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Completeness of ℝ (when ℝ is defined abstractly)

  1. Jan 22, 2012 #1

    Fredrik

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    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.
     
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  3. Jan 22, 2012 #2

    micromass

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    If you define [itex]\mathbb{R}[/itex] 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 [itex]\mathbb{R}[/itex] 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: [itex]\mathbb{R}[/itex] 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 [itex]\mathbb{R}[/itex] is not so unnatural as one might think.

    Does that answer your question or am I way off??
     
  4. Jan 22, 2012 #3

    Fredrik

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    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.
     
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