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Number of points of accumulation of a sequence

up vote 14 down vote favorite

Can a sequence have infinitely many points of accumulation i.e. we can extract infinitely many subsequences from it s.t. they all converge to their respective point of accumulation? I have the feeling it would mean that the period of repetition of something could be infinitely big.

2 Answers

up vote 4 down vote accept

Start with $0,1$. Then travel backwards to $0$ in steps of $1/2$, so $1/2,0$. Then travel forwards to $1$ in steps of $1/4$, so $1/4,2/4,3/4,1$. Then travel backwards to $0$ in steps of $1/8$, so $7/8$, $6/8$, $5/8$, and so on. Continue.

Every real between $0$ and $1$ is an accumulation point of our sequence.

up vote 3 down vote

Yes, this is possible. For example consider the sequence $a_n$ for $n \ge 2$ defined as the smallest divisor of $n$ greater than $1$.

The accumulation points are all the prime numbers. Subsequences witnessing them are for instance the $p$-th powers for each prime $p$.


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Number of points of accumulation of a sequence

up vote 14 down vote

Can a sequence have infinitely many points of accumulation i.e. we can extract infinitely many subsequences from it s.t. they all converge to their respective point of accumulation? I have the feeling it would mean that the period of repetition of something could be infinitely big.


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up vote 9 down vote

Start with $0,1$. Then travel backwards to $0$ in steps of $1/2$, so $1/2,0$. Then travel forwards to $1$ in steps of $1/4$, so $1/4,2/4,3/4,1$. Then travel backwards to $0$ in steps of $1/8$, so $7/8$, $6/8$, $5/8$, and so on. Continue.

Every real between $0$ and $1$ is an accumulation point of our sequence.

edit

@MehdiSlimani: You are welcome. If you prefer we could use $0,1,1/2,0,1/3,2/3,1,3/4,2/4,1/4,0,1/5,2/5,\dots$ and then we can quote the result that the rationals in $[0,1]$ are dense in $[0,1]$. - André Nicolas Apr 29 at 16:29

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