Two Spaces of Sequences
A sequence of real numbers can be viewed as a point in a metric space. The choice of metric matters: the sup metric measures the largest coordinate difference, so convergence in this metric requires all coordinates to be close at once. We will use this uniform control to prove completeness for bounded sequences, and then show that the sequences tending to zero form a closed subspace.
All sequences below are indexed by \(\mathbb{N}=\{1,2,\dots\}\). A real sequence \(x=(x_n)\) is bounded if there is a finite real number \(M\geq 0\) such that \(|x_n|\leq M\) for every \(n\).
The supremum defining \(d_\infty(x,y)\) is finite: if \(|x_n|\leq M\) and \(|y_n|\leq L\) for all \(n\), then \(|x_n-y_n|\leq M+L\). The supremum need not be attained by any coordinate. The formula is a metric: it is nonnegative and symmetric; if its value is zero, every \(|x_n-y_n|=0\), so \(x=y\); and the real triangle inequality gives \(|x_n-z_n|\leq |x_n-y_n|+|y_n-z_n|\leq d_\infty(x,y)+d_\infty(y,z)\) for each \(n\). Taking the supremum proves the triangle inequality for \(d_\infty\).
Completeness of \(\ell^\infty\)
Proof. Let \((x^{(j)})_{j\in\mathbb{N}}\) be a Cauchy sequence in \(\ell^\infty\), where \(x^{(j)}=(x_n^{(j)})_{n\in\mathbb{N}}\). By the definition of a Cauchy sequence, for each \(\varepsilon>0\) there is \(J\in\mathbb{N}\) such that \(d_\infty(x^{(j)},x^{(k)})<\varepsilon\) whenever \(j,k\geq J\). For each fixed \(n\), \[ |x_n^{(j)}-x_n^{(k)}|\leq d_\infty(x^{(j)},x^{(k)}), \] so \((x_n^{(j)})_{j\in\mathbb{N}}\) is a Cauchy sequence in \(\mathbb{R}\). Since \(\mathbb{R}\) is complete, it has a limit; write \(x_n=\lim_{j\to\infty}x_n^{(j)}\). This defines a real sequence \(x=(x_n)\).
We first show that \(x\) is bounded. Choose \(J\) so that \(d_\infty(x^{(j)},x^{(J)})<1\) whenever \(j\geq J\). For every \(n\), taking the coordinate limit as \(j\to\infty\) gives \(|x_n-x_n^{(J)}|\leq 1\). Since \(x^{(J)}\) is bounded, there is \(M<\infty\) with \(|x_n^{(J)}|\leq M\) for all \(n\). Thus \(|x_n|\leq M+1\) for all \(n\), and \(x\in\ell^\infty\).
It remains to prove convergence in the sup metric, not just convergence in each coordinate. Fix \(\varepsilon>0\), and choose \(J\) such that \(d_\infty(x^{(j)},x^{(k)})<\varepsilon/2\) whenever \(j,k\geq J\). Fix \(j\geq J\) and \(n\in\mathbb{N}\). Letting \(k\to\infty\) in \(|x_n^{(j)}-x_n^{(k)}|<\varepsilon/2\) gives \(|x_n^{(j)}-x_n|\leq\varepsilon/2\). This bound holds for every \(n\), so \(d_\infty(x^{(j)},x)\leq\varepsilon/2<\varepsilon\). Hence \(x^{(j)}\to x\) in \(\ell^\infty\). \(\square\)
The use of \(\varepsilon/2\) makes the final strict inequality explicit: taking a coordinate limit can replace a strict bound by a non-strict one, but the resulting upper bound \(\varepsilon/2\) is still strictly less than \(\varepsilon\).
Worked Example: Truncating a Sequence in \(c_0\)
Let \(x_n=1/n\), and for each \(m\in\mathbb{N}\) define \(x^{(m)}\) by \(x_n^{(m)}=1/n\) when \(n\leq m\), and \(x_n^{(m)}=0\) when \(n>m\). Each \(x^{(m)}\) has only finitely many nonzero coordinates, so it belongs to \(c_0\). We compute its sup distance to \(x\).
