Proving that $U=\{x=(x_1,\dots,x_n)\in\Bbb R^n : \sum|x_i|<1\}$ is open in $(\Bbb R^n,d)$: Does the metric $d$ matter?

Question

Consider the euclidean space $(\Bbb R^n,d)$ with $d(x,y)=\sqrt{\sum\limits_{i=1}^{n}|x_i-y_i|^2}$ and $U=\{x=(x_1,\dots,x_n)\in\Bbb R^n : \sum|x_i|<1\}$.

To prove that $U$ is open let'e take any $a\in U$ and $r={1\over n}(1-\sum\limits_{i=1}^n|a_i|)$ and let $b\in B(a,r)$

$\sum|b_i|=\sum|b_i-a_i+a_i|\le\sum(|b_i-a_i|+|a_i|)\le\sum d(a,b)+\sum|a_i|\le1+\sum|a_i|-\sum|a_i|=1$

Can we write $|b_i-a_i|\le d(a,b)$ for any distance on $d'$, even if it's not equivalent to $d(x,y)=\sqrt{\sum(x_i-y_i)^2}$?

Answer 1

Metric matters.

Consider the metric space $(\Bbb R,d') $ where $$d'(x,y)=\frac {|x-y|}{1+|x-y|} .$$ Clearly, $d'(x,y)\leq |x-y|$.

Answer 2

$U$ is open with respect to the topology induced by $|| \cdot ||_1$ as it's just the unit ball. As the norms on $\mathbb{R}^n$ are equivalent, $U$ is open with respect to your $d$. Hence you should consider a metric which is not induced by a norm to find an example for $d$ such that $U$ is not open.