Metrics on $\mathbb{R}^k$ that don't behave well with respect to sums

Question

Let $X=\mathbb{R}^k$ ($k\in \mathbb{Z}_{>0}$), endowed with a metric $d$. Let also $(x_n)_n$, $(y_n)_n$ be two sequences in $X$ convergent to $x,y\in X$ respectively, that is $d(x_n,x)\to 0$ and $d(y_n,y)\to 0$ as $n\to\infty$.

If $d$ is induced by a norm, then it is easy to show (using the triangular inequality) that the sequence $(x_n+y_n)_n$ converges to $x+y$, i.e. $d(x_n+y_n,x+y)\to 0$.

While it is clear in the proof where the norm assumption is used, I still have not managed to construct a counterexample for the convergence of $(x_n+y_n)_n$ to $x+y$ when $d$ is not induced by a norm. I think that at least a non-constructive proof of existence of a suitable $d$ should be doable by means of the axiom of choice, but I have managed to prove that either yet. Ideas?

Answer 1

If we put a norm on $\Bbb R^n$ then they are all equivalent, so the metrics from them give the same standard topology on $\Bbb R^n$. And as addition is continuous, we always have, purely from topology, $x_n \to x, y_n \to y \implies x_n + y_n \to x+y$.

If we have a different metric on $\Bbb R^n$ which induces a different topology on $\Bbb R^2$, then we can make examples: e.g. for the river metric (defined here) on $\Bbb R^2$ we have $(1+\frac1n, 0) \to (1,0)$, $(0,1+\frac1n) \to (0,1)$ but $(1+\frac1n, 1+\frac1n)\not \to (1,1)$.

Answer 2

Here is my attempt.

define the metric $d(x,y)$ as follows. $$ d(x,y)= \left\{\begin{array}{lr} |x - y| & \text{for }|x|,|y|<1\\ \, \delta(x-y) & \text{otherwise}\\ \end{array}\right\} $$ where $|\cdot|$ is standard norm in $\mathbb{R}^k$ (root square norm) , $\delta(x-y) = \left\{\begin{array}{lr} 1 & \text{for }x \neq y\\ 0 & \text{for }x = y\\ \end{array}\right\} $

nonnegativity, identity and symmetry (for metric) is clear.

For show transitivity, enough to consider following two cases.

1 . $x,y$ are in 1 - ball and $z$ is at outside :

$d(x,y) = |x-y|<2$ , $d(y,z) = d(x,z) = 1$

2 . $x$ is in 1 - ball and $y,z$ are at outside :

$d(x,y) = d(x,z) = 1$ , $d(y,z)$ is $0$ or $1$.

$$\\$$ (maybe) $d(x,y)$ is really a metric.

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Now, let consider $(x_n) = [\frac{2}{3} - \frac{1}{10 n}]e_1 $.

Then $x_n \to \frac{2}{3}e_1\,\,$ in $(\mathbb{R}^k , d).\quad$ ($|x_n| , |\frac{2}{3}e_1| < 1$)

However, since $2x_n = x_n + x_n = [\frac{4}{3} - \frac{1}{5 n}]e_1$ have radius greater than 1, $$d(2x_n,2x_m) = 1 \quad\forall n\neq m .$$

This means $(2x_n)$ can not converge to any other point.