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Can anyone please help me to find an "Examples of bounded sequence with infinite sub sequential limits."

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  • $\begingroup$ A constant sequence is an easy example. $\endgroup$ – Mr Toad Mar 23 '17 at 5:29
  • $\begingroup$ @MrToad Sir but a constant sequence has only one limit which is equal to all the sub-sequential limit. I am looking for a bounded sequence having infinitely different many sub-sequential limits. Suppose (-1)^n has only two sub-sequential limit namely "+1" and "-1" i.e only two sub-sequential limit but I want infinitely many. $\endgroup$ – Arkaprabha Karmakar Mar 23 '17 at 5:37
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    $\begingroup$ A bijection from $\mathbb{N}$ to $\mathbb{Q}\cap[0,1]$? $\endgroup$ – delt3 Mar 23 '17 at 5:54
  • $\begingroup$ Users delt3, angryavian and Mr Toad already provided nice examples. A much harder example is the sequence $\{ \sin n : n \in \Bbb{N} \}$. The set of limit points is all of the interval $[-1 ,1]$. $\endgroup$ – Sangchul Lee Mar 23 '17 at 6:12
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\begin{array}{} 1 \\ 1&1/2 \\ 1 & 1/2 & 1/3 \\ 1 & 1/2 & 1/3 & 1/4 \\ \vdots & \vdots & \vdots & \vdots & \ddots \end{array} (Reading from left to right)

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Similar to angryavian's answer and expanding on delt3's comment,
You can enumerate all of the rationals in $[0,1]$. Then given any $1/n$ for $n \in \mathbb{Z} ^+$ (of which there are infinitely-many) there exists an infinite subsequence converging to that number.

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