# Does such a subset has a nonempty interior?

Let $(a_n)_{n=1}^\infty$ be a sequence such that $0\leq a_n \leq 1$, $\sum_{n=1}^\infty a_n=1$ and let $card \{a_n: n \in \mathbb{N} \}=\infty$. Let's consider the set $$S=\{ \sum_{n\in I} a_n: I \subset \mathbb{N} \}.$$ It may happens that $S=[0,1]$ for example for $(\frac{1}{2},\frac{1}{2^2},\frac{1}{2^3}...)$, or $(\frac{1}{3}, \frac{1}{3}, \frac{1}{3^2}, \frac{1}{3^2},...)$. Under what condition $S$ contain an interval? Is it then $S=[0,1]$?

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The conditions. when you get the whole interval $[0,1]$ for a non-increasing sequence are known, see here. –  Martin Sleziak Apr 21 '12 at 17:41
Thanks for links. I don't know if I understood correctly. W.l.o.g. we may assume that $(a_n)$ is decreasing because series is absolutely convergent. Then $S=[0,1]$ iff $a_n \leq \sum_{i=n+1}^\infty a_i$ for all $n \in N$. –  Alex Apr 22 '12 at 7:00
Yes, I agree with your comment. Unless, for some reason, the original ordering of the sequence $(a_n)$ is important for you, you can reorder it to get a non-increasing sequence. –  Martin Sleziak Apr 22 '12 at 7:05

I don't think there's going to be an easy clear condition for $(a_n)$. In any case, $S$ can be much smaller than $[0,1]$ - consider your binary example, but with an arbitrary number of powers replaced by their sum. Then the series still adds to $1$, but some intervals from $[0,1]$ are no longer in $S$, yet (assuming only a finite number of powers are combined) there still are complete intervals in $S$.
On the other hand, if you consider $(\frac{8}{9},\frac{1}{10}, \frac{1}{10^2}, ...)$ then it is easy to see that $S$ contains no interval.
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