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Complete metric space

  1. Sep 3, 2009 #1

    why the set of all real numbers is complete metric space with euclidean metric? I know, that metric space is complete iff all sequences in it converges. But 1,2,3,4,... diverges.

  2. jcsd
  3. Sep 3, 2009 #2
    Complete is if all Cauchy sequences converge, not any sequence
  4. Sep 4, 2009 #3


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    It's what we "know" that gets us into trouble!:biggrin:

    As wofsy said, a metric space is complete iff all Cauchy sequences converge, not "all sequences".

    And a sequence, {an} is "Cauchy" if and only if the sequence {an- am} converges to 0 as m and n go to infinity independently. It is easy to show that any Cauchy sequence converges. The rational numbers are not "complete" because there exist Cauchy sequences that do not converge. For example, the sequence, {3, 3.1, 3.14, 3.1415, 3.14159, 3.141592, ...}, where the nth term is the decimal expansion of [itex]\pi[/itex] to n places, is Cauchy because the nth and mth term are identical to the min(n, m) place and so their difference goes to 0 as m and n go to infinity. But the sequence does not converge because, as a sequence in the real numbers, it converges to [itex]\pi[/itex] and [itex]\pi[/itex] is not a rational number.
    Last edited by a moderator: Sep 4, 2009
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