Prove that the subset $X$ of $\mathbb{R}^n$ is open with respect to the metric $d_q$ [closed]

Question: Let $d_q$ and $d_p$ be two equivalent metrics on $R^n$, and let X $\subset R^n$ be an open set with respect to $d_p$. Prove that X is open with respect to $d_q$.

[Hint: two metrics are equivalent if there exist positive scalars a and b such that $a\cdot d_p$(x,y)$ \leq d_q$(x,y)$ \leq b\cdot d_p$(x,y) for any x and y in X.]

I'm new to topology and real analysis and have no idea where to start.

I thought I had to do something with this: The set $X⊆R^n$ is open with respect to $d_q$ if for every $x∈X$, there exists $ϵ>0$ such that the set $y:∥x−y∥q<ϵ⊆X$. But I don't know how to make it a complete proof

Can someone help me out?

Solution 1:

$\newcommand{\R}{\mathbb{R}^n}$ $X$ is open in $\R$ w.r.t. $d_p$ if $$ P_1 :\Leftrightarrow \forall x \in X:\exists \epsilon > 0: \forall y \in \R: (d_p(x,y) < \epsilon \Rightarrow y \in X) $$ Therefore, the goal is to show that $$ P_2 :\Leftrightarrow \forall x \in X:\exists \epsilon > 0: \forall y \in \R: (d_q(x,y) < \epsilon \Rightarrow y \in X) $$ taking $P_1$ as a premise. Let $\epsilon_1$ be one of the values of $\epsilon$ satisfying $P_1$. Now, we have to find another $\epsilon$ such that $d_q(x,y) < \epsilon$ implies $d_p (x,y) < \epsilon_1$.

$$ d_p(x,y) \le \frac{1}{a}d_q(x,y)< \frac{\epsilon}{a} $$

If we choose $\epsilon = a\epsilon_1 $, we have $y \in X$. We've just shown that such $\epsilon$ that makes the conditional in $P_2$ hold exists.