For \(n\leq m\), \(|x_n^{(m)}-x_n|=|1/n-1/n|=0\). For \(n>m\), \(|x_n^{(m)}-x_n|=|0-1/n|=1/n\). Among integers \(n>m\), the largest value \(1/n\) is \(1/(m+1)\), attained at \(n=m+1\). Therefore \(d_\infty(x^{(m)},x)=1/(m+1)\), which tends to zero as \(m\to\infty\). In particular, \(x\in c_0\), and the computed distances verify convergence in the sup metric.
Worked Example: A Cauchy Sequence with a Uniform Limit
For each \(m\in\mathbb{N}\), define \(u^{(m)}\) by \(u_n^{(m)}=1/(m+n)\) for \(n\in\mathbb{N}\). Each \(u^{(m)}\) is bounded and tends to zero in its coordinate index \(n\), so \(u^{(m)}\in c_0\). If \(k>m\), then for every \(n\), \[ |u_n^{(m)}-u_n^{(k)}|=\frac{1}{m+n}-\frac{1}{k+n}>0. \] The difference is largest at \(n=1\): indeed, the function of a positive integer \(n\) is \(1/(m+n)-1/(k+n)\), and the calculation below can also be bounded by its value at \(n=1\) since \[ \frac{1}{m+n}-\frac{1}{k+n} =\frac{k-m}{(m+n)(k+n)} \leq \frac{k-m}{(m+1)(k+1)}. \] Thus \(d_\infty(u^{(m)},u^{(k)})=1/(m+1)-1/(k+1)\). For example, \(d_\infty(u^{(2)},u^{(5)})=1/3-1/6=1/6\).
For every \(n\), \(|u_n^{(m)}-0|=1/(m+n)\leq 1/(m+1)\), with equality at \(n=1\). Hence \(d_\infty(u^{(m)},0)=1/(m+1)\to0\). This verifies directly that the sequence of points \((u^{(m)})\) converges uniformly in the sup metric to the zero sequence. It is consequently Cauchy as well: for any \(\varepsilon>0\), choose \(J\) with \(1/(J+1)<\varepsilon/2\); for \(j,k\geq J\), the triangle inequality gives \(d_\infty(u^{(j)},u^{(k)})\leq 1/(j+1)+1/(k+1)<\varepsilon\).
The Subspace \(c_0\) Is Closed and Complete
Proof. First we show that the complement of \(c_0\) is open. Let \(x\in\ell^\infty\setminus c_0\). Since \(x_n\) does not tend to zero, the definition of convergence gives some \(\varepsilon>0\) such that for every \(N\in\mathbb{N}\), there is \(n\geq N\) with \(|x_n|\geq\varepsilon\). Let \(y\in c_0\). There is \(N\) such that \(|y_n|<\varepsilon/2\) for all \(n\geq N\). Choose \(n\geq N\) with \(|x_n|\geq\varepsilon\). Then \[ |x_n-y_n|\geq |x_n|-|y_n|>\varepsilon/2, \] so \(d_\infty(x,y)>\varepsilon/2\). This holds for every \(y\in c_0\); therefore the open ball of radius \(\varepsilon/2\) about \(x\) misses \(c_0\). Every point of the complement has such a neighborhood, so the complement is open and \(c_0\) is closed.
