# Showing $(l^1(\mathbb{N}), d_1)$ is complete

Let $$l^1(\mathbb{N}) = \left\{(x_n)_n \mid \sum_{n=0}^{\infty} |x_n| \ \text{converges} \right\}.$$ Prove that $(l^1(\mathbb{N}), d_1)$ with $d_1(x,y) = \sum_{n=0}^{\infty} |x_n - y_n|$ is a complete metric space.

My proof: I'd appreciate any feedback.

Let $(x_n)_n$ be a Cauchy sequence in $(l^1(\mathbb{N}), d_1)$. We need to prove that $(x_n)_n$ converges to a limit in $l^1$. Let $\epsilon > 0$. For all $n,m \in \mathbb{N}$ we have that $$| x_n - x_m | \leq \sum_{n=0}^{N} |x_n - x_m|.$$ Since $(x_n)$ is Cauchy, there exists a $n_0 \in \mathbb{N}$ such that for all $n, m \geq n_0$ we have $$| x_n - x_m | \leq \lim_{N \to \infty} \sum_{n=0}^{N} |x_n - x_m | = d_1(x_n, x_m) < \epsilon/2.$$ This shows $(x_n)$ is a Cauchy sequence in $\mathbb{R}$ for the usual metric, and since $\mathbb{R}$ is complete, $(x_n)$ converges. Denote the limit by $(y_n)_n$. So $(x_n) \rightarrow (y_n)$ for the $d_1$ metric (can I claim that at this point?) We now show that $(y_n) \in l^1$. From the triangle inequality it follows that $$\sum_{k=n_0 + 1}^{N} |y_k| \leq \sum_{k=n_0 + 1}^{N} |y_k - x_k | + \sum_{k=n_0 + 1}^{N} |x_k|.$$ If we now take the limit $N \to \infty$ we find that $$\sum_{k=n_0 + 1}^{\infty} |y_k| \leq \epsilon/2 + \epsilon/2 = \epsilon.$$ This shows $(y_n)$ is an absolutely convergent series, and so $(y_n) \in l^1$. So $(l^1, d_1)$ is complete.

• I am confused about your notation. By $(x_n)_n$, do you mean a Cauchy sequence of sequences that absolutely converge? If so, why do you denote the limit of $(x_n)$ by $(y_n)_n$ ? – spiritfire Jul 27 '16 at 10:57
• Yes. I denote the limit by $(y_n)$ because the limit is itself a sequence? What else would be a better notation? – Kamil Jul 27 '16 at 11:02
• your sequence should have two indices: One for the sequence $\{x_n\}$ of elements in $\ell^1$, while each $x_n$ is of the form $x_n = (x_n^1, x_n^2, x_n^3, \cdots)$. – user99914 Jul 27 '16 at 11:05
• It is also more clear to use $\| \cdot\|$ to denote the $\ell^1$-norm and the usual $| \cdot |$ to denote the absolute value. – user99914 Jul 27 '16 at 11:05
• You can refer to the proof of Riesz-Fischer theorem in Stein's book. Let $(x_n^m)$ be a Cauchy sequence in $l^1$, first find the limit componentwise: define $y_m=limx_n^m$, then prove that $(x_n^m)$ converges to $y_m$ in the metric. – spiritfire Jul 27 '16 at 11:21