There is of course the standard construction of the reals by considering the set of sequences that are Cauchy with respect to the standard metric and taking the quotient by sequences that converge to 0, but in many other cases of the completion of a ring or group, we can complete by taking an inverse limit of factor rings/groups corresponding to some predetermined ideals/subgroups. For example, the $\mathfrak{p}$-adic integers are constructed using precisely this type of inverse limit construction, where the ideals are the powers of $\mathfrak{p}$, where $\mathfrak{p}$ is the ideal $(p)$ for $p$ a prime.


Is there any way to construct the reals using a similar inverse limit construction? 


It seems like this might be related to the so-called "infinite prime", but I don't know if there is really a way to complete with respect to an "infinite prime", which doesn't seem like it is strictly a prime integer or ideal, so that doesn't really seem to help..


(If anyone can think of better tags, please go ahead and change the ones I've used. I wasn't really sure how to classify this).



Completion of the rationals to the reals as an inverse limit construction?
Although your question is not at all vague, there are a few completely different ways to interpret what a good answer would be. Knowing a bit about your personal preferences, I suspect the following is not at all what you wanted, but it is still of interest to the community.


The fact that $\mathbb{Q}\_p$, for example, can be seen as an inverse limit boils down to the fact that the series representations
$$\sum\_{n \in \mathbb{Z}} a\_n p^{-n}$$
give a homeomorphism between $\mathbb{Q}\_p$ and the subspace of $\{0,1,\dots,p-1\}^{\mathbb{Z}}$ of eventually zero sequences (on the positive side). At first glance, it would seem that the same works for $\mathbb{R}$ since every element of $[0,\infty)$ has a very similar binary expansion
$$\sum\_{n \in \mathbb{Z}} a\_n 2^n$$
where $(a\_n)\_{n \in \mathbb{Z}}$ is again in the space of all eventually zero sequences in $\{0,1\}^{\mathbb{Z}}$. Unfortunately this is not a bijection. In fact, there is no way to select such a sequence $(a\_n)\_{n\in\mathbb{Z}}$ continuously — binary rationals are always points of discontinuity no matter how the selection is made.


There is a funny way to remedy this which is commonly used in Computable Analysis and Reverse Mathematics. The idea is to represent real numbers again with series of the form
$$\sum\_{n \in \mathbb{Z}} a\_n 2^n$$
but where $(a\_n)\_{n \in \mathbb{Z}}$ is now an eventually zero sequence from the extended set $\{-1,0,1\}^{\mathbb{Z}}$. The obvious cost is that no number has a unique representation in this form, but the side benefit is that the binary rationals are no longer 'special' in this way. In the end, this representation is extremely well behaved compared to ordinary binary expansions. This can be seen from the fact that any *invariant* continuous function between representations (see note) gives rise to a continuous function $\mathbb{R}\to\mathbb{R}$, and every continuous function $\mathbb{R}\to\mathbb{R}$ admits such a lifting.


The conclusion to draw from this is that, up to some very nice blurring, $\mathbb{R}$ is indeed an inverse limit just like $\mathbb{Q}\_p$.


Note: The topology on eventually zero sequences is not exactly the product topology, it is given by the metric
$$d(\vec{x},\vec{y}) = \inf\{2^n : \forall m \geq n\,({x\_m = y\_m})\}.$$ The *invariance* of a function $f$ between eventually zero sequences from $\{-1,0,1\}^{\mathbb{Z}}$ means that if $\vec{x}$ and $\vec{y}$ represent the same real number then so do $f(\vec{x})$ and $f(\vec{y})$.


