Existence of convergent subsequences for all values in range? - MathOverflow

Existence of convergent subsequences for all values in range?

2010-03-15T15:06:59Z

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)

Answer by Gerald Edgar for Existence of convergent subsequences for all values in range?

2010-03-15T15:19:18Z

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.

Answer by Liran Shaul for Existence of convergent subsequences for all values in range?

2010-03-16T10:46:00Z

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 $a_n$.