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For each $x = \{x_n\}$ and $y = \{y_n\}$ in $X$, define $d(x,y) = \sum_{n=0}^\infty \frac{1}{2^n} \frac{d_n(x_n,y_n)}{1+d_n(x_n,y_n)}$.

b. Show that $(X,d)$ is a complete metric space iff each $(X_n, d_n)$ is complete.

c. Show that $(X,d)$ is a compact metric space iff each $(X_n, d_n)$ is compact.

I just need a hint for b and c. Thank you in advance!

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closed as off-topic by Marios Gretsas, qwr, choco_addicted, Claude Leibovici, mechanodroid Sep 29 '17 at 8:05

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    $\begingroup$ What are your thoughts? Also, the definition of $d$ seems incorrect. Indeed, taking $X_n=\mathbb{R}$, $x_n=0$, and $y_n=1$ for every $n$, then $d(x,y)=\sum_{n=0}^\infty 1/2$ does not converge. $\endgroup$ – John Griffin Sep 28 '17 at 23:04
  • $\begingroup$ @JohnGriffin maybe there must be a sequence $a_n$ in the summation such that $\Sigma |a_n| < \infty$ $\endgroup$ – Marios Gretsas Sep 28 '17 at 23:07
  • $\begingroup$ @Lo12 ...as JohnGriffin pointed out,the first statement might be wrong..and secondly you do not offend me but this site $\endgroup$ – Marios Gretsas Sep 28 '17 at 23:10
  • $\begingroup$ I suggest getting a bounty for this one. $\endgroup$ – Dionel Jaime Sep 28 '17 at 23:19
  • $\begingroup$ @JohnGriffin Yes it seems I forgot a piece of the definition. Let me edit it $\endgroup$ – Lo12 Sep 28 '17 at 23:19
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It will be useful to first prove that a sequence $(x_n)$ of elements in $X$ converges to $x$ if and only if $(x_n(k))_n$ converges to $x(k)$ for every $k\in\mathbb{N}$.

From here (b) should follow from the definition of completeness and (c) should follow using the fact that a metric space is compact iff it is sequentially compact.

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