# Complete sequence metric space proof

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.

• Why did you choose $j_{\epsilon, n}$ to depend on $n$? Mar 12, 2021 at 7:30
• @P.J. My bad. I've edited it accordingly. Mar 12, 2021 at 7:51

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 $$n_0$$.
• @Kraftsman Choose $N$ such that the second term is less than $\epsilon /2$. Then choose $j_0$ such that $j >j_0$ implies that each of the terms in the first sum is less than $\epsilon /{2N}$. Mar 12, 2021 at 7:54
Given any $$e>0,$$ take $$m$$ large enough that $$2^{-m} You have shown that each $$(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|. And let $$k=\max_{n\le m}j_n.$$ Now we have $$j\ge k \implies d(x^j,x)=\sum_{n=1}^m 2^{-n}\frac {|x^j_n-x_n|}{1+|x^j_n-x_n|}+\sum_{n=m+1}^{\infty}2^{-n}\frac {|x^j_n-x_n|}{1+|x^j_n-x_n|}\le$$ $$\le\sum_{n=1}^m2^{-n}(e/2)+\sum_{n=m+1}^{\infty}2^{-n}<$$ $$
• The idea is that if $m$ is large enough then the sum of the terms for $n\ge m+1$ in the series for $d(x^j,x)$ will be arbitrarily small, regardless of the values of $j$ and $x^j$ and $x,$ so we then attend to a finite set of series $\{(x^j_n)_j:n\le m\}.$ Mar 12, 2021 at 8:39
• The topology on $X=\Bbb R^{\Bbb N}$ generated by $d$ is the (Tychonoff) product topology. We can use $d$ to show $X$ is separable by showing that the set of all $(y_n)_n\in X$ such that (i) every $y_n\in \Bbb Q$ and (ii) $\{n\in\Bbb N : y_n\ne 0\}$ is finite, is a countable dense subset of $X$. Mar 12, 2021 at 11:24