Dedekind-Like construction of p-adic numbers

I don't know too much about the $p$-adic numbers, but I know about Pontryagin duality, so I'm going to swing that hammer! We can build the ring of $p$-adic integers as the dual of a structure distilled from the rational numbers.

Duality? Does this answer your question? Maybe it doesn't sound like it's analogous to the construction of the real numbers using Dedekind cuts. So here's a recap of the Dedekind construction which is biased to make future comparisons easier:

1. Represent each rational number $x\in \mathbb Q$ as the open ray $(-\infty,x)\subset\mathbb Q$.
2. The collection of all sets $(-\infty,x)\subset\mathbb Q$ doesn't get us anywhere, since the supremum recovers $x$.
3. The collection of all sets that look like $(-\infty,x)$ -- the nonempty, downward-closed, proper subsets without greatest element -- is a larger collection, which we turn into $\mathbb R$.
4. $\mathbb R$ is an ordered field. Multiplication is a pain, but eventually we get there.

Now, how exactly did this construction use the binary order relation $<$? The ray $(-\infty,x)$ is the set of all $a$ such that $a<x$. Subsets are equivalent to unary relations, so we have effectively represented $x$ by the unary relation $\bullet<x$. In a single formula, the Dedekind embedding of $\mathbb Q$ into $\mathbb R$ is generated by $x\mapsto(\bullet<x)$.

Well, if we're looking to extend $\mathbb Q$ in a way that doesn't respect the order relation, what other binary symbols do we have to work with? There's addition, $x\mapsto(\bullet+x)$. Let's try it:

1. Represent each rational number $x\in\mathbb Q$ as the translation $(\bullet+x):\mathbb Q\to\mathbb Q$ defined by $a\mapsto a+x$.
2. The collection of all "translations" of the group $(\mathbb Q, +)$ doesn't get us anywhere, since the image of $0$ recovers $x$. (This could probably be made precise, but it's aside from the point.)
3. Give up. The number $0$ is a problem, and it's not easy to get rid of it without destroying the whole structure.

So let's try multiplication, $x\mapsto(\bullet\times x)$.

1. Represent each rational number $x\in\mathbb Q$ as the additive-group endomorphism $a\mapsto a\times x$.
2. The collection of all endomorphisms of the group $(\mathbb Q, +)$ doesn't get us anywhere, since the image of $1$ recovers $x$. (Not obvious, but true.)
3. Since $1$ is a problem, annihilate it! Consider instead the endomorphism ring of $\mathbb Q/\langle1\rangle = \mathbb Q/\mathbb Z$. This ring is a more interesting structure, which we identify as $\hat{\mathbb Z}$.
4. Oh, you wanted a field? Well, that's where you have a choose a prime number $p$. We have $\hat{\mathbb Z}\cong\prod_p\mathbb Z_p$, where each $\mathbb Z_p$ is an integral domain, so we can form its field of fractions, $\mathbb Q_p$.

That's the analogy! A slightly more direct way to get $\mathbb Z_p$ is to take the endomorphism ring of the Prüfer $p$-group $\mathbb Z[1/p]/\mathbb Z$, which is the group of rational numbers with $p$-power denominators, modulo $1$. A nice property of this construction is that each individual $p$-adic integer and $p$-adic number is represented as a hereditarily countable set:

• Explicitly, a $p$-adic number is a ratio of endomorphisms of a subgroup of a quotient of the rationals.

This might not be quite as simple as a real number being a subset of the rationals, but it's similar. By contrast, in the Cauchy completion, each number is represented as an uncountable collection of sequences.

Finally, to explain why I mentioned duality, $\mathbb Z_p$ is the Pontryagin dual of the (discrete) Prüfer $p$-group. The dual is essentially defined as the group of homomorphisms $\mathbb Z[1/p]/\mathbb Z\to\mathbb R/\mathbb Z$, and the image of every homomorphism lies in $\mathbb Z[1/p]/\mathbb Z$ anyway, so we recover the group of endomorphisms, which we might as well equip with its natural ring structure.

(I suspect that these hand-waving contortions look horribly backwards to an expert. If so, I apologize; I'm just trying to make the construction fit the analogy as well as possible, not to make it especially elegant or straightforward.)

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.

Here’s yet another construction. I’m going to exhibit $\mathbb Z_p$ as the endomorphism ring of a certain abelian group, $T$ (for “torsion”), which I’ll describe in two ways. First definition of $T$ is that it’s $\mathbb Z[1/p]\,/\,\mathbb Z$, and by $\mathbb Z[1/p]$ I mean the rationals with only powers of $p$ in the denominator. You can see that this is a divisible group (for every $z\in T$ and every $m>0$, there are $z'\in T$ with $mz'=z$).

The other definition of $T$ is just the group of all $p$-power roots of unity in $\mathbb C$. Now $T$ is written multiplicatively, of course, and you can get an isomorphism between the two constructed groups by taking the rational number $r/p^n$ to $e^{2ri\pi/p^n}$.

Maybe it’s easier to think of $T$ as the rationals with only $p$-powers in the denominator, modulo $\mathbb Z$. Each subgroup $C_n=(p^{-n}\mathbb Z)/\mathbb Z$ is characteristic and must be mapped into itself by any endomorphism of $T$, so we get something that gives consistent endomorphisms of these cyclic groups of order $p^n$. But the endomorphism ring of a cyclic group $C_n$ of order $p^n$ is just $\mathbb Z/(p^n)$. You look at things, and see that an endomorphism of $T$ is just an element of $\text{projlim}_n(\mathbb Z/(p^n))$, with respect to the natural maps $\mathbb Z/(p^{n+1})\to\mathbb Z/(p^n)$

There’s no question that this is harder to see than the Cauchy-sequence construction, but you stare at things for a while, and see that they’re the same.

Expanding the comment of Bruno Joyal, in ${\Bbb R}$ the usual topology (open sets defined via the balls $(x-\epsilon,x+\epsilon)$ in the usual distance) and the order topology (open intervals (a,b) form a basis of the topology) are obviously the same.

EDIT: maybe not very enlighting, but different. As inverse limit (universal property/abstract nonsense): http://math.arizona.edu/~gradprogram/workshops/integration/2003/p-adicsProject.pdf.

And in the Dedekind cuts the order is essential...