Complete Metrics Question: for $x, y \in \Bbb R$ let $d_1(x,y) = max |x_j-y_j|$ for $j=1,2,...,k$
Show that $d_1$ is a complete metric for $\Bbb R^k$. 
I have already shown that it is a metric, and now I am confused about the 
"complete" notion. My textbook's definition of a complete metric is that every cauchy sequence in my space, in this case $\Bbb R^k$ defined with this metric, converges to an element within the space. Any hints appreciated.
 A: Let $(x^{(n)})_{n\in\mathbb N}$ be a Cauchy sequence in $\mathbb R^k$ with respect to the metric $d_1$. By definition, this means that for every $\varepsilon>0$, there exists some $N\in\mathbb N$ such that for every pair integers $m,n\geq N$, one has that $$d_1(x^{(m)},x^{(n)})=\max_{j\in\{1,\ldots,k\}}\left|\,x_j^{(m)}-x_j^{(n)}\right|<\varepsilon.$$ This implies that $$\left|\,x_j^{(m)}-x_j^{(n)}\right|<\varepsilon\quad\text{for each $j\in\{1,\ldots,k\}$}$$ whenever $m,n\geq N$. Hence, $(x_j^{(n)})_{n\in\mathbb N}$ is a Cauchy sequence in $\mathbb R$ with respect to the ordinary Euclidean metric (difference of absolute values) for each $j\in\{1,\ldots,k\}$.
Now, the space of real numbers is complete with respect to the ordinary Euclidean metric. This implies that for each $j\in\{1,\ldots,k\}$, there exists some $x_j\in\mathbb R$ such that the sequence $(x_j^{(n)})_{n\in\mathbb N}$ converges to $x_j$ with respect to the Euclidean metric. As a consequence, for each $\varepsilon>0$, there exist some $N_j\in\mathbb N$ such that for any integer $n\geq N_j$, one has $$\left|\,x_j^{(n)}-x_j\right|<\varepsilon.$$
I claim that the sequence $(x^{(n)})_{n\in\mathbb N}$ in $\mathbb R^k$ converges to $x\equiv(x_1,\ldots,x_k)$ with respect to the metric $d_1$, which will complete the proof. Let $N\equiv\max\{N_1,\ldots,N_k\}\in\mathbb N$. If $n$ is an integer such that $n\geq N$, then $n\geq N_j$ for each $j\in\{1,\ldots,k\}$ by the definition of $N$. By the preceding argument, this yields that $$\left|\,x_j^{(n)}-x_j\right|<\varepsilon$$ for every $j\in\{1,\ldots,k\}$, whence $$d_1(x^{(n)},x)=\max_{j\in\{1,\ldots,k\}}\left|\,x_j^{(n)}-x_j\right|<\varepsilon.$$
