Different p-adic topologies on $\mathbb{Q} $ are homeomorphic Endow $\mathbb{Q}$ (the rationals) with the p-adic (resp. q-adic) topology, where p,q are primes. Are these topological spaces homeomorphic?
I know the norms are not equivalent, therefore do not induce the same topology.
What I am asking is if the two distinct topologies are homeomorphic to one another. The reason I am asking is that I know that $\mathbb{Q}_p$ and $\mathbb{Q}_q$ are homeomorphic to each other.
 A: Here's an alternative to Lee Mosher's argument, using a different general theorem from topology.  Any two countable metric spaces with no isolated points are homeomorphic.  In particular, this implies the $p$-adic and $q$-adic topologies on $\mathbb{Q}$ give homeomorphic spaces, and in fact these spaces are also homeomorphic to $\mathbb{Q}$ with the usual topology.
(Actually, one way to prove this theorem is by going through facts about the Cantor set similar to the argument in Lee Mosher's answer.  You can find a proof along these lines as well as some other proofs in this nice paper.)
A: Yes this is true. Here's a somewhat roundabout proof, by application of two theorems about Cantor sets (and my guess is that any route to a direct proof will, more or less, repeat the elements of the proofs of these two theorems).
The completion of $\mathbb{Q}$ with respect to the $p$-adic norm is the complete metric space denoted $\mathbb{Q}_p$, and the completion of $\mathbb{Q}$ with respect to the $q$-adic norm is $\mathbb{Q}_q$. 
Neither of the metric spaces $\mathbb{Q}_p$ and $\mathbb{Q}_q$ is compact, so let me denote their 1-point compactifications as $1pc(\mathbb{Q}_p)$ and $1pc(\mathbb{Q}_q)$. Those two spaces are each compact, metrizable, totally disconnected, and have no isolated points. Therefore, by a general theorem of metric spaces, the two spaces $1pc(\mathbb{Q}_p)$ and $1pc(\mathbb{Q}_q)$ are both homeomorphic to the Cantor middle thirds set $C$.
In the $p$-adic norm topology, $\mathbb{Q}$ is a countable dense subset of $1pc(\mathbb{Q}_p)$. In the $q$-adic norm topology, $\mathbb{Q}$ is a countable dense subset of $1pc(\mathbb{Q}_q)$. By another general theorem of metric spaces, for any two countable dense subsets $X,Y \subset C$ there is a homeomorphism $f : C \to C$ such that $f(X)=Y$; hence, by restriction of $f$, we get a homeomorphism from $X$ to $Y$. Putting this together with the previous paragraph, we get a homeomorphism from $\mathbb{Q}$ with the $p$-adic norm topology to $\mathbb{Q}$ with the $q$-adic norm topology.
A good source for this material is Chapter 12, "Homeomorphisms between Cantor sets", in the book "Geometric topology in dimensions 2 and 3" by Moise.
