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Consider the following:

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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?

  • $x\mapsto xc$ and $y\mapsto cy$ are uniformly continuous – Hagen von Eitzen Jul 06 '20 at 22:38
  • But we need jointly uniformly continuity? @HagenvonEitzen –  Jul 06 '20 at 22:40
  • We want to get a multiplication map $\Bbb{Q}_p \times \Bbb{Q}_p \to \Bbb{Q}_p$ extending the multiplication map on $\Bbb{Q}$. –  Jul 06 '20 at 22:40
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    The extended map $\Bbb Q_p \times \Bbb Q_p \rightarrow \Bbb Q_p$ is NOT uniformly continuous. However, this map can be defined, as $\Bbb Q \times \Bbb Q \rightarrow \Bbb Q$ is uniformly continuous on any ($p$-adically) bounded set. – WhatsUp Jul 06 '20 at 23:00
  • @Whatsup Mind giving some more details? –  Jul 06 '20 at 23:08
  • Fix a real number $B > 0$. Consider the open set $S_p = {x \in \Bbb Q_p: |x|_p < B}$ and let $S$ be the intersection $S_p \cap \Bbb Q$. Then $S$ is dense in $S_p$ and the multiplication map $S\times S \rightarrow \Bbb Q$ is uniformly continuous, if we equip both sides with the topology induced from $\Bbb Q_p$. It follows that this extends to a map $S_p \times S_p \rightarrow\Bbb Q_p$. – WhatsUp Jul 06 '20 at 23:14
  • And I guess we define the map $\Bbb{Q}_p\times \Bbb{Q}_p\to \Bbb{Q}_p$ then by gluing these maps together? Well-definedness follows because of the uniqueness of the extension? And the way we glue together these maps ensures that we end up with a continuous map. Is this the idea? –  Jul 06 '20 at 23:23
  • I've usually seen these justifications done (somewhat laboriously) by showing that multiplication of [equivalence classes of] Cauchy sequences is well-defined. A little tedious, sure, but it gets the job done. Looking for the slickest possible justification for a result (e.g. a one-line proof based on uniform continuity) can be counterproductive because a lot of math really is just messy, and rejecting a correct solution on grounds of insufficient elegance isn't how math gets done. – Rivers McForge Jul 11 '20 at 18:47
  • Given $B$, let $m_B$ be the extension of the multiplication from $S\times S$ to $S_p\times S_p$ from WhatsUp’s comment above. If $C>B$ then the restriction $m_{C}|\operatorname{dom} m_B$ of $m_{C}$ onto the domain $\operatorname{dom} m_B$ of $m_B$ is a continuous map, coinciding with $m_B$ on a set $\operatorname{dom} m_B\cap \Bbb Q\times\Bbb Q$, which is dense in $\operatorname{dom} m_B$. – Alex Ravsky Jul 16 '20 at 11:51
  • Since the space $\Bbb Q_p\times \Bbb Q_p$ is Hausdorff, it easily follows (see, for instance, Theorem 1.5.4 in “General topology” by Ryszard Engelking (Heldermann Verlag, Berlin, 1989)) that $m_{B’}|\operatorname{dom} m_B=m_B$. That is, the maps $m_B$ are compatible and so there exists their combination $\nabla m_B$, see this quotation from “General topology”. Since the family ${\operatorname{dom} m_B: B>0 }$ is an open cover of the space $\Bbb Q_p\times \Bbb Q_p$, the combination $\nabla m_B$ is continuous by Proposition 2.1.11. – Alex Ravsky Jul 16 '20 at 11:51

1 Answers1

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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 $Y$.

Alex Ravsky
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