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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.

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  • $\begingroup$ Why did you choose $j_{\epsilon, n}$ to depend on $n$? $\endgroup$ – P. J. Mar 12 at 7:30
  • $\begingroup$ @P.J. My bad. I've edited it accordingly. $\endgroup$ – SupremePickle Mar 12 at 7:51
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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 $$ j\ge k \implies d(x^j,x)=\sum_{n=1}^m 2^{-n}\frac {|x^j_n-x_n|}{1+|x^j_n-x_n|}+\sum_{n=m+1}^{\infty}2^{-n}\frac {|x^j_n-x_n|}{1+|x^j_n-x_n|}\le$$ $$\le\sum_{n=1}^m2^{-n}(e/2)+\sum_{n=m+1}^{\infty}2^{-n}<$$ $$<e/2+2^{-m}<e.$$

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  • $\begingroup$ 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 $\{(x^j_n)_j:n\le m\}.$ $\endgroup$ – DanielWainfleet Mar 12 at 8:39
  • $\begingroup$ 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 $X$. $\endgroup$ – DanielWainfleet Mar 12 at 11:24
  • $\begingroup$ Ah this makes sense. Thank you. $\endgroup$ – SupremePickle Mar 12 at 18:17
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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 $n_0$.

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  • $\begingroup$ I'm struggling to put this down formally.. Any chance you could elaborate? $\endgroup$ – SupremePickle Mar 12 at 7:51
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    $\begingroup$ @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 $\epsilon /{2N}$. $\endgroup$ – Kavi Rama Murthy Mar 12 at 7:54

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