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Discrete metric

  1. Aug 24, 2009 #1
    how is discrete metric space given by d((x1,x2,....xn)(y1,y2,....yn))=0 if xi=yi else 1
    disc
    is complete
     
  2. jcsd
  3. Aug 24, 2009 #2

    HallsofIvy

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    Have you thought about what convergence of a sequence means in a discrete space?

    A metric space is "complete" if and only if every Cauchy sequence converges. And, of course, a Cauchy sequence is one where [itex]\lim_{m,n\rightarrow \infty} d(a_n,a_m)= 0[/itex]. Since d(x,y)= 1 for [itex]x\ne y[/itex], in order for that to happen the sequence must be "eventually constant", i.e. for some N, if n,m> N, [itex]a_n= a_m[/itex] and it is easy to show that such a sequence converges.
     
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