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I am particularly interested on the sequences $a_n = n \alpha - [n \alpha]$ and $b_n = \sin (\pi n \alpha)$ where $\alpha \notin \mathbb{Q}$. We wish to find the set of limit points of $(a_n)$ and $(b_n)$

Try:

def of limit point in my book: $x$ is a limit point of $(x_n)$ if $\exists$ a subsequence $(x_{n_k})$ such that $\lim_{k \to \infty} x_{n_k} = x$.

which I believe can be translated as follows:

$x$ is limit point of $(x_n)$ if $\exists$ index $k>0$ so that $\forall \epsilon > 0 \exists N>0$ so that $n_k > N \implies |x_{n_k} - x | <\epsilon $

Evidently for sequence $(a_n)$ we may see that any multiple of $\alpha$ is a limit point.

the image of $(a_n)$ is the interval $[0,1)$. Is the set of all limit points this interval?

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Theoneandonly
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    Assuming that your $[x]$ is my $\lfloor x\rfloor$, the integer part of $x$, this answer should take care of the first sequence. Note that it’s not possible for all multiples of $\alpha$ to be limit points, because $0\le a_n<1$ for every $n$. I’ve not thought about the second sequence, but I suspect that something very similar happens. – Brian M. Scott May 17 '20 at 00:33
  • In the translation should be noted that ${n_k}$ is a sequence of distinct natural numbers. – Alex Ravsky May 21 '20 at 06:48
  • I would like to clarify is [ . ] the floor function? – matcha_ May 24 '20 at 11:23

1 Answers1

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Clearly $a_n\in [0,1]$ and $b_n\in [-1,1]$ for each natural $n$. On the other hand, the set $\{a_n\}$ is dense in $[0,1]$ and the set $B=\{b_n\}$ is dense in $[-1,1]$ (so the sets of the limit points of $\{a_n\}$ and $\{b_n\}$ are $[0,1]$ and $[-1,1]$, respectively). Both these facts are well-known. For the first, see, for instance, my detailed recent answer here. To prove the second, note that for each $n$, $b_n=\sin 2\pi a'_n$, where $a'_n$ is $a_n$ with $\alpha$ replaced by (an other irrational number) $\alpha/2$. Since by the first fact a set $A=\{a'_n\}$ is dense in $[0,1]$ and $B$ is an image of the set $A'$ by a continuous map onto $[-1,1]$, the set $B$ is dense in $[-1,1]$.

Alex Ravsky
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