Proving that Euclidean Space is a complete metric space

I start my proof with supposing that I have a Cauchy sequence $x^\Bbb{R^n}_k$ in $\Bbb{R^n}$- that is, $x^\Bbb{R^n}_k=(x_1, x_2,...,x_N,...,x_m,..,x_n...)$ ,where $x_i=(x^i_1,x^i_2,x^i_3,...,x^i_n)$ e.g. $x^i_2$ denotes the second element of the $i$th entry in $x^\Bbb{R^n}_k$.

Since $x^\Bbb{R^n}_k$ is a Cauchy sequence for any $\epsilon>0$: $\exists N \in\Bbb{N}$: $\forall m,n>N$: $d(x_m,x_n)<\epsilon$, where $d$ denotes the Euclidian metric in $\Bbb{R^n}$.

Some proofs I come across assumes/shows that the fact that $x^\Bbb{R^n}_k$ is a Cauchy is extended to the elements of $x^\Bbb{R^n}_k$ - that is $x_i=(x^i_1,x^i_2,x^i_3,...,x^i_n)$ is Cauchy $\forall i$ (e.g. https://proofwiki.org/wiki/Euclidean_Space_is_Complete_Metric_Space), and I do not understand how. Any clue or, of course, a full proof is appreciated.

Since your sequence is a Cauchy sequence in $\mathbb{R}^n$, each $(x_k^i)_{k\in\mathbb N}$ is a Cauchy sequence of real numbers. Therefore, it converges to some $y_i$. So, your sequence converges to $(y_1,y_2,\ldots,y_n)$.
Note: Each $(x_k^i)_{k\in\mathbb N}$ is a Cauchy sequence because, if $\varepsilon>0$, there is a natural $p$ such that$$q,r\geqslant p\implies\bigl\|(x_q^1,x_q^2,\ldots,x_q^n)-(x_r^1,x_r^2,\ldots,x_r^n)\bigr\|<\varepsilon$$and$$|x_q^i-x_r^i|\leqslant\bigl\|(x_q^1,x_q^2,\ldots,x_q^n)-(x_r^1,x_r^2,\ldots,x_r^n)\bigr\|.$$Therefore,$$q,r\geqslant p\implies|x_q^i-x_r^i|<\varepsilon.$$
• Yes, this is what I have been seeing/hearing. However, I do not understand why this is the case. Why does every element of $x_k$ also have to be Cauchy? – Larx Oct 28 '17 at 11:33
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