Componentwise Convergence for sequences in $\mathbb{R^{n}}$ I'm trying to prove that a sequence in $\mathbb{R^{n}}$ converges, if and only if it converges componentwise, using $N- \epsilon$ definition.
Let $(x_{k})$ be a sequence in $\mathbb{R^{n}}$ that converges to $a$.  I have to show that for all $j = 1, 2,..,k$ $$x_{k,j} \rightarrow a_{j} \ \ \text{as} \ \ k \rightarrow \infty$$
We know,
$$\forall \epsilon > 0 : \exists N \in \mathbb{N} : \forall k > N: \lVert x_{k} - a \rVert < \epsilon$$
which means $\lVert (x_{k,1} - a_{1}, x_{k,2} - a_{2},...,x_{k,n} - a_{n}) \rVert < \epsilon$
Fix $1 \leq j \leq n$ I have to prove $\lVert x_{k,j} - a_{j} \rVert < \epsilon$
How do I continue?
 A: Hint
1. $$|x_{k,i}-a_i|\leq\Vert x_k - a\Vert$$
2. $$\Vert x_k - a \Vert \leq \sum_{i=1}^n |x_{k,i}-a_i|$$
You can figure out which of the two you need for which case. I hope this helps.

Edit (explanation)
1. For two vectors $x=(x_1,...,x_n)$ and $y=(y_1,...,y_n)$ we have: $$|x_i -y_i |=\sqrt[]{(x_i-y_i)^2}\leq \sqrt[]{(x_1-y_1)^2+...+(x_n-y_n)^2}=\Vert x-y\Vert$$
2. For two vectors $x=(x_1,...,x_n)$ and $y=(y_1,...,y_n)$ we have: 
\begin{align}
\Vert x-y\Vert &= \Vert (x_1-y_1,...,x_n-y_n)\Vert\\
&= \Vert  (x_1-y_1,0,...,0) +  (0,x_2-y_2,0,...,0)+...+ (0,0,...,0,x_n-y_n)\Vert \\
&\leq \Vert  (x_1-y_1,0,...,0)\Vert + \Vert (0,x_2-y_2,0,...,0)\Vert+...+ \Vert(0,0,...,0,x_n-y_n)\Vert\\
&= |x_1-y_1| + |x_2-y_2| + ... |x_n-y_n|\\
&= \sum_{i=1}^n|x_i-y_i|
\end{align}
To arrive the inequality we have used the triangle inequality. You see that those are properties for any vectors $x$ and $y$ so this applies to your sequence too.
A: HINT
$$|x_k^{(j)}-a_j|=\sqrt{(x_k^{(j)}-a_j)^2} \leq \sqrt{(x_k^{(1)}-a_1)^2+...+(x_k^{(j)}-a_j)^2+...+(x_k^{(n)}-a_n)^2}$$ $$=||x_n-a||_2 \to 0$$ where $x_k=(x_k^{(1)},...,x_k^{(n)})$ and $a=(a_1,a_2....a_n)$
$||.||_2$ is the Euclideian norm in $\Bbb{R}^n$.
Every other norm in $\Bbb{R}^n$ is equivalent to this norm,  because in a finite dimensional space,all nomrs are equivalent.
