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.

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.