Consider $k$ with $\frac 12≤ k < 1$
Define the set $A = \{k^n\}_{n=1}^{\infty}$
How can we prove that if $x\in (0,1)$ then there is a finite or infinite sum of the numbers in A (repetitions not permitted) which equals x.
The only way I can think of is using $$\sum_{i=0}^\infty k^i = \frac{k}{1-k}$$ But I guess this doesn't cover all numbers in $(0,1)$
Algorithm: choose $x\in (0,1)$. Let $x_1=k^i$ where $i=i_1\in \mathbb N$ such that $$k^i≤x< k^{i-1}$$ Similarly define $x_n=x^{i_n}$ where $i_n\in \mathbb N $is such that $$k^{i_n}≤x-\sum_{l=0}^{n-1}x_{l}< k^{i_n-1}$$
Claim 1: The exponents $\{i_j\}$ are strictly increasing.
Pf: It is enough to show that $i_1<i_2$. But suppose otherwise. Then we must have $$x-k^{i_1}≥k^{i_1}$$ But $$x-k^{i_1}≥k^{i_1}\implies x≥2k^{i_1}≥\frac 1k k^{i_1}=k^{i_1-1}$$ a contradiction.
Note: this is where we have used the fact that $k≥\frac 12$. Indeed, the desired result is false without this.
Claim 2: $$x=\sum_{l=1}^{\infty}x_l$$
Pf: Let $S$ denote that sum. Clearly the sum converges so $S$ is some real value. Let $S_n$ denote the partial sums. By definition, $x-S_n≥0$ for each $n$ so $S$ can not exceed $x$. Suppose $S<x$. Then there is some $k^{i_*}< x-S$. But at some point all the $i_l>i_*$ This is absurd, as clearly at any stage we could have subtracted $k^{i_*}$ . Thus $S=x$ and we are done.