Why is $U ⊂ \mathbb{R}^n$ open with respect to metric $d_p$ iff it is open with respect to metric $d_q$ for $q ∈ [1, ∞)$?

Question

Let's say that for any $p ∈ [1, ∞)$ we have a distance function on $\mathbb{R}^n$ given by $$d_p(x, y) := \left(\sum^n_{j=1}|x_i - y_i|^p\right)^{\frac{1}{p}}$$

How would I show that a set $U ⊂ \mathbb{R}^n$ is open with respect to the metric $d_p$ if and only if it is open with respect to the metric $d_q$ for any $q ∈ [1, ∞)$?

Answer 1

Hint: if two distances are equivalent, they induce the same topology (see here).

So you can either prove that such distances are equivalent, by hand, or you can more generally prove that all norms on a finite dimensional space are equivalent and notice your distances are induced by norms.