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This sequence has no convergent subsequence?

  1. May 3, 2007 #1
    1. The problem statement, all variables and given/known data

    Let {x_n} be a sequence in a metric space such that the distance between x_i and x_{i+1} is epsilon for some fixed epsilon > 0 and for all i. Can it be shown that this sequence has no convergent subsequence?

    2. Relevant equations


    3. The attempt at a solution

    I'm afraid I have no clue on this one. I'm not 100 % sure that it's always true either, but it seems intuitive that it should be true. Any hints would be greatly appreciated.
    Last edited: May 3, 2007
  2. jcsd
  3. May 3, 2007 #2

    Tom Mattson

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    My analysis isn't very good, so I'm just thinking out loud here, but...

    Isn't it the case that sequences converge in a metric space iff they are Cauchy? And isn't it the case that no subsequence of this sequence can be Cauchy, as there is a fixed minimum distance between elements in the sequence?
  4. May 4, 2007 #3

    matt grime

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    Sorry, but this question falls easily to JUST DOING IT, and you're going to kick yourselves (assuming I've not made some trivial misreading of the question).

    Let e be epsilon.

    R is a metric space. So let's start with a sequence that goes 0,e, now what can come next? I.E. what numbers in R are e away from e?
    Last edited: May 4, 2007
  5. May 4, 2007 #4


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    Actually, your first statement is incorrect. In a complete metric space, all Cauchy sequences converge but not in an incomplete metric spaces. For example, in the set of rational numbers, with the usual topology, the sequence 3, 3.1, 3.14, 3.141, 3.1415, 3.14159,... with the obvious continuation, is a Cauchy sequence but does not converge.

    It is, however, true that in any metric space, any convergent sequence is a Cauchy sequence. However, that doesn't help here because we are talking about subsequences, not the entire sequence.

    Consider the sequence (-1)n, in the set of real numbers with the usual topology and [itex]\epsilon= 1[/itex].
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