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A friend of mine challenged me to find a sequence such that the set of its limit points is exactly $\mathbb N$.

He has not found any, and neither have I.

Do you have any idea?

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2 Answers 2

up vote 2 down vote accepted

$1, 1, 2, 1, 2, 3, 1, 2, 3, 4, 1, 2, 3, 4, 5 \ldots$ I suppose you get the idea

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you ommitted $0$ but that's fine. +1 –  G.T.R May 10 at 11:58
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I did this on purpose, to me $0$ is not natural, but that's just a question of one's personal taste ;) –  mm-aops May 10 at 11:59
    
Formalizing: $$a_n=n+1-\frac{k(k+1)}{2},$$with $$k=\lfloor\frac{\sqrt{8n+1}-1}{2}\rfloor.$$ –  enzotib May 10 at 12:26
    
How did $\sqrt{8n+1}$ appear? –  Berci May 10 at 12:35
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If $U \subset \Bbb{R}$, then isn't a limit point of $U$ a point $x \in \Bbb{R}$ such that given any deleted neighborhood $N \subset \Bbb{R}$ around $x$, $N \cap U \neq \varnothing$? If so, then the answer above has no limit points; though it does have convergent subsequences. –  Tom May 10 at 12:48
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$0-\frac12,0-\frac13,1-\frac13,0-\frac14,1-\frac14,2-\frac14,...$ has the set of nonnegative integers as the set of limit points.

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