Defining p-adic numbers via formal competion: Question about continuity of multiplication Consider the following:

Why does the multiplication function $$\cdot: \Bbb{Q} \times \Bbb{Q} \to \Bbb{Q}$$ extend to the completion? I thought the universal property of completion says that uniformly continuous extend to the completion? I don't think the multiplication map on $\Bbb{Q}$ w.r.t. the p-adic norm is uniformly continuous, so why does such an extension exist?
 A: I expect WhatsUp’s approach from the comments works as discussed.
But, if we want a direct proof, we can use the following
Lemma 2. (see this answer).  Let  $X$   be a Fréchet-Urysohn topological space, $Y$  be a regular topological space, and $D$  be a dense subset of the space $X$.  A continuous mapping   $f : D\to Y$  can be extended to a continuous mapping $\hat f : X\to Y$ iff for each sequence $\{x_n\}$ of points of the set $D$ convergent in $X$, a sequence  $\{ f(x_n)\}$ is also convergent.
Remark that Lemma 2 holds when both spaces $X$ and $Y$ are metric, because metric spaces are Fréchet-Urysohn and regular. In our particular case, $X=\Bbb Q_p\times \Bbb Q_p$, $Y=\Bbb Q_p$, and $D=\Bbb Q\times \Bbb Q$. Moreover, since $Y$ is complete, it suffices to check that the multiplication map $f$ keeps Cauchy sequences, when $X$ is endowed with a metric $d$, compatible with its topology. For instance, we can put $$d((x,y),(x’,y’))=|x-x’|_p+|y-x’|_p$$ for any $(x,y), (x’,y’)\in X$.
So, let $\{(x_k, y_k)\}$ be a Cauchy sequence in $X$. That is for each $\varepsilon>0$ there exists a natural number $N$ such that
$d((x_m, y_m), (x_n, y_n))<\varepsilon$ for each $m,n\ge N$. In particular, sequences $\{x_k\}$ and  $\{y_k\}$ are bounded, that is there exist a constant $M$ such that $|x_k|_p\le M$ and $|y_k|_p\le M$ for each $k$
So we have
$$|x_my_m- x_ny_n |_p=$$
$$|x_my_m-x_my_n+x_my_n - x_ny_n|_p\le$$
$$|x_my_m-x_my_n|_p+|x_my_n - x_ny_n|_p=$$
$$|x_m|_p|y_m-y_n|_p+|y_n||x_m - x_n|_p\le $$
$$|x_m|_p\varepsilon+|y_n|_p\varepsilon\le 2M\varepsilon.$$
Thus $\{x_ky_k\}$ is a Cauchy sequence in $Y$.
