Multiplication in $\Bbb{Q}$ w.r.t. p-adic norm is uniformly continuous. Consider the multiplication map $. : \Bbb{Q} \times \Bbb{Q} \to \Bbb{Q}$. Give $\Bbb{Q}$ the p-adic norm $| \cdot |_p$. I want to show that this multiplication map is uniformly continuous.
Given $x_0,y_0 \in \Bbb{Q}$, I tried:
$$d_p(xy, x_0y_0) = |xy-x_0y_0|_p \leq |x|_p|y-y_0|_p + |y_0|_p |x-x_0|_p$$
But this depends on the chosen points in the estimate. How can I estimate better?
 A: To take this off the "unanswered" list: The multiplication map $\mathbb Q \times \mathbb Q \rightarrow \mathbb Q$ is not uniformly continuos w.r.t. the $p$-adic metric on the right and the product of the $p$-adic metrics on the left.
Namely, e.g. let $\epsilon =1/2$, then if it were uniformly continuous, there would be a $\delta > 0$ such that for all $(x_0,y_0) \in \mathbb Q \times \mathbb Q$ and for all $(x,y)$ with $\lvert x-0\rvert_p < \delta$ and $\lvert y-p^{-n}\rvert < \delta$ it would follow that $\lvert xy -0 \rvert_p < 1/2$. But there exists $n_0$ with $p^{n_0} < \delta$, and setting $x_0:=0, y:=y_0:=p^{-n_0}, x:=p^{n_0}$ we have $\lvert xy\rvert_p=1$.
Note however that on each bounded subset $B \subset \mathbb Q \times \mathbb Q$
(say, $B \subset \{x \in \mathbb Q_p: \lvert x \rvert_p < C\} \times \{y \in \mathbb Q_p: \lvert x \rvert_p < C\}$ which by triangle inequality implies also $\max( \lvert x_1 -x_2\rvert_p, \lvert y_1-y_2\rvert_p) < C$ for all $(x_1,y_1), (x_2,y_2) \in B$),
the map is uniformly continuous and your proof will work: W.l.o.g. we have $C > 0$ and thus for a given $\epsilon$ can choose $\delta < \dfrac{\epsilon}{C}$.
