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Let $s:=\mathcal{F}(\mathbb{N})=\{x = (x_j)_{j\in \mathbb{N}}|x+j\in \mathbb{C} \ \forall j\in \mathbb{N}\}$ be the space of all scalar sequences. And $d(x,y):=\sum_{j=1}^\infty 2^{-j}\frac{|x_j-y_j|}{1+|x_j-y_j|},$ provided it is a metric.

Now I want to prove $x\rightarrow y$ in $d(\cdot,\cdot)$ is equivalent to $x\rightarrow y$ componentwise, namely $x_j\rightarrow y_j$ for all $j\in\mathbb{N}$.

'$\Rightarrow$' is easy by formulating the contraposition.

How to prove "$\Leftarrow$"? Because pointwise convergence does not imply convergence in metric, I am stuck here.

Thank you in advance.

1 Answers 1

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Hint 1: Notice that $x\mapsto \frac{x}{1+x}$ is $0$ for $x=0$, it is positive for positive value of $x$ and moreover it is increasing on $[0,+\infty)$. Now use the fact that, for any $j\in\mathbb N$, $2^{-j}\frac{|x_j-y_j|}{1+|x_j-y_j|}\leq d(x,y).$

Hint 2: Notice that the sum $\sum_{j=1}^{+\infty}2^{-j}$ is convergent. Thereefore, for any $\varepsilon>0$ there is $j_0(\varepsilon)$ such that $\sum_{j=j_0(\varepsilon)}^{+\infty}2^{-j}<\varepsilon.$ Morever $x\mapsto \frac{x}{1+x}<1,\;\forall x\in\mathbb R_{\geq 0}.$ Now you are left with finitely may terms of the series to control.