This sequence has no convergent subsequence?

jdstokes
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Homework Statement



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?

Homework Equations



None.

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.
 
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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?
 
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?
 
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Tom Mattson said:
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?

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 \epsilon= 1.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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