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My professor challenge me to give a sequence with limit points from zero to one including 0 and 1?

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Try $$ a_n=\sin n. $$ Also, you can enumerate $\mathbb Q\cap [0,1]$: $$ 0,1,\frac{1}{2},\frac{1}{3},\frac{2}{3},\frac{1}{4},\frac{3}{4},\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5},\frac{1}{6}\cdots. $$

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  • $\begingroup$ sin(n) gives only 360 terms $\endgroup$ Feb 21 '15 at 15:17
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    $\begingroup$ $\sin n$, with $n$ in radians not degrees! $\endgroup$ Feb 21 '15 at 18:39
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The rationals are dense in the interval [0,1]. So the sequence mapping N into $Q\cap [0,1]$ will give a sequence where if L is the set of limit points, min(L) = 0 and max(L) = 1. Andre's explicit constructions above are nice examples, although we can come up with others. Can you?

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If you can find a sequence such that:

  • $\limsup a_n=1$ and $\liminf a_n=0$
  • $\lim (a_{n+1}-a_n) = 0$

The for this sequence the set of limit points will be the interval $[0,1]$. (See: Set of cluster points of a bounded sequence and If a sequence satisfies $\lim\limits_{n\to\infty}|a_{n+1} - a_n|=0$ then the set of its limit points is connected. But if you come up with an example of such sequence, it will probably be easier to verify what the set of limit points of that particular sequence looks like than to check the proof of the general result valid for all sequences with these properties.)

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