Completeness in Familiar Settings
The previous tutorial defined completeness by requiring every Cauchy sequence to converge to a point in its own metric space. It also established useful ways to transfer completeness: closed subsets of complete spaces are complete, and a surjective isometry preserves completeness. We now use these results to examine concrete examples and prove completeness for two important spaces that are not finite-dimensional.
The examples illustrate two recurring methods. One is to place a space inside a known complete space and show that it is closed. The other is to start with a Cauchy sequence, construct a candidate limit, and verify both that the limit belongs to the space and that convergence occurs in the given metric. The second method is essential for spaces of functions and sequences, where coordinatewise or pointwise limits alone do not settle the question.
Examples from Finite Dimensions and Closed Subsets
The Completeness of Finite-Dimensional Normed Spaces Theorem from Tutorial 625 immediately gives a large family of examples. In particular, \(\mathbb{R}^n\) is complete under every norm on \(\mathbb{R}^n\), including the Euclidean norm. The choice of norm changes the distances, but not the fact of completeness in finite dimensions.
Worked Example: A Cauchy Sequence in the Euclidean Plane
Let \((z_n)\) be a Cauchy sequence in \(\mathbb{R}^2\) with the Euclidean metric, and write \(z_n=(a_n,b_n)\). For any \(m,n\), $$ |a_m-a_n|\leq \sqrt{(a_m-a_n)^2+(b_m-b_n)^2} =d_2(z_m,z_n), $$ and likewise \(|b_m-b_n|\leq d_2(z_m,z_n)\). Thus both coordinate sequences are Cauchy in \(\mathbb{R}\). Since \(\mathbb{R}\) is complete, there are \(a,b\in\mathbb{R}\) such that \(a_n\to a\) and \(b_n\to b\). The Coordinate Characterization of Norm Convergence in Tutorial 624 now gives \(z_n\to(a,b)\) in the Euclidean norm. This verifies directly how coordinate limits describe convergence here; the finite-dimensional completeness theorem applies in any finite dimension.
Closed subsets give another broad class of examples. By the Closed Subset of a Complete Space Theorem, a subset of a complete metric space is complete whenever it is closed, using the restricted metric. In particular, a closed interval in \(\mathbb{R}\) is complete. The conclusion does not require a separate analysis of every Cauchy sequence in the interval: the ambient completeness and closedness together suffice.
Worked Example: A Closed Ray Is Complete
Consider \(A=[2,\infty)\) with the usual metric. Its complement \((-\infty,2)\) is open in \(\mathbb{R}\), so \(A\) is closed. Since \(\mathbb{R}\) is complete, the Closed Subset of a Complete Space Theorem shows that \(A\) is complete. More explicitly, if \((a_n)\) is Cauchy in \(A\), it is Cauchy in \(\mathbb{R}\), so it has a real limit \(a\). Every \(a_n\geq2\); if \(a<2\), then the positive number \((2-a)/2\) would eventually satisfy \(|a_n-a|<(2-a)/2\), forcing \(a_n<2\), a contradiction. Hence \(a\geq2\), and the limit belongs to \(A\).
Bounded Continuous Functions
For a nonempty metric space \(E\), let \(C_b(E)\) denote the set of bounded continuous real-valued functions on \(E\). Its supremum norm is $$ \|f\|_\infty=\sup_{x\in E}|f(x)|. $$ The induced metric is \(d(f,g)=\|f-g\|_\infty\). A Cauchy sequence in this metric satisfies a uniform estimate: after some index, the values of any two functions differ by less than a prescribed amount at every point of \(E\). We will show that its limit is still bounded and continuous, and that convergence to it is in the supremum norm.
