Completeness of Finite-Dimensional Spaces

From Cauchy Coordinates to a Limit

The previous tutorial showed that norm convergence in a finite-dimensional space is equivalent to convergence of each coordinate in a fixed basis. Completeness asks a related question: if the terms of a sequence become arbitrarily close to one another, must they converge to a vector in the space? In finite dimensions, the answer is yes. The key is that a Cauchy sequence has Cauchy coordinate sequences, and the real numbers are complete.

Fix a real normed vector space \(V\) of dimension \(d\geq1\), a basis \(e_1,\ldots,e_d\), and a norm \(N\) on \(V\). Write \(x=\sum_{i=1}^d a_i e_i\), and use the coordinate Euclidean norm

|x|_2=\left(\sum_{i=1}^d a_i^2\right)^{1/2}.

By the Proposition (A Norm Is Comparable to a Coordinate Norm) from the tutorial Proof of Equivalence of Norms, there are constants \(m,C>0\) such that

m|x|_2\leq N(x)\leq C|x|_2 \qquad\text{for every }x\in V.

These inequalities let us pass between distances measured by \(N\) and differences between coordinates. We first make this precise for Cauchy sequences.

Theorem (Coordinate Characterization of Cauchy Sequences): Let \(x_n=\sum_{i=1}^d a_{n,i}e_i\) be a sequence in \(V\). The sequence \((x_n)\) is Cauchy in \(N\) if and only if, for each \(i=1,\ldots,d\), the real sequence \((a_{n,i})\) is Cauchy.

Proof. Suppose first that \((x_n)\) is Cauchy in \(N\). Let \(\varepsilon>0\). By the Cauchy property, there is an index \(n_0\) such that \(N(x_n-x_k)<m\varepsilon\) whenever \(n,k\geq n_0\). The lower comparison bound gives

|x_n-x_k|_2\leq\frac{1}{m}N(x_n-x_k)<\varepsilon.

For every coordinate \(i\), its absolute difference is at most the Euclidean norm of the full coordinate difference. Thus

|a_{n,i}-a_{k,i}|\leq |x_n-x_k|_2<\varepsilon \qquad(n,k\geq n_0).

Each coordinate sequence is therefore Cauchy in \(\mathbb{R}\).

Conversely, suppose each coordinate sequence is Cauchy. Given \(\varepsilon>0\), choose, for each \(i\), an index \(n_i\) such that

|a_{n,i}-a_{k,i}|<\frac{\varepsilon}{C\sqrt d} \qquad\text{whenever }n,k\geq n_i.

There are finitely many coordinates, so \(n_0=\max\{n_1,\ldots,n_d\}\) is an index. For \(n,k\geq n_0\), every coordinate difference satisfies the displayed bound. The upper comparison bound now gives

N(x_n-x_k) \leq C|x_n-x_k|_2 =C\left(\sum_{i=1}^d|a_{n,i}-a_{k,i}|^2\right)^{1/2} < C\left(d\frac{\varepsilon^2}{C^2d}\right)^{1/2} =\varepsilon.

Hence \((x_n)\) is Cauchy in \(N\). \(\square\)

The finite maximum of the coordinate thresholds is essential: it gives one index after which all coordinates are controlled at once. The same finite-coordinate feature will allow us to assemble scalar limits into a vector limit.

Worked Example: A Cauchy Sequence in a Weighted Norm on \(\mathbb{R}^3\)

On \(\mathbb{R}^3\), define \(N(x_1,x_2,x_3)=2|x_1|+|x_2|+4|x_3|\), and let

x_n=\left(1+\frac1n,\;2-\frac1n,\;\frac{1}{n}\right).

The coordinates converge respectively to \(1\), \(2\), and \(0\), so the Coordinate Characterization of Norm Convergence from the previous tutorial gives \(x_n\to(1,2,0)\) in \(N\). The norm error can also be calculated directly:

N\bigl(x_n-(1,2,0)\bigr) =2\left|\frac1n\right|+\left|-\frac1n\right|+4\left|\frac1n\right| =\frac7n\longrightarrow0.

This also verifies the Cauchy property without using the coordinate Cauchy theorem as a black box. For \(n,k\geq1\), the triangle inequality gives

N(x_n-x_k) \leq N\bigl(x_n-(1,2,0)\bigr)+N\bigl(x_k-(1,2,0)\bigr) =\frac7n+\frac7k.

For any \(\varepsilon>0\), choosing \(n,k>14/\varepsilon\) makes this sum less than \(\varepsilon\). Thus the sequence is Cauchy and its limit belongs to \(\mathbb{R}^3\).

Every Finite-Dimensional Normed Space Is Complete

Recall that a normed space is complete if every Cauchy sequence in it converges to an element of the space. The result below is the main completeness theorem for finite-dimensional spaces. It uses the coordinate characterization just proved and the completeness of \(\mathbb{R}\).

Theorem (Completeness of Finite-Dimensional Normed Spaces): Every finite-dimensional real normed vector space is complete.

Proof. First consider the zero-dimensional space \(V=\{0_V\}\). Every sequence in \(V\) is constant at \(0_V\), so it converges. Now suppose \(V\) has dimension \(d\geq1\), and let \((x_n)\) be a Cauchy sequence in \(V\). Fix a basis \(e_1,\ldots,e_d\) and write

x_n=\sum_{i=1}^d a_{n,i}e_i.

