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  1. Determine at least three limit points for the set {$\sin(n)$: n a positive integer}.

  2. How many limit points does the set {$\sin(n)$: n a positive integer} have?

Our professor gave us a definition to use for limit point in order to differentiate between a cluster point. The definition is as follows:

Let $S$ be a nonempty set of $\mathbb{R}$ where $S \subseteq \mathbb{R}$

Let $x \in \mathbb{R}$

We say that $x$ is a limit pont of $S$ if:

For each $\epsilon > 0 $ there is an element of $S$ in $(x-\epsilon$, $x+\epsilon)$.


With our first question and the given definition, wouldn't the numbers $\sin(1)$, $\sin(2)$, $\sin(3)$ work?

Choose the $x = \sin(1)$

We then have:

$\sin(1)-\epsilon < \sin(1) < \sin(1) + \epsilon$

Which seems trivially true. The same argument would follow for the next two points. However, the next part is where myself and a few of my other classmates are completely lost.

A few questions for 2.
What direction should I head for the second half? The professor mentioned Kronecker's theorem. If possible, could someone give a breakdown of Kronecker's theorem and it's applications?

Kronecker's Theorem: http://mathworld.wolfram.com/KroneckersApproximationTheorem.html

How can I find the cardinality of the set of limit points?

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You might be able to work out the answers after having a look at math.stackexchange.com/questions/63526/… –  Gerry Myerson Feb 10 '12 at 4:44
    
I adjusted my definition. I incorrectly transposed it from my notes. –  user17366 Feb 10 '12 at 5:04
    
Consider irrational rotation on the circle. That is, consider $R_r: S^1\rightarrow S^1$, $R_r(x)= (r + x)(\mathrm{mod} 1)$, where $S^1$ is the unit circle given by $[0,1]/\sim$, where ~ is the equivalence relation identifying $0$ with $1$ (you simply wrap the interval $[0, 1]$ into a circle, gluing the ends). Now for $r$ try some nice irrational values, like $\pi/2$, $\pi/4$, $\pi/6$, and start applying $R_r$ repetitively to zero: $R_r(0)$, then $R_r(R_r(0))$, and so on... (hint: $\{x_n\}_{n\in\mathbb{N}}$ is dense in $[0,1]$, where $x_n = R_r\circ\cdots\circ R_r(0)$ $n$-times. –  William Feb 10 '12 at 5:08
    
I was just reading this en.wikipedia.org/wiki/Irrational_rotation –  user17366 Feb 10 '12 at 5:09
    
@arete: Well, think about it. If it still doesn't make sense, comment, and I'll make my comment into an answer. –  William Feb 10 '12 at 5:52

2 Answers 2

Per OP's request, I am making my comment into an answer.

Notice that Kronecker's theorem, as stated here, is essentially equivalent to what I suggested in the comment above.

Define $S^1 = [0,1)$, and we can think of $S^1$ as a circle of unit circumference. Let us define rotation on the circle by $\alpha\in[0,1)$ as $R_\alpha(x) = (x + \alpha)\mod 1$. Then $n$-fold application of $R_\alpha$ to $x\in[0,1]$ is

$$ R_\alpha^n(x) := \underbrace{R_\alpha\circ\cdots\circ R_\alpha}_{n-\text{ times }}(x), $$

and has the value $(x + n\alpha)\mod 1$, or (as in the theorem of Kronecker referenced above) $(x + n\alpha) - \lfloor x + n\alpha\rfloor$. On the other hand, irrational rotation on $S^1$ is minimal, meaning that for any $x\in S^1$, if $\alpha$ is irrational, then the sequence $\{x_n\}_{n\in\mathbb{N}}$ given by $x_n = R_\alpha^n(x)$ fills $S^1$ densely.

After looking for a quick reference on the internet, I didn't find anything, so let us just prove it here quickly.

Suppose for some irrational $\alpha\in(0,1)$ and some $x\in S^1$, the sequence $\{x_n\}$, as defined above, is not dense in $S^1$. But then there exist open intervals in $S^1$ not containing $x_n$, for any $n\in\mathbb{N}$. Let $I$ be such an interval of maximal length. Obviously $R^n_\alpha(I)$ is also such an interval, of the same length as $I$, for any $n\geq 1$. If $R^n_\alpha(I)$ intersects another such interval, say $J$, and $R_\alpha(I)\neq J$, then we obtain an interval $J\cup R_\alpha^n(I)$ not containing any $x_n$, and of length greater than $I$, which by maximality of $I$ cannot happen. On the other hand, if $R_\alpha^n(I) = J$ and $J = I$, then for rational $q\in I$, we have $q = q + n\alpha \mod 1$, or $\alpha$ is rational, contradicting irrationality of $\alpha$. In other words, the sequence $\{R_\alpha^n(I)\}_{n\in\mathbb{N}}$ is a sequence of disjoint subintervals of $S^1$, each of the same positive length as $I$. But this, of course, cannot happen, since $S^1$ has finite length. We're done.

Now, say $\alpha\in (0,1)$ is irrational (say $\alpha = \pi/2$, for instance). Then the sequence $\{n\alpha\mod 1\}_{n\in\mathbb{N}}$ is dense in $[0, 1)$. In particular, there is a strictly increasing subsequence $\{n_i\}$ of natural numbers such that

$$ \lim_{i\rightarrow \infty}(n_i\alpha\mod 1) = 0. $$

It follows that

$$ \lim_{i\rightarrow\infty}\left|n_i\alpha - n_i\right| = 0. $$

This implies that

$$ \lim_{i\rightarrow\infty}\left|\sin(n_i) - \sin(n_i\alpha)\right| = 0. $$

Recall that $\alpha = \pi/2$. I think you can take it from here.

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Per the Wikipedia defintion "In mathematics, a limit point (or accumulation point) of a set S in a topological space X is a point x ( which is in X, but not necessarily in S ) that can be "approximated" by points of S in the sense that every neighbourhood of x with respect to the topology on X also contains a point of S other than x itself" so you cannot use the fact that $\sin(1)$ is in a neighborhood of $\sin(1)$. You need to show that there is another $n$ such that $\sin(n)$ is near $\sin(1)$, which is true.

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Note that the OP is speaking about limit points of a set, not a sequence, which means that we're using a (usually) more forgiving definition than the one you reproduce. –  Henning Makholm Feb 10 '12 at 5:04
    
@HenningMakholm: OP has added some text that answers my complaint of the lacking part of the sentence. You are right that this changes the answer. I missed the original that said it was a limit point of a set. –  Ross Millikan Feb 10 '12 at 5:15
    
On the other hand I missed the title of the question which still calls it a sequence. Shall we just agree that the question is ambiguous? –  Henning Makholm Feb 10 '12 at 5:17
    
@HenningMakholm: I can buy that easily. I have updated my answer hoping that it will help. –  Ross Millikan Feb 10 '12 at 5:22

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