Complete sequence metric space proof

Question

Suppose $X = \{\{x_n\}_{n \in \mathbb{N}} : x_n \in \mathbb{R}$ $\forall n\geq 1\}$. Given that $d: X \times X \rightarrow \mathbb{R}$ such that $d(\{x_n\}_{n \in \mathbb{N}}, \{y_n\}_{n \in \mathbb{N}}) = \sum_{n = 1}^{\infty} 2^{-n}\frac{|x_n - y_n|}{1 + |x_n - y_n|}$ is a metric, show that $(X, d)$ is a complete metric space.

My attempt:

Consider $\{x^j\}_{j\geq 1}$ Cauchy in $d$. Fix $n \geq 1$. Then for all $0 < \epsilon < \frac{1}{2}$, there is an $j_{\epsilon} \in \mathbb{N}$ so that $d(x^{(j)}, x^{(k)}) < \frac{\epsilon}{2^n}$ for all $j, k \geq j_{\epsilon}$. So for $j, k \geq j_{\epsilon}$ we have

$\sum_{m = 1}^{\infty} 2^{-m}\frac{|x_m^{(j)} - x_m^{(k)}|}{1 + |x_m^{(j)} - x_m^{(k)}|} < \frac{\epsilon}{2^n}$

$\implies 2^{-m}\frac{|x_m^{(j)} - x_m^{(k)}|}{1 + |x_m^{(j)} - x_m^{(k)}|} < \frac{\epsilon}{2^n}$

$\implies |x_m^{(j)} - x_m^{(k)}| < \frac{\epsilon}{1-\epsilon} \leq 2\epsilon$

So $\{x^{(j)}\}_{j\geq 1}$ is cauchy $\implies x_n^{(j)} \rightarrow x_n$ as $x \rightarrow \infty$.

Is this correct so far? I have come this far but I'm not sure how to complete the proof. Any assistance is much appreciated.

Answer 1

Given any $e>0,$ take $m$ large enough that $2^{-m}<e/2.$ You have shown that each $(x^j_n)_j\to x_n$ as $j\to\infty.$ So for $n\le m,$ take $j_n $ such that $(j\ge j_n\implies |x^j_n-x_n|<e/2)$. And let $k=\max_{n\le m}j_n.$ Now we have $$ j\ge k \implies d(x^j,x)=\sum_{n=1}^m 2^{-n}\frac {|x^j_n-x_n|}{1+|x^j_n-x_n|}+\sum_{n=m+1}^{\infty}2^{-n}\frac {|x^j_n-x_n|}{1+|x^j_n-x_n|}\le$$ $$\le\sum_{n=1}^m2^{-n}(e/2)+\sum_{n=m+1}^{\infty}2^{-n}<$$ $$<e/2+2^{-m}<e.$$

Answer 2

Your $j_{\epsilon,n}$ should be just $j_{\epsilon}$. Take $m=n$ in your argument. Rest is fine.

You now have $x_n=\lim_{j \to \infty} x_n^{j}$ for each $n$. Now let $x=(x_n)$. Then $d(x,x^{j}) \leq \sum_{m=1}^{N} 2^{-m} \frac {|x_m^{j}-x_m|} {1+|x_m^{j}-x_m|} + \sum_{m=N+1}^{\infty} 2^{-m} \frac {|x_m^{j}-x_m|} {1+|x_m^{j}-x_m|}$. The second term here is at most $\sum_{m=N+1}^{\infty} 2^{-m} <\epsilon$ if $N$ is large enough. The first term is a finite sum and each term tends to $0$. I will let you write out a proof in terms of $\epsilon$ and $n_0$.