How would I show that there exists some set of open balls with rational radius and rational centre such that they are a subset of the reals.That is,
$\exists p,q\in \mathbb{Q} $ and $ r,x \in \mathbb{R} $ such that $B_p(q) \subset B_r(x)$
It seems pretty trivial to me if I use the density of the rationals in the reals. But I honestly feel as if I'm missing something. I'm using the euclidean metric $d_E$ on $\mathbb{R}^n$
My attempt: Let $U\in O(d_E)$. This means that $\forall x\in U$ $\exists \epsilon_x >0$ s.t. $B_{\epsilon_x}(x) \subset U$. Because we know that $\forall a,b\in\mathbb{R}$ with $a<b$ $\exists r\in \mathbb{Q} $ s.t $a<r<b$ then if we show that $\exists p_x > 0 $ s.t $B_{p_x}(q) \subset B_r(x)$
I think what you want to show is
$$\forall r,x \in \mathbb{R} \quad \textrm{with}\quad r>0 \quad \exists p,q \in \mathbb{Q} \quad \textrm{with}\quad p>0 \quad \textrm{such that} \quad B_p(q) \subset B_r(x).$$
Ineed, choose $q \in B_{r/2}(x) \cap \mathbb{Q}$ and $p \in (0,r/2) \cap \mathbb{Q}$ which exist by density of $\mathbb{Q}$ in $\mathbb{R}$.