By the Coordinate Characterization of Cauchy Sequences, for each \(i\) the real sequence \((a_{n,i})\) is Cauchy. Completeness of \(\mathbb{R}\) gives a real number \(a_i\) such that \(a_{n,i}\to a_i\). Since there are only finitely many coordinates, the vector

x=\sum_{i=1}^d a_i e_i

is well-defined and belongs to \(V\). Every coordinate of \(x_n\) converges to the corresponding coordinate of \(x\). By the Coordinate Characterization of Norm Convergence from the previous tutorial, \(x_n\to x\) in \(N\). Thus every Cauchy sequence in \(V\) converges in \(V\), proving that \(V\) is complete. \(\square\)

The proof gives a practical route for finding the limit: take the real limit of each coordinate, then use those finitely many limits as the coordinates of the vector. It also proves completeness for any choice of norm on \(V\), because the coordinate comparison constants exist for every norm. This agrees with the Completeness Is Invariant Under Equivalent Norms theorem from the tutorial Banach Spaces.

Worked Example: An Arbitrary Norm on \(\mathbb{R}^2\)

Let \(N\) be any norm on \(\mathbb{R}^2\); no explicit formula for \(N\) is needed. Consider

y_n=\left(\frac1n,\;2+\frac{1}{\sqrt n}\right).

In the standard basis, the coordinates converge to \(0\) and \(2\). Therefore \(y_n\to(0,2)\) in \(N\) by the Coordinate Characterization of Norm Convergence. To see the role of the norm comparison directly, choose \(C>0\) such that \(N(u)\leq C|u|_2\) for all \(u\in\mathbb{R}^2\). Then

N\bigl(y_n-(0,2)\bigr) \leq C\left(\frac{1}{n^2}+\frac1n\right)^{1/2} \leq \frac{C\sqrt2}{\sqrt n}\longrightarrow0,

where the second inequality follows from \(1/n^2\leq1/n\) for \(n\geq1\). In particular, the triangle inequality yields, for \(n,k\geq1\),

N(y_n-y_k) \leq N\bigl(y_n-(0,2)\bigr)+N\bigl(y_k-(0,2)\bigr) \leq C\sqrt2\left(\frac1{\sqrt n}+\frac1{\sqrt k}\right).

The right-hand side becomes arbitrarily small when both indices are sufficiently large, confirming that \((y_n)\) is Cauchy in the chosen norm. The example illustrates that the argument requires neither a familiar norm formula nor special geometric properties of its unit ball.

Worked Example: Completeness of a Space of Trigonometric Functions

Let \(W\) be the span of the functions \(\sin t\) and \(\cos t\) on \([0,\pi/2]\), equipped with the supremum norm. These two functions are linearly independent: if \(a\sin t+b\cos t=0\) for every \(t\) in the interval, evaluating at \(t=0\) gives \(b=0\), and evaluating at \(t=\pi/2\) gives \(a=0\). Thus \(W\) is two-dimensional. Define

f_n(t)=\left(2-\frac1n\right)\sin t+\left(3+\frac1n\right)\cos t.

The coefficient sequences converge to \(2\) and \(3\), so the completeness theorem—or directly the coordinate convergence theorem—gives the limit \(f(t)=2\sin t+3\cos t\) in \(W\). Indeed, for \(0\leq t\leq\pi/2\), both \(|\sin t|\leq1\) and \(|\cos t|\leq1\), and hence

|f_n(t)-f(t)| =\left|-\frac1n\sin t+\frac1n\cos t\right| \leq\frac{|\sin t|+|\cos t|}{n} \leq\frac2n.

Taking the supremum over the interval gives \(\|f_n-f\|_\infty\leq2/n\to0\). This is a finite-dimensional function space: although its elements are functions, each is determined by just two real coefficients. Its completeness follows from the same finite-coordinate argument as for vectors in \(\mathbb{R}^2\).

Why the Finite-Dimensional Hypothesis Matters

The proof does not depend on the particular basis or on a special formula for the norm. A basis converts each vector into a finite list of real numbers, and norm comparison ensures that Cauchy behavior in the space matches Cauchy behavior in those coordinates. Each coordinate then has a real limit, and the finite list of limits defines a vector in the original space.

The finite number of coordinates is not a cosmetic detail. With infinitely many coordinates, there is no finite maximum of all the coordinate thresholds used in the proof, and coordinate limits need not automatically produce a limit in the space under consideration. Thus the theorem should not be extended to infinite-dimensional spaces without additional hypotheses.

A useful distinction is that completeness concerns Cauchy sequences, not just bounded sequences. The Bolzano–Weierstrass Theorem in Finite Dimensions from the previous tutorial says bounded sequences have convergent subsequences. The completeness theorem proved here instead starts from the stronger condition that all sufficiently late terms are close to each other and proves that the entire sequence converges. These are different statements with different hypotheses.

Finally, the limit is required to lie in \(V\). In the proof, this is ensured by writing the limit using the original basis vectors: \(x=\sum_{i=1}^d a_i e_i\). This step matters because in a general metric space, a Cauchy sequence may have a limit only after the space is enlarged. In finite-dimensional real normed spaces, the coordinate construction places the limit directly in the space.

Takeaway: In a finite-dimensional real normed space, a sequence is Cauchy exactly when all its coordinates in a fixed basis are Cauchy. Completeness of the real numbers supplies the coordinate limits, and finitely many such limits assemble into a vector in the space.

Check Your Understanding

Use the coordinate characterization and the completeness theorem to answer the following questions.

Why does a Cauchy sequence in the norm \(N\) have Cauchy coordinate sequences?
Where is finite dimensionality used to turn coordinatewise Cauchy behavior into a Cauchy sequence in \(N\)?
How does the proof construct a vector limit from the limits of the coordinate sequences?
Why does the completeness theorem hold for any norm on a finite-dimensional space?
What distinction separates the Bolzano–Weierstrass result from completeness?