Let $X$ a metric space and $d, d'$ two metrics on $X$ such that $d \le d'.$ Why does any ball on the matric $d'$ is contained on other ball on $d-$metric?
I just cannot see the reason. For example, over $\mathbb{R}^n$ we have that $\|\cdot\|_{\infty} \le \|\cdot\|,$ where the first metric is the sup metric and the second the standard metric of $\mathbb{R}^n$. So, why does follow that the ball of the euclidean standard metric is contained on the ball on the sup norm?
Let $B_{d'}(x,\epsilon)$ be a ball on the metric $d'$. Then, $B_{d'}(x,\epsilon) \subseteq B_d(x,\epsilon)$.
To see that, let $y\in B_{d'}(x,\epsilon)$. Since $d\leq d'$, we have: $d(x,y)\leq d'(x,y)<\epsilon$. Hence, $y\in B_d(x,\epsilon)$
I think you just got disturbed by the fact that $d\leq d'$ implies $B_{d'} \subseteq B_d$ which is the other way around. It's just due to the fact that if a metric is smaller than another one, the same ball with the same radius will "appear" to be bigger on the smaller metric because there are points that were not contained in the first ball that will end up in the second one, since the distance is smaller.