Proof. Let \((f_n)\) be a Cauchy sequence in \(C_b(E)\). For each fixed \(x\in E\) and all \(m,n\), $$ |f_m(x)-f_n(x)|\leq\|f_m-f_n\|_\infty. $$ The Cauchy condition in the supremum norm therefore makes \((f_n(x))\) a Cauchy sequence in \(\mathbb{R}\). Completeness of \(\mathbb{R}\) gives a real limit for each \(x\). Define \(f:E\to\mathbb{R}\) by $$ f(x)=\lim_{n\to\infty}f_n(x). $$
We first prove uniform convergence. Given \(\varepsilon>0\), choose \(N\) such that \(\|f_m-f_n\|_\infty<\varepsilon/2\) whenever \(m,n\geq N\). Fix \(n\geq N\) and \(x\in E\). For every \(m\geq N\), $$ |f_n(x)-f_m(x)|\leq\|f_n-f_m\|_\infty<\varepsilon/2. $$ Letting \(m\to\infty\) gives \(|f_n(x)-f(x)|\leq\varepsilon/2\). This bound holds for every \(x\), so $$ \|f_n-f\|_\infty\leq\varepsilon/2<\varepsilon \qquad(n\geq N). $$ In particular, \(f_n\) converges uniformly to \(f\).
Next, \(f\) is bounded. Choose \(N\) so that \(\|f_N-f\|_\infty\leq1\), which is possible by the uniform convergence just proved. For every \(x\in E\), $$ |f(x)|\leq|f_N(x)|+|f(x)-f_N(x)| \leq\|f_N\|_\infty+1. $$ The right-hand side is finite and independent of \(x\), so \(f\) is bounded.
Finally, \(f\) is continuous. Fix \(x_0\in E\) and \(\varepsilon>0\). Uniform convergence gives an index \(n\) such that \(\|f-f_n\|_\infty<\varepsilon/3\). Since \(f_n\) is continuous at \(x_0\), there is \(\delta>0\) such that \(d_E(x,x_0)<\delta\) implies \(|f_n(x)-f_n(x_0)|<\varepsilon/3\). For such \(x\), the triangle inequality yields $$ |f(x)-f(x_0)| \leq |f(x)-f_n(x)|+|f_n(x)-f_n(x_0)|+|f_n(x_0)-f(x_0)| <\varepsilon. $$ Thus \(f\) is continuous at every point of \(E\). We have shown that \(f\in C_b(E)\) and that \(\|f_n-f\|_\infty\to0\). Every Cauchy sequence in \(C_b(E)\) therefore converges in that space, proving completeness. \(\square\)
The proof uses uniform control at two key points. It ensures that the pointwise limit is bounded, and it allows continuity of the approximating functions to pass to the limit. Pointwise convergence alone would not provide either uniform estimate in this argument.
Worked Example: A Cauchy Sequence in \(C[0,1]\)
Every continuous real-valued function on the compact interval \([0,1]\) is bounded, by the Extreme-Value Theorem. Hence \(C[0,1]=C_b([0,1])\), with the supremum norm, and the theorem shows that this function space is complete.
For a specific sequence, define \(f_n(x)=x+1/n\) for \(x\in[0,1]\). Each \(f_n\) is continuous and bounded. For positive integers \(m,n\), $$ \|f_n-f_m\|_\infty =\sup_{x\in[0,1]}\left|\frac1n-\frac1m\right| =\left|\frac1n-\frac1m\right| \leq\frac1n+\frac1m. $$ If \(m,n\geq N\), this is at most \(2/N\), which tends to zero as \(N\) increases. Thus \((f_n)\) is Cauchy in the supremum norm. Its limit is \(f(x)=x\), because $$ \|f_n-f\|_\infty=\sup_{x\in[0,1]}\frac1n=\frac1n\longrightarrow0. $$ The limit is continuous and belongs to \(C[0,1]\), as completeness requires.
The Space of Bounded Sequences
The same uniform-limit idea applies to bounded sequences of real numbers. Let \(\ell^\infty\) be the set of all bounded real sequences \(x=(x_1,x_2,\ldots)\), with norm $$ \|x\|_\infty=\sup_{j\geq1}|x_j|. $$ The induced distance is \(\|x-y\|_\infty\). A Cauchy sequence in this space consists of bounded sequences whose corresponding coordinates become uniformly close over all coordinate indices. The uniformity over every coordinate is what makes convergence hold in this norm.
