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 $S=[0,1]$?
Does such a subset has a nonempty interior?
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analysis
metric-spaces
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0Yes, 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. – 2012-04-22
1 Answers
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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.