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Cauchy Sequence in Metric Space

  1. Feb 20, 2010 #1
    For a metric space (X,d), prove that a Cauchy sequence {xn} has the property d(xn-xn+1)--->0 as n--->\infty

    In working this proof, is it really as simple as letting m=n+1?
  2. jcsd
  3. Feb 21, 2010 #2


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    Yes, it really is that easy! The definition of "Cauchy sequence" is that [itex]d(x_n, x_m)[/itex] goes to 0 as m and n go to infinity. Since that is true, in particular, [itex]d(x_n, x_{n+1})[/itex] must go to 0 as n goes to infinity.

    Notice, however, that the converse is not true. If [itex]d(x_n, x_{n+1})[/itex] goes to 0, it does NOT follow that [itex]d(x_n,x_m)[/itex] goes to 0.
  4. Feb 21, 2010 #3
    Thanks for the input.

    An example of a sequence in which [tex] d(x_n,x_{n+1})\rightarrow 0 [/tex] , but [tex] d(x_n,x_m)\not\rightarrow 0 [/tex] would be let [tex]x_n=ln(n)[/tex]
    Last edited: Feb 21, 2010
  5. Feb 21, 2010 #4


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    I'll have to think about that! I was thinking of
    [tex]a_n= \sum{i= 1}^n \frac{1}{i}= \frac{1}{n}+ \frac{2}{n}+ \frac{3}{n}+ \cdot\cdot\cdot+ \frac{1}{n}[/tex]

    Then [itex]|a_n- a_{n-1}= \frac{1}{n}[/itex] goes to 0 as n goes to infinity but the series is not Cauchy since it is well known that the harmonic series does not converge.

    (After about 30 seconds of thought I see that ln(n)- ln(n+1)= ln(n/(n+1)) so, in fact, your seires is a variation of mine.
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