Finding a Sequence with Limit Set [0,1]

[PingPong]

Homework Statement

Find a sequence whose set of subsequential limits is the interval [0,1].

Homework Equations

If the sequence does not repeat itself, then any subsequential limit is a cluster point.

The Attempt at a Solution

I've an idea that $|\sin n|$ is a solution to this, but I'm not sure how to prove this.

What I need to show is that, since the sequence I've chosen is non-repeating, then every point in the interval [0,1] is a cluster point. That is, if I take any number in [0,1] and am given an epsilon>0, can I always find an infinite amount of members in the sequence in an epsilon neighborhood around the number? This seems to intuitively work since $|\sin n|$ is all over the interval [0,1] (though not necessarily touching every point in the interval), but can this argument be made more rigorous? Any suggestions?

 

[morphism]

What can the rationals in [0,1] do for you?

 

[Dick]

|sin(n)| does almost certainly work. But I don't know an obvious proof of this. Maybe think about an easier example though. How about n*sqrt(2) mod 1? That should be 'all over', too, and maybe easier to deal with. Sorry, but I was just going zzzzz and found your post, so I can't give hints right now.

 

[Dick]

What can the rationals in [0,1] do for you?

Works also, and even easier. Nice.

 

[PingPong]

Thanks Dick and Morphism. I feel really silly now - I was trying to make it much too complicated. Obviously, since the closure of $Q\cap [0,1]$ is again [0,1], then the subsequential limits of a sequence of rational numbers in that interval (say, in the order given by the Cantor diagonalization process, throwing away those outside the interval) is the interval itself.

As a side note, in my search for information on cluster points, I stumbled upon an article (on JStor, which my university subscribes to, so I won't post the url since most people won't be able to access) that gives a proof that the cluster points of sin n is the interval [-1,1].