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Let $S=\{0,1,2,...,N-1\}$ be the set of N first natural numbers. We define $X = \sum$, where $x \in \sum$ if $x=x_1x_2x_3...x_n...$ where $x_n \in S$. We define $$d(x,y) = d(x_1x_2x_3...,y_1y_2y_3...) = \sum_{i=1}^{\infty}\frac{|x_i-y_i|}{(N+1)^i}.$$

Show that $(\sum,d)$ is a complete metric space.

I was able to show that (X, d) is metric space. Now I need to show that X is complete.

Let $ (x_k)_{k\in\mathbb {N} } $ be an arbitrary Cauchy sequence in X. Then $\forall \epsilon >0 \: \exists \:\delta >0$ $\:$ such that $$d(x_k,x_l)<\epsilon, \:\: \forall \: k,l>\delta.$$

I got stuck here.

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Note that $x_k$ is a member of $\Sigma$, i.e. is itself a sequence of members of $S$. I'll use $x_{k,j}$ for the $j$'th element in the sequence $x_k$.

Hint: for any fixed $j$, show that $x_{k,j}$ is eventually constant, i.e. there exists $K$ such that $x_{k,j} = x_{k',j}$ for all $k, k' > K$.

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  • $\begingroup$ but what can I conclude from this? $\endgroup$ – Guilherme de Loreno Dec 8 '17 at 16:42
  • $\begingroup$ The $j$'th element of the limit of the Cauchy sequence. Then prove that it is the limit. $\endgroup$ – Robert Israel Dec 8 '17 at 19:43

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