# Number of points of accumulation of a sequence

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.

• Since there are countably many rational numbers, there is a sequence that includes them all. The accumulation points of this sequence is the entire set of real numbers. Apr 29, 2016 at 17:29
• I like your example a lot ! Apr 29, 2016 at 18:15
• There's a classic theorem that if you walk around a circle in discrete steps of $a$ radians, where $a/\pi$ is irrational, then the set of points that you visit is dense in the circle. This implies that $(\sin(an))_{n\in\mathbb N}$ is dense in $[-1, 1]$, and you can take $a=1$ for a cute example of a sequence with infinitely many accumulation points. Apr 29, 2016 at 23:16
• See also this question. Some of the posts linked there might be of interest, too. Apr 30, 2016 at 8:23

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.

• Your sequence is bounded AND has infinitely many points of accumulation. Thank you for answering :) Apr 29, 2016 at 16:28
• @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 $[0,1]$. Apr 29, 2016 at 16:29
• I'm a bit confused. How can subsequences of your sequence converge if each member is in a different space? Maybe I don't have the necessary knowledge yet Apr 29, 2016 at 16:36
• For every real $x$ between $0$ and $1$, there is a subsequence of our sequence that has limit $x$. For in our back and forth travels, we get within $1$ of $x$, then within $1/2$ of $x$, then within $1/4$ of $x$, and so on. With the modified version in which we take steps $1$, then $1/2$, then $1/3$, then $1/4$, then $1/5$, and so on, our travels bring us to every rational between $0$ and $1$, so again there is a subsequence that converges to $x$. Apr 29, 2016 at 16:39
• Ok, I hadn't understood it well. Thank you again Apr 29, 2016 at 16:43

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 $p$.

• Simple and nice example, thank you a lot :) Apr 29, 2016 at 16:24

The rationals are a countable set. We can define a 1-to-1 function $f:N\to Q$ with $Q=\{f(n):n\in N\}.$ Consider the sequence $S=(f(n))_{n\in N}.$ Every real number is a limit point of a subsequence of $S.$

All the answers have uncountably many accumulation points. If you only want countably many, consider the sequence:

$1,1,2,1,2,3,1,2,3,4,1,2,3,4,5,\cdots$

Every positive integer is an accumulation point, and nothing else.

If you further want it to be bounded:

$\frac11,\frac11,\frac12,\frac11,\frac12,\frac13,\frac11,\frac12,\frac13,\frac14,\cdots$

Suppose each row of the infinite matrix below converges to a different real number

\begin{bmatrix} a_{11} & a_{12} & a_{13} & a_{14} & \dots \\ a_{21} & a_{22} & a_{23} & a_{24} & \dots \\ a_{31} & a_{32} & a_{33} & a_{34} & \dots \\ a_{41} & a_{42} & a_{43} & a_{44} & \dots \\ \vdots & \vdots & \vdots & \vdots & \end{bmatrix}

Then the sequence

$a_{11}, a_{21}, a_{12}, a_{31}, a_{22}, a_{13}, a_{41},a_{32}, a_{23}, a_{14}, a_{51}, \dots$

contains an infinite number of convergent subsequences.

• Some authors define accumulation point not so strictly, namely that it is enough that a subsequence converges to it. Apr 30, 2016 at 9:37