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Analysis question - given a sequence $\{a_n\}_{n=1}^\infty$, how many limit points can $\{a_n\}$ have? Initially I thought only $\aleph_0$, or countably many, because there are only countably many terms in such a sequence.

But then I thought about the sequence where $a_n = \sin(2^n)$, which looks like this for the first 1000 terms.

enter image description here

This sequence has no repeating terms, or else $\pi$ is rational. However I do not think every real number in $[0,1]$ appears in this sequence, as then the reals are countable, an absurd conclusion. It would not surprise me if this sequence has limit points, but how many? I do not immediately see any reason why $\sin(2^n)\notin (r-\epsilon,r+\epsilon)\subset[0,1]$, for $n$ large enough, because $r$ is arbitrary. This seems to imply that this sequence could have every point in $[0,1]\subset \mathbb R$ as a limit point, which slightly frightens me. So,

  • How many limit points does $\{\sin(2^n)\}_{n=1}^\infty$ have?

And in general,

  • How many limit points can a sequence with countably many terms have? What is an example of a sequence with the maximum number of limit points?

I would really like to see a proof of the answer to this last statement. I can easily come up with a sequence with $\aleph_0$ limit points, but I can't prove that is the upper bound.

Thanks for any help and tips on this topic.

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The irrationality of $\pi$ shows that $\sin(n)$ is dense in $[0,1]$... –  N. S. Oct 23 '12 at 2:46
    
@N.S. How about a proof of that? –  jlv Oct 23 '12 at 2:52
    
It is based on a theorem by Diricheclet: If $a$ is irrational, then $na+m$ is dense in $\mathbb R$... Use it for $\pi$, and then use $a=\pi$. –  N. S. Oct 23 '12 at 2:57
    
@N.S. OK, so here is a detailed explanation of why $na+m$ is dense in $\mathbb R$. But why does that show $\{\sin(2^n): n\in \mathbb N\}$ is dense in $[0,1]$? The elements of the sequence are not integer multiples of some other element, so I don't quite follow. –  jlv Oct 23 '12 at 3:07
    
@JimboBimbo, he did not claim that $\sin (2^n)$ is dense in the unit interval, just $\sin n.$ Maybe there is some trick involving logarithm for that, but I do not immediately see it. –  Will Jagy Oct 23 '12 at 3:17

2 Answers 2

The positive rationals are countable. meaning there is a sequence $x_n$ that includes every one. The set of limit points of this sequence is the entire positive real line. I'll see if I can dig up a picture for the ordering, it is this standard diagonal back and forth thing.

Oh, well. It is like the Cantor pairing function, except each pair $(x,y)$ is regarded as the rational number $\frac{x}{y}$ when $y \neq 0,$ plus you ignore any pair where $\gcd (x,y) > 1$ because you have already represented that fraction in lowest terms.

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OH JESUS you are indeed right - let $\{r_n\}_{n=1}^\infty$ be an ordering of the rationals in $[0,1]$. Let $\{s_n\}_{n=1}^\infty$ be a sequence of rationals in $[0,1]$ converging to $r\in[0,1]\subset \mathbb R$. Clearly $\{s_n\}\subset \{r_n\}$. Nice. Now how about the number of limit points on that sin sequence I have? –  jlv Oct 23 '12 at 2:49

There is a simple argument for showing that $\exp(2^n i)$ is dense in the unit circle in $\mathbb{C}$: suppose that, for any $n$ big enough, $\exp(2^n i)$ does not belong to a certain interval $I$ in the unit circle. Then, for any $n$ big enough, the sequence does not belong to $$ I^2 = \left\{z^2:z\in I\right\}, $$ neither to $I^{2^k}$ for any $k\in\mathbb{N}$. This is clearly a contradiction, since $$ \mu(I^2)= 2\cdot\mu(I). $$ This argument immediately shows that the sequence $\sin(2^n)$ is dense in $[-1,1]$.

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Except that $z\notin I$ does not imply that $z^2\notin I^2$ since $z\mapsto z^2$ is not one-to-one. –  Did Oct 23 '12 at 12:48

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