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Completeness of R^m

  1. Jul 13, 2012 #1
    I'm reading through Pugh's Real Mathematical Analysis and one of the theorems he puts down states that "R^m is complete". I assume the superscript refers to it being metric as that's what he's talking about right before this theorem. Anyway, the proof seems -to me- circular. I was hoping someone could show me why it isn't or perhaps that I'm expecting too much.

    It says:

    let p be a Cauchy sequence in R^m. express p in component form p={p1,p2,p3,...pmn}. Because p is Cauchy, each component sequence is Cauchy. Completeness of R implies that the component sequences converge, and therefore the vector sequence converges.


    I dunno, but for some reason, and I've read it over 20+ times, that seems like circular reasoning.

    (Sorry for being to lazy to post in Latex format)
     
  2. jcsd
  3. Jul 13, 2012 #2
    I think [itex]\mathbb R^m[/itex] refers to an m-dimensional real vector space. If so then he's proving the m-dimensional space to be complete using the fact that the 1-dimensional real line is complete.
     
  4. Jul 13, 2012 #3
    Ah, you pointing that out and me re-reading the proof makes it more clear. Thank you.
     
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