Existence of convergent subsequences for all values in range?

Consider sequence $s(n) = \sin{nx}$. Are there values of $x$ for which the following holds: For every $y \in [-1,1]$ there is a subsequence of $s(n)$ converging to $y$? (Or perhaps just for the open interval...) Someone hypothesised that the answer is yes, and further that every $x$ that is relatively irrational with $\pi$ has this property.

The question I am more interested in is the generalised version of this to arbitrary sequences. What are necessary and sufficient conditions for a sequence having subsequences converging to any point in the set of values the sequence visits? Does it have anything to do with properties like the function $f(n)$ being ergodic or mixing?

(suggestions for tags welcome in comments)

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 The answer to the question in your 1st paragraph is yes. As for $f$ being 'mixing' or 'ergodic'... notice those two notions do not apply in their usual form to functions $\mathbb N\to [-1,1]$, say. – Mariano Suárez-Alvarez Mar 15 2010 at 15:18 Yes, I meant properties "like" mixing and ergodic. Sequences with the property I'm talking about seem to "bounce all over the place forever" in much the same way mixing functions do... – Seamus Mar 15 2010 at 15:37

More conventional language: Are there values of $x$ such that the sequence $\sin(nx)$ is dense in the interval $[-1,1]$. The answer is yes, almost all $x$ have this property, in particular all $x$ such that $x/\pi$ is irrational.

See Weyl's Criterion http://en.wikipedia.org/wiki/Weyl%27s_criterion for something (equidistributed) that implies much more than merely dense. And $nx$ mod 1 is equidistributed in $[0,1]$ if $x$ is irrational.

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Is being dense in the interval enough to guarantee that there is a subsequence that converges to any point in the interval? I wouldn't have thought so... – Seamus Mar 15 2010 at 15:34
@Seamus: what do you think the definition of "dense" is? [I suppose without Axiom of Choice you could give alternate definitions. But in this case we can always do a "choose the first $n$ such that..." argument and avoid it.] – Gerald Edgar Mar 15 2010 at 15:37
There is a nice proof of this result using Fourier Analysis. See T.W. Korner's Fourier Analysis, which I highly recommend in it's own right. – Tony Huynh Mar 15 2010 at 17:59
I had an incorrect comment earlier. Sorry. I misread the condition as that every value of the sequence was a limit of a subsequence, as opposed to every value in [-1,1] being a limit of a subsequence. – Douglas Zare Mar 16 2010 at 3:02

Here is a typical sufficient condition: If for a sequence $a_n$, one has $\lim a_{n+1}-a_n = 0$, then every number between $\lim inf a_n$ and $\lim sup a_n$ is a limit of $a_n$. Thus, if a sequence satisfies this condition, and in addition, $inf a_n = \lim inf a_n$, and $sup a_n = \lim sup a_n$, then every element of $a_n$ is a limit of $a_n$.

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