TITLE: Componentwise Convergence in $\mathbb R^n$
QUESTION [7 upvotes]: I came across the following question while preparing an exercise for basic analysis:
Suppose $d$ is some arbitrary metric on $\mathbb R^N$ and $(x^n) \subset \mathbb R^N$ converges to $x\in \mathbb R^N$ with respect to $d$.
Question: Does $(x^n)$ converge component-wise with respect to the usual norm? (And if not: Are there simple counter-examples?)
Up to now, I know the basics:

The case is solved if $d$ is induced by a norm, because all norms on $\mathbb R^N$ are equivalent and $(x^n)$ converges componentwise iff it converges with respect to any $p$-norm.
The reverse of my question is not true. If for example $d$ is the discrete metric or the TGV metric, convergence in norm need not to imply convergence in metric.
If $\mathbb R^N$ is viewed as a topological space $(\mathbb R^N,\tau)$, then convergence with respect to $\tau$ need not to imply convergence in norm. (E.g. $\tau = \{\emptyset,\mathbb R^N\}$, which is not induced by a metric.)
All examples for metrics I considered so far (the usual examples) posess the desired property (and the proofs are easy, as these metrics are defined by using componentwise distances $|x_k-y_k|$). One could say, these metrics are just too normy, but I don't know, if there are not so normy examples or if my question can only be solved in a more abstract way.

My question arises naturally, so maybe there is a simple answer I overlooked up to now.  Thank you in advance for your help.

REPLY [5 votes]: The answer is no.  For instance, let $g:\mathbb{R}^N\to\mathbb{R}^N$ be any bijection at all, and define $d(x,y)=d_0(g(x),g(y))$, where $d_0$ is the Euclidean metric.  Then $(x^n)$ converges to $x$ with respect to $d$ iff $g(x^n)$ converges to $g(x)$ with respect to $d_0$.  Writing $f=g^{-1}$, your question thus becomes: if $f:\mathbb{R}^N\to\mathbb{R}^N$ is a bijection, then does $x^n\to x$ imply $f(x^n)\to f(x)$ (all with respect to the usual topology)?  That is, is every bijection $\mathbb{R}^N\to\mathbb{R}^N$ continuous?
The answer is obviously no, since (for instance) a bijection could swap two points and be the identity everywhere else.