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. 
 A: 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 $p$.
A: 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.
A: 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 $S.$
A: 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:

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

A: 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.
