Completeness of R^n

Completeness Through Cauchy Sequences

In “Coordinatewise Convergence,” convergence of a sequence in \(\mathbb{R}^n\) was reduced to convergence in each of its finitely many coordinates. A related question is whether a sequence is Cauchy: do its terms eventually become arbitrarily close to one another, even before a limit is specified? In this tutorial, we characterize that property coordinate by coordinate and use it to understand completeness. The earlier theorem “Completeness of \(\mathbb{R}^n\)” established that Euclidean \(\mathbb{R}^n\) is complete; here we use that result to study subsets and the standard norms without reproving it.

A Cauchy sequence is controlled by distances between pairs of sufficiently late terms. This differs from the definition of convergence, which compares each sufficiently late term with a proposed limit. Completeness asserts that the internal control expressed by the Cauchy condition is enough to guarantee a limit inside the space.

The Cauchy Criterion in Coordinates

For vectors, the Euclidean distance is determined by the coordinate differences. In one direction, each coordinate difference is bounded by the Euclidean distance. In the other, small differences in all finitely many coordinates make the Euclidean distance small. This yields a coordinatewise test for the Cauchy property.

Proof. Suppose first that \((x^{(k)})\) is Cauchy in the Euclidean metric. For every coordinate \(i\) and every \(k,\ell\), we have \[ |x_i^{(k)}-x_i^{(\ell)}|\leq\|x^{(k)}-x^{(\ell)}\|_2, \] since the square of the Euclidean distance includes the square of this coordinate difference. Given \(\varepsilon>0\), choose \(K\) so that \(\|x^{(k)}-x^{(\ell)}\|_2<\varepsilon\) whenever \(k,\ell\geq K\). The displayed inequality shows that \(|x_i^{(k)}-x_i^{(\ell)}|<\varepsilon\) for every such pair. Thus every coordinate sequence is Cauchy.

Conversely, suppose each coordinate sequence is Cauchy, and fix \(\varepsilon>0\). For every \(i\in\{1,\ldots,n\}\), there is an index \(K_i\) such that \[ |x_i^{(k)}-x_i^{(\ell)}|<\frac{\varepsilon}{\sqrt{n}} \qquad\text{whenever }k,\ell\geq K_i. \] There are only finitely many coordinates, so \(K=\max\{K_1,\ldots,K_n\}\) is a positive integer. If \(k,\ell\geq K\), all the coordinate estimates hold at once. Therefore \[ \|x^{(k)}-x^{(\ell)}\|_2 =\left(\sum_{i=1}^{n}|x_i^{(k)}-x_i^{(\ell)}|^2\right)^{1/2} <\left(\sum_{i=1}^{n}\frac{\varepsilon^2}{n}\right)^{1/2} =\varepsilon. \] This is exactly the Euclidean Cauchy condition, proving the converse. \(\square\)

The common index \(K\) is essential: the coordinate sequences may become Cauchy at different rates, but finitely many coordinate-specific indices have a maximum. This is the same finite-dimensional synchronization principle used for coordinatewise convergence. The argument does not establish the corresponding claim for infinitely many coordinates.

Completeness of the Standard Norms

The coordinate criterion connects Cauchy behavior to the real numbers: each real Cauchy sequence converges by completeness of \(\mathbb{R}\). Together with the Coordinate Criterion for Convergence, this provides a coordinate-level way to understand the earlier result that Euclidean \(\mathbb{R}^n\) is complete. We will use that established result, rather than repeat its proof, to see that completeness does not depend on choosing only the Euclidean norm among the standard norms.

Proof. The Euclidean case is the earlier theorem “Completeness of \(\mathbb{R}^n\).” For the other two norms, use the established comparison of the standard norms: \[ \|v\|_\infty\leq\|v\|_2\leq\sqrt{n}\,\|v\|_\infty, \qquad \|v\|_2\leq\|v\|_1\leq\sqrt{n}\,\|v\|_2. \] Suppose \((x^{(k)})\) is Cauchy in the \(1\)-norm. For every \(\varepsilon>0\), sufficiently late pairs satisfy \(\|x^{(k)}-x^{(\ell)}\|_1<\varepsilon\). Since \(\|v\|_2\leq\|v\|_1\), those pairs also satisfy \(\|x^{(k)}-x^{(\ell)}\|_2<\varepsilon\). Thus the sequence is Euclidean Cauchy. By Euclidean completeness, there is \(a\in\mathbb{R}^n\) such that \(x^{(k)}\to a\) in the Euclidean norm. The inequality \(\|v\|_1\leq\sqrt{n}\|v\|_2\) then gives \[ \|x^{(k)}-a\|_1\leq\sqrt{n}\,\|x^{(k)}-a\|_2\longrightarrow 0. \] So the sequence converges in the \(1\)-norm.

