# Complete metric on the space of sequences

Let $S$ be the set of all real sequences $x=\{x_n\}$, $d: S\times S \rightarrow \mathbb R$ be defined by:

$$d(x,y)=\sum_{n=1}^{\infty} \frac{|x_n-y_n|}{2^{n}[1+|x_n-y_n|]}.$$

Show that $(S,d)$ is a complete metric space.

I want to say we can use the Cantor Intersection property or an isometry here to prove completeness. However, I am having trouble getting started. Any suggestions?

• Hint: a sequence $(x^k)_{k\in\mathbb N} = (x_n^k)_{k,n\in\mathbb N}$ in $S$ converges to $(x_n)_{n\in\mathbb N}$ if and only $x_n^k \to x_n$ for $k\to\infty$ and every $n\in\mathbb N$. – user251257 Sep 14 '15 at 3:14

Hint: Pick a Cauchy sequence $\{x_n\}$ in $S$. Pick $\varepsilon>0,k\in\mathbb{N}\setminus\{0\}$.
\begin{align} \exists N:m,n\ge N&\implies d(x_{m},x_n)<{\varepsilon\over2^k(\varepsilon+1)}\\ &\implies\sum_{j=1}^\infty{|x_{m,j}-x_{n,j}|\over2^{j}[1+|x_{m,j}-x_{n,j}|]}<{\varepsilon\over2^k(\varepsilon+1)}\\ &\implies {|x_{m,k}-x_{n,k}|\over2^{k}[1+|x_{m,k}-x_{n,k}|]}<{\varepsilon\over2^k(\varepsilon+1)}\\ &\implies |x_{m,k}-x_{n,k}|<\varepsilon \end{align} Therefore $y_n=x_{n,k}$ is a Cauchy sequence in $\mathbb{R}$ converging to the limit $x^*_{k}$. Prove that $x^*$ is the limit of $\{x_n\}$ and you're done.
Note: If $x_n\in S$, $x_n$ is a sequence and by $x_{n,k}$ we mean its $k$-th term.
• That is ,if $(\;(x_{n.i})_{n\in N}\;)_{i \in N}$ is a Cauchy sequence in $S$ then $(x_{n,i})_{i\in N}$ is a real Cauchy sequence for each $n\in N$, converging to some $y_n$. So prove that the sequence in $S$ converges to $(y_n)_{n\in N}.$ – DanielWainfleet Feb 5 '16 at 22:44