# Sums of reciprocals of subsets of natural numbers

There exists such a subset $A$ of the reciprocals of natural numbers $\{\frac{1}{n} \ |\ n \in \mathbb N\}$ that any real number $x$ on the interval $[0,1]$ can be expressed as sum of members of some subset $B_x$ of $A$ in such a way that all elements of $B_x$ are distinct, i.e. without any repeating terms in the sum. Call this as the property.

An example of a set $A$ having the property would be the inverses of powers of $2$. This is easy to see, as every number $x$ on the interval $[0,1]$ can be expressed in base two, using any power of $2$ only once, and these numbers form the set $B_x$.

Thus the set of reciprocals of any subset of $\mathbb N$, that has the powers of $2$ as a subset, also has the property.

There are also subsets of reciprocals of natural numbers that do not have the property. Take inverses of powers of $3$ as an example. If we try to express $\frac{1}{5}$ in this way, we note that $$\sum _{n=2}^{\infty } \frac{1}{3^n} = \frac{1}{6}<\frac{1}{5}$$ and $$\frac{1}{3}>\frac{1}{5}.$$ We conclude that we cannot avoid repeating a term.

Question 1) Consider the reciprocals of prime numbers. Does this set have the property?

Question 2) Clearly the sum over all the members of $A$ has to be greater or equal to $1$ for the set $A$ to have the property. Is this enough? What are the necessary and sufficient conditions for a set to have the property?

(2) it is not sufficient because if A includes the reciprocals of {2,3,4} and then reciprocals of powers of 1000 then there are number less than $\frac{1}{4}$ which are not expressible. I conjecture that being able to express just $1$ as the sum of elements of $A$ is sufficient, if $A$ is infinite and you express $1$ as an infinite sum.