# 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?

• What does "closer" mean without some notion of a metric? The norms are the most common metrics, you don't have to choose them, but it's a very reasonable choice. – Adam Hughes Nov 24 '16 at 3:10
• Is $(3,-1)$ or $(-3,1)$ closer to $(0,0)$? Norms, metrics, and other abstract concepts precisely express what closeness mean... – user251257 Nov 24 '16 at 3:17
• Your question is in a sense circular. In $\mathbf R^n$, convergence in both the norm $||\cdot||_2$ and in the norm $||\cdot||_\infty$ are both equivalent to what you're asking about: componentwise convergence. In fact every vector space norm on $\mathbf R^n$ has its convergence equivalent to componentwise convergence. So it's all a moot point. The situation changes completely if you work in the space $C[0,1]$ of continuous functions $[0,1] \rightarrow \mathbf R$, where the analogues of $||\cdot||_\infty$ and $||\cdot||_2$ do not define the same notion of convergence. – KCd Nov 24 '16 at 3:22
• Since the norms are equivalent on $\mathbb{R}^n$, the choice is usually one of convenience. On infinite dimensional space (such as the space of sequences) the choice of norm makes a big difference. – copper.hat Nov 24 '16 at 3:27

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.
2. 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 $n$.