Prove that a sequence of vectors $(x^{(k)})$, where $x^{(k)} = (x^{(k)}_1,...,x^{(k)}_n)$, converges to $x = (x_1,...,x_n)$ in $(\mathbb{R}^n, ||.||_1)$ iff each coordinate sequence $(x^{(k)}_i)$ converges to $x_i$ in $\mathbb{R}$.
I haven't started the proof yet, because I was confused about the notation. Namely what the $k$ signifies. Is $k$ being used to differentiate between each separate vector? If so, then I assume each vector $x^{(k)}$ has $n$ components--$(x^{(k)}_1,...,x^{(k)}_n)$. But the last sentences state iff each coordinate sequence $(x^{(k)}_i)$ converges to $x_i$. I thought it would be $x^{(i)}$ converges to $x_i$. What I mean is that, say a vector $x^{(3)}$ would converge to $x_3$ (the sequence of the third vector converges to the 3rd component of $x$_). So it with this assumed understanding, I am confused about the wording of the last sentence.
You have a sequence of vectors $x^{(1)}, x^{(2)}, x^{(3)} , \ldots$ They are a sequence in $\mathbb{R}^n$. We claim that they converges to $x$ which is another vector in $\mathbb{R}^n$ when measured in $L_1$ norm if and only if when we inspect the sequence of each component, they converge to $x_i$.
$k$ is the index for the sequence and $i$ refers to the $i$th component of the $n$-tuple.
For example, let $x^{(k)}=\begin{bmatrix} \frac1{k} \\ 1+\frac1{k^2}\end{bmatrix}$, then $x^{(k)}_1=\frac1k$ and $x^{(k)}_2=1+\frac1{k^2}$. $x_1=0, x_2=1$.