Proof. Let \((x^{(n)})\) be Cauchy in \(\ell^\infty\), and write \(x^{(n)}=(x^{(n)}_1,x^{(n)}_2,\ldots)\). For each fixed coordinate \(j\), $$ |x^{(m)}_j-x^{(n)}_j|\leq\|x^{(m)}-x^{(n)}\|_\infty. $$ Thus \((x^{(n)}_j)\) is Cauchy in \(\mathbb{R}\). Define \(x_j=\lim_{n\to\infty}x^{(n)}_j\) for every \(j\), and let \(x=(x_1,x_2,\ldots)\).
We check first that \(x\) is bounded. Choose \(N\) so that \(\|x^{(n)}-x^{(N)}\|_\infty<1\) for all \(n\geq N\). For each \(j\), taking the limit as \(n\to\infty\) in \(|x^{(n)}_j-x^{(N)}_j|<1\) gives \(|x_j-x^{(N)}_j|\leq1\). Therefore $$ |x_j|\leq |x^{(N)}_j|+1\leq\|x^{(N)}\|_\infty+1 \qquad\text{for every }j. $$ So \(x\in\ell^\infty\).
It remains to prove convergence in the supremum norm. Given \(\varepsilon>0\), choose \(N\) such that \(\|x^{(m)}-x^{(n)}\|_\infty<\varepsilon/2\) whenever \(m,n\geq N\). Fix \(n\geq N\). For each coordinate \(j\), letting \(m\to\infty\) in \(|x^{(n)}_j-x^{(m)}_j|<\varepsilon/2\) gives \(|x^{(n)}_j-x_j|\leq\varepsilon/2\). Taking the supremum over \(j\) yields $$ \|x^{(n)}-x\|_\infty\leq\varepsilon/2<\varepsilon \qquad(n\geq N). $$ Thus \(x^{(n)}\to x\) in \(\ell^\infty\), proving completeness. \(\square\)
Worked Example: Truncations Converge in \(\ell^\infty\)
For each positive integer \(n\), define the bounded sequence \(x^{(n)}\) by $$ x^{(n)}_j= \begin{cases} 1/j,&1\leq j\leq n,\\ 0,&j>n. \end{cases} $$ Let \(x=(1,1/2,1/3,\ldots)\), which is bounded. For \(j\leq n\), \(x^{(n)}_j=x_j\). For \(j>n\), \(x^{(n)}_j=0\), so \(|x^{(n)}_j-x_j|=1/j\). Consequently, $$ \|x^{(n)}-x\|_\infty=\sup_{j>n}\frac1j=\frac1{n+1}\longrightarrow0. $$ This also verifies the Cauchy condition directly: for \(m>n\), the sequences \(x^{(m)}\) and \(x^{(n)}\) agree through coordinate \(n\), and their difference at every coordinate \(j>n\) is at most \(1/(n+1)\). Hence $$ \|x^{(m)}-x^{(n)}\|_\infty\leq\frac1{n+1}. $$ For indices \(m,n\geq N\), this distance is at most \(1/(N+1)\), so the sequence of truncations is Cauchy and converges in \(\ell^\infty\) to \(x\).
What These Examples Have in Common
These examples use different routes to the same conclusion. Finite-dimensional spaces are complete by the theorem from Tutorial 625. Closed subsets inherit completeness from a complete ambient space. In \(C_b(E)\) and \(\ell^\infty\), the proofs construct a candidate limit from real limits and then use uniform estimates to show that it remains in the space and is reached in the prescribed metric.
A common pitfall is to establish only that each coordinate, or each pointwise function value, has a limit. That does not by itself prove convergence in a supremum norm: the estimates must control all coordinates or all points at once. Nor is membership automatic. In the two completeness proofs, boundedness and, for functions, continuity are verified before concluding that the limit belongs to the space.
Check Your Understanding
Use the definitions and results in this tutorial to answer the following questions.
- Which earlier theorem implies that every closed subset of \(\mathbb{R}\) is complete?
- In the proof for \(C_b(E)\), why is the limit bounded, and why is it continuous?
- Why does pointwise convergence alone not supply the supremum-norm estimate needed for completeness?
- How does the proof that the coordinatewise limit belongs to \(\ell^\infty\) use a fixed term of the Cauchy sequence?
- For the truncations in the last example, why is the supremum error \(1/(n+1)\)?