Now let \((y^{(j)})_{j\in\mathbb{N}}\) be Cauchy in \(c_0\). It is also Cauchy in \(\ell^\infty\), because the metric is the same, so completeness of \(\ell^\infty\) gives a limit \(y\in\ell^\infty\). To check that this limit belongs to \(c_0\), fix \(\varepsilon>0\). Choose \(j\) such that \(d_\infty(y^{(j)},y)<\varepsilon/2\). Since \(y^{(j)}\in c_0\), there is \(N\) such that \(|y_n^{(j)}|<\varepsilon/2\) for all \(n\geq N\). For those \(n\), \[ |y_n|\leq |y_n-y_n^{(j)}|+|y_n^{(j)}|<\varepsilon/2+\varepsilon/2=\varepsilon. \] Thus \(y_n\to0\), so \(y\in c_0\). The convergence \(y^{(j)}\to y\) in \(\ell^\infty\) is therefore convergence in \(c_0\), proving completeness. \(\square\)
Worked Example: The Constant-One Sequence Is Not in \(c_0\)
Let \(\mathbf{1}\) be the bounded sequence with \(\mathbf{1}_n=1\) for every \(n\). It belongs to \(\ell^\infty\), but it does not tend to zero, so it is not in \(c_0\). We can compute its distance from \(c_0\) exactly.
For any \(y\in c_0\), \(y_n\to0\), and therefore \(|1-y_n|\to1\). Since \(d_\infty(\mathbf{1},y)=\sup_n|1-y_n|\), this supremum is at least \(1\): it is an upper bound for every term, so it is at least the limit of those terms. For the zero sequence \(y=0\in c_0\), every coordinate difference is \(1\), giving \(d_\infty(\mathbf{1},0)=1\). Thus the distance from \(\mathbf{1}\) to \(c_0\), defined as \(\inf_{y\in c_0}d_\infty(\mathbf{1},y)\), is exactly \(1\). This also shows concretely that \(c_0\) is a proper subspace of \(\ell^\infty\).
Why Coordinatewise Convergence Is Not Enough
Common pitfall. A coordinatewise limit does not by itself give a limit in the sup metric. For \(m\in\mathbb{N}\), let \(v^{(m)}_n=1\) if \(n\leq m\), and \(v^{(m)}_n=0\) if \(n>m\). For each fixed \(n\), \(v^{(m)}_n\to1\) as \(m\to\infty\), so the coordinatewise limit is \(\mathbf{1}\in\ell^\infty\). But for every \(m\), when \(n>m\) we have \(|v_n^{(m)}-\mathbf{1}_n|=|0-1|=1\), while for \(n\leq m\) the difference is \(0\). Hence \(d_\infty(v^{(m)},\mathbf{1})=1\) for every \(m\), not a quantity tending to zero. The same calculation with \(k>m\) gives \(d_\infty(v^{(m)},v^{(k)})=1\), because at \(n=m+1\) the difference is \(1\), and no coordinate difference exceeds \(1\). Thus this sequence is not Cauchy in the sup metric. The completeness proof succeeds because the Cauchy condition controls all coordinates with one bound, uniformly.
The central distinction is that coordinate limits first produce a candidate sequence, while the Cauchy condition supplies uniform control that proves both boundedness of the candidate and convergence to it in \(d_\infty\). Closedness then ensures that a uniform limit of sequences tending to zero still tends to zero. These facts make \(\ell^\infty\) and \(c_0\) useful complete metric spaces, while keeping their different membership conditions visible.
Check Your Understanding
Use the sup metric throughout, and justify any claim of convergence by estimating the supremum over all coordinates.
- For \(x_n=1/n\), compute the sup distance between \(x\) and the sequence that agrees with \(x\) through coordinate \(m\) and is zero afterward.
- Explain why a Cauchy sequence in \(\ell^\infty\) has a limit in each coordinate, and why coordinate limits alone would not prove convergence in \(d_\infty\).
- Show that the sequence \(\mathbf{1}\) does not belong to \(c_0\), and determine its distance from \(c_0\).
- For \(u_n^{(m)}=1/(m+n)\), compute \(d_\infty(u^{(m)},0)\) and use it to prove \(u^{(m)}\to0\) in \(c_0\).
- Which part of the proof that \(c_0\) is closed uses the defining tail condition \(y_n\to0\)?