This sequence has no convergent subsequence?

In summary: This sequence is not convergent, but the sequence (-1)+n is convergent, for all n.In summary, the homework statement asks if it can be shown that a sequence has no convergent subsequence. It is not clear if this question is asking about a sequence in a metric space or not. However, it is true that any sequence in a metric space converges if it is Cauchy. Additionally, no subsequence of a convergent sequence can be Cauchy.
  • #1
jdstokes
523
1

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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  • #2
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?
 
  • #3
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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  • #4
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 [itex]\epsilon= 1[/itex].
 

FAQ: This sequence has no convergent subsequence?

1. What does it mean for a sequence to have no convergent subsequence?

When a sequence has no convergent subsequence, it means that the sequence does not have any subsequences that converge to a single limit. In other words, the terms in the sequence do not get closer and closer to a specific number as the sequence progresses.

2. Why is it important for a sequence to have a convergent subsequence?

A convergent subsequence is important because it allows us to determine the behavior and properties of the original sequence. It also helps us to understand the limit of the sequence and whether it approaches a specific number or diverges.

3. Can a sequence have more than one convergent subsequence?

Yes, a sequence can have more than one convergent subsequence. This means that there can be multiple subsequences within the original sequence that converge to different limits.

4. What are some examples of sequences that have no convergent subsequence?

One example is the sequence {1, -1, 2, -2, 3, -3, ...} where the terms alternate between positive and negative integers. Another example is the sequence {1, 2, 3, 4, ...} which diverges to infinity and does not have a finite limit.

5. How can we prove that a sequence has no convergent subsequence?

To prove that a sequence has no convergent subsequence, we can use the Bolzano-Weierstrass theorem which states that a bounded sequence must have at least one convergent subsequence. Therefore, if we can show that the sequence is unbounded, then we can conclude that it has no convergent subsequence.

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