# Cluster Points in a Sequence

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• Mar 6th 2010, 01:50 AM
Markeur
Cluster Points in a Sequence
Is there any sequence that has infinitely many cluster points?If there is,what are the sequences that have this property?
• Mar 6th 2010, 02:28 AM
Enrique2
Since $\displaystyle \mathbb{Q}$ is countable, you can consider the sequence $\displaystyle (q_n)_n$ formed by all the rationals, obtaining then a sequence such that its cluster points are ALL the real numbers!
• Mar 6th 2010, 03:55 AM
Markeur
Quote:

Originally Posted by Enrique2
Since $\displaystyle \mathbb{Q}$ is countable, you can consider the sequence $\displaystyle (q_n)_n$ formed by all the rationals, obtaining then a sequence such that its cluster points are ALL the real numbers!

Hmm I don't quite understand.It would be helpful if you could give some concrete examples.

Apart from the sequences formed by rational numbers,is there any sequence in $\displaystyle \mathbb{R}$ that has this property as well?
• Mar 6th 2010, 05:17 AM
Enrique2
It is a concrete example!!! $\displaystyle \mathbb{Q}$ is countable and dense. Countable means that there exists a bijective aplication $\displaystyle \pi:\mathbb{N}\to \mathbb{Q}$. Hence $\displaystyle (\pi(n))_n$ is a sequence containing ALL the rational numbers (not formed by rational numbers). Since $\displaystyle \mathbb{Q}$ is dense in $\displaystyle \mathbb{R}$, every real point is a cluster point of this sequence.
• Mar 6th 2010, 05:25 AM
Markeur
Quote:

Originally Posted by Enrique2
It is a concrete example!!! $\displaystyle \mathbb{Q}$ is countable and dense. Countable means that there exists a bijective aplication $\displaystyle \pi:\mathbb{N}\to \mathbb{Q}$. Hence $\displaystyle (\pi(n))_n$ is a sequence containing ALL the rational numbers (not formed by rational numbers). Since $\displaystyle \mathbb{Q}$ is dense in $\displaystyle \mathbb{R}$, every real point is a cluster point of this sequence.

Okay,I get it.Thanks very much.
• Mar 6th 2010, 06:09 AM
Opalg
If you want a simple constructive formula then you could take $\displaystyle x_n = \sin(\ln n)$, which oscillates slowly between –1 and +1, so that every point of the interval [–1,1] is a cluster point. That also provides an answer to the question in your other long-running thread Limit of a Sequence.