Defining p-adic numbers via a formal completion. Consider following fragment of the definition of p-adic numbers in "A course in abstract harmonic analysis" by Folland:

So we have that $+: \Bbb{Q} \times \Bbb{Q} \to \Bbb{Q}$ is continuous w.r.t. the metric $|\cdot|_p$. Why does this map extend to a map $+: \Bbb{Q}_p \times \Bbb{Q}_p \to \Bbb{Q}_p$?
I know that every map continuous $\Bbb{Q} \to Z$ lifts to a map $\Bbb{Q}_p \to Z$ by the universal property of the completion but I don't see why this holds for this 'product-domain' as well. Is there a way to see this without going to the proofs of the existence of the completion?
 A: This is immediate since $+$ is uniformy continuous on $\Bbb{Q}$ and then you can invoke the universal property of completion to give you the extension you want. What is less trivial is that the multiplication extends to the completion, because this is NOT uniformly continuous anymore.
Here is one way to see  this. The routine details are left to the reader. If anybody wants a gap to be filled in, feel free to ask.
Let us prove that we can extend the multiplication map $m: \Bbb{Q}\times \Bbb{Q} \to \Bbb{Q}$ to a map $m_p: \Bbb{Q}_p \times \Bbb{Q}_p \to \Bbb{Q}_p$.
Given $x,y \in \Bbb{Q}_p$, choose sequences $(x_n)_n, (y_n)_n$ in $\Bbb{Q}$ with $x_n \to x$ and $y_n \to y$. Then we define
$$m_p(x,y) := \lim_n m(x_n, y_n)$$
The limit exists since $(m(x_n,y_n))_n$ is easily shown to be a Cauchy sequence and $\Bbb{Q}_p$ is complete. Moreover, one can show that the result does not depend on the choices of $(x_n)_n, (y_n)_n$ . That is, we get a well-defined function
$$m_p: \Bbb{Q}_p \times \Bbb{Q}_p \to \Bbb{Q}_p$$
Choosing constant sequences, it is clear that $m_p$ extends $m$. Moreover, all properties the multiplication $m_p$ should have still hold, because they hold for $m$.
Finally, once all that is settled, the standard argument shows that $m_p$ is continuous.