If \((x^{(k)})\) is Cauchy in the infinity norm, then \(\|v\|_2\leq\sqrt{n}\|v\|_\infty\) shows that it is Euclidean Cauchy as well. It therefore converges to some \(a\in\mathbb{R}^n\) in the Euclidean norm. Finally, \(\|v\|_\infty\leq\|v\|_2\) implies \[ \|x^{(k)}-a\|_\infty\leq\|x^{(k)}-a\|_2\longrightarrow 0. \] Thus the sequence converges in the infinity norm. All three metrics are complete. \(\square\)

The proof uses both directions of the norm comparisons: one converts a Cauchy sequence for a given norm into a Euclidean Cauchy sequence, and the other converts Euclidean convergence back to convergence in that norm. A comparison in only one direction would not, by itself, justify both steps.

Complete Subsets of Euclidean Space

A subset \(A\) of a metric space inherits the metric by restricting distances to pairs of points in \(A\). A Cauchy sequence in \(A\) is therefore Cauchy in the ambient space, but its limit there need not belong to \(A\). In Euclidean space, closedness is precisely the condition that prevents this limit from escaping.

Proof. First suppose \(A\) is closed. Let \((a^{(k)})\) be a Cauchy sequence in \(A\). It is also Cauchy in \(\mathbb{R}^n\), because the restricted metric uses the same Euclidean distances. By the earlier theorem “Completeness of \(\mathbb{R}^n\),” there is \(x\in\mathbb{R}^n\) such that \(a^{(k)}\to x\). Since \(A\) is closed and the sequence lies in \(A\), its limit \(x\) belongs to \(A\). Thus every Cauchy sequence in \(A\) converges in \(A\), proving that \(A\) is complete.

Conversely, suppose \(A\) is complete. Let \(x\in\overline{A}\), the closure of \(A\) in \(\mathbb{R}^n\). For every positive integer \(k\), the open ball \(B_{1/k}^{(2)}(x)\) meets \(A\), by the definition of closure. Choose \(a^{(k)}\in A\cap B_{1/k}^{(2)}(x)\). Then \(\|a^{(k)}-x\|_2<1/k\), so \(a^{(k)}\to x\) in \(\mathbb{R}^n\). Every convergent sequence is Cauchy, so \((a^{(k)})\) is Cauchy in \(A\). Completeness of \(A\) gives a point \(a\in A\) such that \(a^{(k)}\to a\) in the restricted metric, hence also in \(\mathbb{R}^n\). Limits in the Euclidean metric are unique, so \(a=x\). Therefore \(x\in A\). We have shown \(\overline{A}\subseteq A\), and consequently \(A\) is closed. \(\square\)

What Completeness Does and Does Not Say

Completeness concerns limits of Cauchy sequences, not whether a set is bounded or whether every sequence in it has a convergent subsequence. For example, \(\mathbb{R}^n\) is complete but is unbounded. The criterion for subsets gives a useful practical test in Euclidean space: a subset is complete exactly when it contains all of its ambient limit points.

A common error is to observe that a Cauchy sequence in a subset converges in \(\mathbb{R}^n\) and conclude immediately that the subset is complete. The ambient limit might lie outside the subset, as the open-half-space example demonstrates. One must verify that the limit belongs to the space in which convergence is being claimed. Closedness supplies exactly that guarantee for subsets of \(\mathbb{R}^n\).

The coordinate criterion gives another practical route: a vector sequence is Euclidean Cauchy precisely when each of its finitely many coordinate sequences is Cauchy. The real coordinates then have limits, and the earlier completeness theorem for \(\mathbb{R}^n\) ensures the resulting vector is an ambient limit. The subset criterion adds the final question: does that limit remain in the subset?