# Any continuous non-constant map from $Q_p$ to $R$

Consider $$Q_p$$ and $$R$$ where $$Q_p$$ is $$p-$$adic numbers with $$p-$$adic topology. If $$Q_p\to R$$ is a ring homomorphism, then certainly it must be trivial as $$Z\to Z$$ inducing $$Q\to Q$$ but infinite series of p-adic element does not converge in $$R$$ in general.

$$\textbf{Q:}$$ It is natural to ask whether one can classify all continuous maps $$Hom_{Cts}(Q_p,R)$$ besides constant maps. It is certainly possible to get just set theoretical embedding by axiom of choice. Does $$Hom_{Cts}(Q_p,R)$$ have non-constant functions? I do not see an obvious choice of candidate for non-constant maps.

There are many continuous, non-constant functions; for example, the $$p$$-adic norm is one such (as is any positive power of the $$p$$-adic norm).
• (Topologically, $\mathbb{Q}_p$ is a countable disjoint union of $\mathbb{Z}_p$ which looks like the Cantor set, so there are of continuous maps.) – hunter Sep 11 at 18:15
• Oh, in that case, I was dumb. I forgot the obvious $p-$adic norms. Why cantor sets are homeomorphic to $Z_p$ here? It looks like $Z_p$ is countable but cantor set is not. Do I need axiom of choice? – user45765 Sep 11 at 18:16
• @user45765 In fact, $\mathbb{Z}_p$ is uncountable. For the homeomorphism to the Cantor set, no AC is needed. You can see a nice explicit proof here: personal.psu.edu/axk29/TOPOLOGY/p-adic.pdf – hunter Sep 11 at 18:19