Let $\mathcal{F}\_N$ be the set of all strictly increasing sequences of positive integers. For every two $F\_1, F\_2\in\mathcal{F}\_N$, if we use $\delta(F\_1,F\_2)$ to denote the first $n$-th coordinate where $F\_1(n)\neq F\_2(n)$, then $d(F\_1, F\_2) = \exp[-\delta(F\_1, F\_2)]$ defines a metric on the space $\mathcal{F}\_N$. One can observe that $d$ is an ultra-metric because, for each $F\_1, F\_2, F\_3\in\mathcal{F}\_N$,
$$
d(F\_1, F\_3)\leq \max\Big[ d(F\_1, F\_2), d(F\_2, F\_3) \Big].
$$
Now fix $r\in\big(\frac{1}{2}, 1\big)$. My questions are:


1. Will there exists a finite set of positive integers $M$ such that $\sum\_{i\in M}r^i = 1$? Does the existence of such a finite set depend on $r$?
2. Notice that we can view $\mathcal{F}\_N$ as the family of all **infinite** subset of $\mathbb{N}$ (viewing each element in $\mathcal{F}\_N$ as a sequence is more compatible to the given metric). For the fix $r\in\big(\frac{1}{2}, 1\big)$, one can observe that the set $\Big\{ \sum\_{i\in F}r^i\,\vert\, F\in\mathcal{F}\_N \Big\}$ is bounded below by $0$ (which is also the infimum) and bounded above by $\frac{r}{1-r}$. Is the following function surjective?
$$
S\_r:\mathcal{F}\_N\rightarrow \Big(0, \frac{r}{1-r}\Big], \quad F\mapsto \sum\_{i\in F}r^i
$$
In particular, I wonder, when **part 1)** is not true for all $r\in\big(\frac{1}{2}, 1\big)$, if it is true that, for each $r\in\big(\frac{1}{2}, 1\big)$, I can always find $F\_r\in\mathcal{F}\_N$ such that $S\_r(F\_r)=1$. One can easily check that $S\_r$ is continuous for each $r\in\big(\frac{1}{2}, 1\big)$. If I instead let $\mathcal{F}\_N$ be the set of all **non-negative** integers, then I suppose $S\_{\frac{1}{n}}$ will be surjective for each $n\in\mathbb{N}$; however, under this assumption, I do not know if $S\_r$ will be surjective even when $r\neq\frac{1}{n}$ for each $n\in\mathbb{N}$, but my current biggest concern is on the those two questions above.


Any hints or thoughts will be appreciated! The same question is posted in **MS** and I would like thank [Ryszard Sszwarc](https://math.stackexchange.com/users/715896/ryszard-szwarc) for his help with this question. He proved that numbers that meet the condition in **part 1)** must be algebraic and irrational. However, for a fixed algebraic and irrational number, whether or not a necessary condition that guarantees the existence of such a finite set exists remain unclear.



Could the range of $\sum_{k\geq 1}r^{n(k)}$ for $r\in \big(\frac{1}{2}, 1\big)$ be continuous?
For (1), I'm not sure you're going to find a better condition than "$r$ is an algebraic number which is the root of some polynomial with coefficients $0$ and $1$"; I certainly don't think there's an intrinsic necessary and sufficient condition there.


For (2), the answer is yes, the function is always surjective. You could prove this with a greedy algorithm. For a fixed $x \in (0, \frac{r}{1-r}]$, define $x\_0 = x$, and for all $n > 0$, define $x\_{n}$ to be $0$ if $x\_{n-1} = 0$, and if $x\_{n-1} > 0$, define $x\_n = x\_{n-1} - r^m$, where $m = m\_n$ is the minimal positive integer for which $r^m \leq x\_{n-1}$.


Since $m$ was minimal, $x\_{n-1} < r^{m-1}$, and so $x\_n < x\_{n-1}(1-r)$. Therefore, $x\_n \rightarrow 0$. This means that $x = \sum\_n r^{m\_n}$. Clearly
$m\_0 > 0$, since $x < 1$. So it remains only to check that the $m\_n$ are strictly increasing. But this is easy; if $m\_{n+1} \leq m\_n$, then $x\_{n-1} \geq r^{m\_n}$ and $x\_n = x\_{n-1} - r^{m\_n} \geq r^{m\_{n+1}}$ , so $x\_{n-1} \geq r^{m\_n} + r^{m\_{n+1}} 
\geq 2r^{m\_n} \geq r^{m\_n-1}$, contradicting minimality of $m\_n$. Therefore $m\_n$ is strictly increasing, and you have a representation $x = \sum\_n r^{m\_n}$.


By the way, this proof works even if there is a finite expansion, it will just end at that point (when $x\_n = 0$).


Another point is that even when your (1) is true, i.e. $1$ has a finite expansion, you can still find an infinite expansion of $1$! This is because, if $1 = r^{n\_1} + \ldots + r^{n\_k}$, then $1 - r^{n\_k} = r^{n\_1} + \ldots + r^{n\_{k-1}}$. But
$1 + r^{n\_k} + r^{2n\_k} + r^{3n\_k} + \ldots = \frac{1}{1 - r^{n-k}}$. Multiplying these yields


$1 = (r^{n\_1} + \ldots + r^{n\_{k-1}})(1 + r^{n\_k} + r^{2n\_k} + \ldots) = 
r^{n\_1} + \ldots + r^{n\_{k-1}} + r^{n\_1 + n\_k} + r^{n\_2 + n\_k} + \ldots + r^{n\_{k-1} + n\_k} + r^{n\_1 + 2n\_k} + r^{n\_2 + 2n\_k} + \ldots + r^{n\_{k-1} + 2n\_k} + \ldots$.


So, you don't have to have the clause "when (1) is not true"; for every $r \in (1/2, 1)$, $1$ can be written as $S\_r(F)$ for some $F \in \mathcal{F}\_N$.


Also, something you said is confusing; you said if you allow $0$ exponents, then $S\_{1/n}$ would always be surjective, but this is clearly false. For instance, the range of $S\_{1/3}$ would be the set of numbers with ternary expansion (including a possible $(1/3)^0 = 1$ term) containing only $0$s and $1$s, which is clearly a Cantor set and not an interval.


