Completeness of ℝ (when ℝ is defined abstractly)

[Fredrik]

Suppose that we define ℝ abstractly instead of by explicit construction, i.e. we just say that ℝ is any Dedekind-complete* ordered field. Can we now prove that ℝ is a complete metric space? Does the question even make sense? I mean, the definition of "metric space" refers to ℝ. What ℝ is that anyway, the abstract one or one defined by explicit construction (Dedekind cuts)?

*) By that I mean that every set that's bounded from above has a least upper bound.

 

[micromass]

If you define \mathbb{R} simply as a Dedekind-complete ordered field, then it carries no natural structure of a metric space yet. You must first define the metric before you can talk about completeness.

It does carry a natural topology though: every linearly ordered set can be topologized with the order topology: http://en.wikipedia.org/wiki/Order_topology

So, we can say that \mathbb{R} is completely metrizable: that is, we can say that it has a metric that makes it complete. But there are many such metrics.

Then again: \mathbb{R} with the order topology is a topological vector space, even a locally convex topological vector space. And it can be shown that it's topology can be induced by a norm. So the norm on \mathbb{R} is not so unnatural as one might think.

Does that answer your question or am I way off??

 

[Fredrik]

That's a good answer. I see now that I had a bit of a brain malfunction when I asked the question. I was worried about a few things that it doesn't make sense to worry about. You helped clear that up, and i appreciate that. Thank you.