Let $\{(X_n, d_n)\}$ be a sequence of metric spaces, and let $X=\prod X_n$. [closed]

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!

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

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• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Marios Gretsas, qwr, choco_addicted, Claude Leibovici, mechanodroid
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• 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. – John Griffin Sep 28 '17 at 23:04
• @JohnGriffin maybe there must be a sequence $a_n$ in the summation such that $\Sigma |a_n| < \infty$ – Marios Gretsas Sep 28 '17 at 23:07
• @Lo12 ...as JohnGriffin pointed out,the first statement might be wrong..and secondly you do not offend me but this site – Marios Gretsas Sep 28 '17 at 23:10
• I suggest getting a bounty for this one. – Dionel Jaime Sep 28 '17 at 23:19
• @JohnGriffin Yes it seems I forgot a piece of the definition. Let me edit it – Lo12 Sep 28 '17 at 23:19

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 $k\in\mathbb{N}$.