Every metric space $M$ has a completion $\widehat M$, that is, it can be embedded as a dense subset in a complete metric space. When I first came across this theorem, I thought "well, that's amazing, we don't need equivalence classes of Cauchy sequences or Dedekind cuts anymore, we can use this to construct the real line!" It took me some time to realize that the whole theory of metric spaces is built around the concept of real numbers, so no, you can't use this to build the real line.

Now, can one build up the theory of metric spaces (not all, but just enough to squeeze real numbers out of the rationals) without needing the reals first? "Oh, that's easy, just build the reals from the rationals using Cauchy equivalence classes, that's the basic theory you want." Ok, but can it be generalized? Banach completion is close to what I want, but (i) it's still kind of limited (only applies to vector spaces); and (ii) it works for vector spaces over the field of.... the real numbers! So, again, we need to build the reals first.

Is it just that the real line is our canonical example/definition of "complete" or "continuum", so that everything "complete" or "continuum" needs to be defined on top of it? I would like to know what kind of theorems and/or arguments exist around the idea of completing "incomplete" spaces from scratch.

  • 1
    $\begingroup$ I don't know, whether this is relevant (don't know anything about metric spaces), but taking $\mathbb{R}$ to be the codomain / range of a metric is sort of canonical, since it is the maximal archimedian ordered field. (i.e. every other achimedian ordered field is a subfield of $\mathbb{R}$). Non-archimedian ordered fields on the other hand don't really capture our intuition of distance. On a side note: You can also construct the reals from the surreals ;) . $\endgroup$ – Stefan Perko Feb 28 '15 at 8:07
  • 1
    $\begingroup$ Am I mistaken if I say that $\mathbb Q$ is the source of an infinite number of normed spaces, where the p-adic norms are defined $||x||_p$ for $p=$ primes and even for $p=\infty$ which is the easiest case: $||x||_{\infty}=|x|$. So you have norms over $\mathbb Q$. Why don't you use the Banach compl theorem. Or does it require that the norm be real? I've not recall if the theorem at stake uses that the norm is real valued. I just know that, p.ex. in the case of $\mathbb Q$ the norms are $\mathbb Q$ valued. $\endgroup$ – zoli Feb 28 '15 at 11:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.