the Contraction Mapping Theorem to maps whose domain is \(\mathbb{R}^n\).

Proof. Let \((x^{(k)})_{k\geq1}\) be a Cauchy sequence in \((\mathbb{R}^n,d_2)\), and write \(x^{(k)}=(x_1^{(k)},\ldots,x_n^{(k)})\). For any coordinate \(i\),

Since the sequence is Cauchy in \(d_2\), this inequality shows that \((x_i^{(k)})_{k\geq1}\) is a Cauchy sequence in \(\mathbb{R}\). Completeness of \(\mathbb{R}\) gives a real limit \(a_i\) for each \(i\). Because there are finitely many coordinates, \(a=(a_1,\ldots,a_n)\) is a point of \(\mathbb{R}^n\). Let \(\varepsilon>0\). For each \(i\), coordinate convergence gives an index \(N_i\) such that \(k\geq N_i\) implies \(|x_i^{(k)}-a_i|<\varepsilon/\sqrt n\). Choose \(N\) to be the largest of the finitely many indices \(N_1,\ldots,N_n\). Then, for \(k\geq N\),

It follows that \(d_2(x^{(k)},a)<\varepsilon\). Thus \(x^{(k)}\to a\) in the Euclidean metric, proving completeness. \(\square\)

Why the Euclidean Structure Matters

The same set \(\mathbb{R}^n\) can be equipped with different metrics, and metric properties should not be confused with the underlying coordinates. Here, Euclidean distance provides a particular quantitative measure of separation. The comparison with the maximum metric shows that it gives the familiar product topology: a set is open in Euclidean space precisely when its points have sufficiently small coordinate neighborhoods inside the set. Yet the numerical values of distances depend on the chosen metric.

The completeness theorem is also a statement about the metric, not merely the coordinates. Its proof relies on both the completeness of the real line and the fact that there are only finitely many coordinates. The finite choice of a common index \(N\) matters: the argument does not automatically extend to an infinite product with the same distance formula. For the finite-dimensional space considered here, Euclidean distance, product topology, and coordinatewise convergence fit together in a precise and useful way.