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Let $X$ be the space of all $n-$tuples $x=(x_1,x_2,\ldots,x_n)$ of real numbers. Define $$d(x,y)=\max_i |x_i-y_i|, \qquad \text{where } y=(y_1,y_2,\ldots,y_n).$$ Show that $(X,d)$ is complete.

Edited: Here is my conclusion

Consider a Cauchy sequence $\{x_m\}$ in $\Bbb R^n$, $$x_m=(\xi_1^{(m)},\cdots, \xi_n^{(m)})$$ Since $\{x_m\}$ is Cauchy, for every $\varepsilon>0$ there is an $N$ such that \begin{equation} d(x_m,x_r)=\max_i|\xi_{i}^{(m)}-\xi_{i}^{(r)}|<\varepsilon\qquad (m,r>N)\tag{1} \end{equation} Then we have $$|\xi_{i}^{(m)}-\xi_{i}^{(r)}|<\varepsilon$$ This shows for each fixed $i$, ($1\le i\le n$), the sequence \begin{equation} \{\xi_i^{(1)},\xi_i^{(2)},\ldots\}\tag{2} \end{equation} is a Cauchy sequence of real numbers. Since $\Bbb R$ is complete the sequence in $(2)$ converges say, $\xi_i^{(m)}\to\xi_i$ as $m\to \infty$. Using these $n$ limits, we define $x=(\xi_1,\ldots,\xi_n)\in\Bbb R^n$.

From $(1)$, with $r\to\infty$, $$d(x_m,x)<\varepsilon$$ This shows that $\Bbb R^n$ is complete w.r.t $d$.

What do you say?

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    $\begingroup$ Where are you stuck? $\endgroup$ – user228113 Apr 9 '16 at 14:34
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    $\begingroup$ It's equivalent to $d=\sqrt{\sum |x_i-y_i|^2}$ so it's equally complete. $\endgroup$ – Zelos Malum Apr 9 '16 at 14:36
  • $\begingroup$ @G.Sassatelli: I don't know how to show a Cauchy sequence in $\Bbb R^n$ converges in the metric $d(x,y)=\max_i|x_i-y_i|$. $\endgroup$ – marya Apr 9 '16 at 14:41
  • $\begingroup$ @Sucre I guessed as much, since that's what "showing $(\Bbb R^n,d)$ is complete" means. At what point of the proof are you stuck? $\endgroup$ – user228113 Apr 9 '16 at 14:44
  • $\begingroup$ @ZelosMalum: you are right but I don't want the proof in that way. $\endgroup$ – marya Apr 9 '16 at 14:45
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Hint: You can check that $(x^{n})_{n}$ being Cauchy in $\mathbb{R}^{n}$ implies that $(x_{i}^{n})_{n}$ is Cauchy in $(\mathbb{R},|\cdot|)$ for each $i\in\{1,\ldots,n\}$. Can you use the completeness of $(\mathbb{R},|\cdot|)$ to finish the proof?

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  • $\begingroup$ are you saying being Cauchy in $\Bbb R^n$ implies Cauchy in $\Bbb R$ but $\Bbb R$ is complete so each coordinate conveges to a point in $\Bbb R$. Therefore $\Bbb R^n$ is complete. $\endgroup$ – marya Apr 9 '16 at 14:58
  • $\begingroup$ Yes, that's the general idea $\endgroup$ – parsiad Apr 9 '16 at 15:01
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If you want to do that without using metric equivalence then you showi t by showing that a sequence $(x_i)$ with $x_i=(x_i^1,x_i^2,\ldots,x_i^n)\in\mathbb{R}^n$ converges to an element in $\mathbb{R}^n$. This can be done by focus on only 1 coordinate really because it's only the largest deviation that defines the distance. As $(x_i)$ is cauchy we have that for each $j$ that $(x_i^j)$ is also cauchy sequences. And as $x_i^j\in\mathbb{R}$ we have that $(x_i^j)$ converges to a real number $y_j$, that is $x_i^j\to y_j$ and hence we have that $(x_i)\to y=(y_1,y_2,\ldots,y_n)\in\mathbb{R}^n$

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