Consider two metrics $d_{p}=(\sum\limits_{k=1}^n |x_{k}-y_{k}|^p)^{1/p}$ and $d_{q}=(\sum\limits_{k=1}^n |x_{k}-y_{k}|^q)^{1/q}$
Prove that a non-empty subset $A \subset \mathbb R ^n$ is open with respect to $d_{p}$ iff it is open with respect to $d_{p}$
Attempt
So $A$ is open in both those metrics if $d_{p}$ and $d_{q}$ are equivalent.
Then I have to prove that $\alpha d_p(x,y) \leq d_q(x,y) \leq \beta d_p(x,y)$ for $\alpha, \beta$ positive.
How do I start to prove this?
If $p<q$ define $A_i=\frac{|x_i-y_i|^{1/p}}{\sum_{j=1}^n |x_j-y_j|^{1/p}}\le 1$ so $f(x)=\sum_{i=1}^nA_i^x$ has negative derivative and thus $n\ge f(0)\ge f(p/q)\ge f(1)=1$ implies $nd_p(x,y)^{p^2}\ge d_q(x,y)^{q^2} \ge d_p(x,y)^{p^2}$. Now, if $A$ is open under $d_p$, for any $x \in A$ there exists $r$ s.t. $d_p(x,y)<r\Rightarrow y\in A$. Choosing $R=r^{p^2/q^2}$ shows that $d_q(x,y)<R$ implies $d_p(x,y)\le d_q(x,y)^{q^2/p^2}<r$ so $y\in A$ and $A$ is open under $d_q$. The converse follows similarly.