It is not Dedekind-like construction but it is cool, I think you should know (if you don't already do).

First of all, cartesian product of topological spaces or, simply, Tikhonov product $\prod_{\alpha\in A}X_\alpha$.

It is the topological space $(X,\mathcal{T})$ which is the family of topological spaces $\{(X_\alpha,\mathcal{T}_\alpha)\;|\; \alpha\in A\}$

The base of the Tikhonov product is the family of all sets $\pi_{\alpha_1}^{-1}(U_{\alpha_1})\cap\cdots\cap \pi_{\alpha_n}^{-1}(U_{\alpha_n})$, where $\{\alpha_1,\dots,\alpha_n\}\in A$ — is finite collection of elements of $A$, $U_{\alpha_i}$ is an element of the topology $\mathcal{T}_{\alpha_i}$ and $\pi_{\alpha_i}$ is natural projection mapping $(X,\mathcal{T}) \to (X_{\alpha_i},\mathcal{T}_{\alpha_i})$.

Now suppose set $S$, $|S|=n$ and let it be a discrete topological space. Let then $X_n$ be a Tikhonov product of countably many copies of $S$ — just a family of sequences.

And we have that set of p-adic integers $\mathbb Z_p$ is homeomorphic to $X_n$ naturally.

Then, Cantor set — $C$. It contains numbers from $[0,1]$ which possibly have only $0$'s and $2$'s in terms of base 3 notation.

Now we could construct a function $f: X_2 \to C$, which transforms sequence $\{a_1, a_2, \cdots, a_n, \cdots\}$ into $0.a_1a_2\ldots a_n\ldots$. It is a bijection and $f$ and $f^{-1}$ are continuous.

So, $\mathbb Z_2$ is homeomorphic to Cantor set.

Suppose now an infinite tree. It has a root and all other vertices are sorted into disjointed countable union of finite sets $X_1, X_2, \cdots, X_n, \cdots$ — vertices of the first, second, etc. levels.

Let $P$ be the set of all possible paths on this tree from the root which go through one vertice of each level. And let the subsets of paths, where the head of path (of some length) is fixed and the tale is any, be the base of topology on $P$.

Now we get, that $P$ is homeomorphic to $\prod_{i=1}^\infty A_i$ if there are $|A_1|$ edges from the root to first level vertices, exactly $|A_2|$ edges from every vertice of the first level to the second and etc.

It is relatively easy to prove that for any tree (which has at least two edges to the next level from every vertice) $P$ is homeomorphic to Cantor set.

So, as a consequence, any ring $\mathbb Z_p$ of p-adic integers is homeomorphic to Cantor set.

There is also a result that $\mathbb Q_p$ is homeomorphic to Cantor set without a point.

The book **Robert, Alain M. A Course in p-adic Analysis** could be really helpful.

I hope this will be useful for you.