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Let $s$ denote the metric space of all sequences of real or complex numbers with the following metric: $$ d( (\xi_j), (\eta_j) ) := \sum_{j=1}^{\infty} \frac{1}{2^j} \frac{|\xi_j - \eta_j|}{ 1 + |\xi_j - \eta_j| } $$ for any sequences $(\xi_j)$, $(\eta_j)$ of numbers.

Then if $(x_n)$ is a sequence in $s$, where $x_n \colon= (\xi_{j}^{(n)})$ for all $n = 1, 2, 3, \cdots$, and if $x$ is an element of $s$, where $x \colon= (\xi_j)$, then we have to prove that $x_n$ converges to $x$ as $n$ goes to infinity if and only if $\xi_{j}^{(n)}$ converges to $\xi_j$ for each $j = 1, 2, 3, \cdots$. How to?

I've figured out that if $(x_n)$ converges to $x$, then $\xi_{j}^{(n)}$ converges to $\xi_j$ for each $j = 1, 2, 3, \cdots$.

How to prove the converse, I wonder?

share|cite|improve this question
Begin with the easy direction: if $d((\xi_j^{(n)}),(\xi_j))<\epsilon$, how can you estimate $|\xi_j^{(n)}-\xi_j|$ from above? – user53153 Jan 28 '13 at 2:18
Yes, it is easy to go in this direction. How to prove the converse now? – Saaqib Mahmuud Feb 9 '13 at 6:37

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