Why do sequences need to converge with respect to a norm? Suppose there is a sequence $x_n$ in $\mathbb{R}^n$, which converges to $x$, with respect to $\lVert\cdot\rVert_2$, say. Then we can write down vectors $x_1, x_2, x_3, ...$, and so on, and, with each $x_i$, we will be closer and closer to $x$. Now, suppose there is another sequence, $y_n$, which converges to the same point $x$, but now with respect to $\lVert\cdot\rVert_\infty$. Then we can write this sequence down as numbers $y_1, y_2, ...$, and with each $y_i$ we will be closer and closer to $x$.
My point is that why do we need sequences to converge with respect to certain norms if we can just write them down as numbers, one by one, without any norms, and the numbers (or vectors) will just be closer and closer to the point of convergence? I understand that convergence with respect to a norm $\lVert\cdot\rVert$ means that $\lim\limits_{n\to\infty}\lVert x_n-x \rVert=0$. But if we can write down approaching vectors one by one, without any norms, and the vectors will be approaching a certain point, then shouldn't this mean that convergence must be satisfied with respect to any norm, and the norm should not be relevant to convergence at all?
 A: Your intuition is correct, to an extent: in $\mathbb R^n$, all norms are equivalent (i.e., they define the same notion of convergence). In particular, convergence in any norm is equivalent to coordinate-wise convergence, as you say. 
There are two things to consider:


*

*In many applications, the vectors represent certain objects from a model (in physics, economics, etc.) and so the concrete metric used could have a significance in terms of its meaning for the model. 

*Your intuition does not carry to infinite dimension. Consider $\ell^2(\mathbb N)$. This is the set of sequences $x$ such that $\sum_{j=1}^\infty|x(j)|^2<\infty$. Consider the sequence $\{x_n\}$, where each $x_n$ consists of the sequence with all zeroes with the exception of the $n^{\rm th}$ which is $1$. Then it is clear that, pointwise, $x_n\to0$, since $x_n(j)=0$ if $n>j$. But, in the 2-norm, $\|x_n\|_2=1$, so the sequence does not converge to zero. Or to also use an example you give, consider now the sequence $\{y_n\}$, where $$y_n=\left(\overbrace{\frac1{\sqrt n}\,\ldots,\frac1{\sqrt n}}^{n},0,0,\ldots\right).$$ Then $\|y_n\|_2=1$ for all $n$, but $\|y_n\|_\infty=1/{\sqrt n}\to0$. 
Similarly, if 
$$z_n=\left(\overbrace{\frac1n\,\ldots,\frac1n}^{n},0,0,\ldots\right),$$
then $\|z_n\|_2=1/{\sqrt n}$, while $\|z_n\|_1=1$ for all $n$. 
