When is a sequence $(x_n) \subset [0,1]$ dense in $[0,1]$? Weyl's criterion states that a sequence $(x_n) \subset [0,1]$ is equidistributed if and only if
$$\lim_{n \to \infty} \frac{1}{n}\sum_{j = 0}^{n-1}e^{2\pi i \ell x_j} = 0$$
for ever non-zero integer $\ell$. I was wondering if anyone knows of a criterion similar to this one which characterizes when a sequence $(x_n) \subset [0,1]$ is dense in $[0,1]$? I find this formula particularly interesting because of the strong resemblance to Fourier series (which can in fact be used to prove the criterion), so I would be very pleased if anyone knows an answer which keeps this in mind.
Thanks in advance.
 A: Here are some necessary and sufficient conditions:
1.a. The set $\{n\mid a\leqslant x_n\leqslant b\}$ is nonempty for every $0\leqslant a<b\leqslant 1$. 
1.b. The set $\{n\mid a\leqslant x_n\leqslant b\}$ is infinite for every $0\leqslant a<b\leqslant 1$. 
2.a. The sum of the series $\sum\limits_nf(x_n)$ is positive for every continuous nonnegative function $f$ not identically zero.
2.b. The series $\sum\limits_nf(x_n)$ diverges for every continuous nonnegative function $f$ not identically zero.
A: Though it is clear what would count as a positive answer, the question is not precise enough that to make clear what counts as a negative answer.
Nonetheless, I believe that no similar formula could possibly be equivalent to denseness.  The problem is that there are sequences that are dense but that are wildly non-equidistributed.  For example, your formula would have to detect sequences constructed in the following way.  Let $\{q_1,q_2,\ldots\}$ be an enumeration of $\mathbb{Q} \cap [0,1]$, and choose an increasing function $f : \mathbb{N} \rightarrow \mathbb{N}$ that goes to infinity very rapidly.  For instance, $f$ could be a busy beaver function.  Now define a sequence $\{x_n\}$ by the formula
$$x_n = \begin{cases}
q_k & \text{if } f(k)=n \text{ for some }k,\\
0 & \text{otherwise.}
\end{cases}$$
As you can see, $x_n$ is extremely close to the constant sequence $0$.  I can't imagine any formula would be able to detect the tiny blips where it jumps to the various rational numbers.
A: The problem is that equidistribution is a property of sequences, but density is a property of sets. You may wish to prove as an exercise that a countable subset of $[0,1]$ is dense if and only if it can be ordered in such a way as to be equidistributed. So the set $\lbrace\,x_0,x_1,\dots\,\rbrace$ is dense if and only if there is a permutation $\sigma$ such that $$\lim_{n\to\infty}{1\over n}\sum_{j=0}^{n-1}e^{2\pi i\ell x_{\sigma(j)}}=0$$ for every nonzero integer $\ell$. 
