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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?

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(1) Yes, because the sum of all of them diverges. There was a question recently where it was shown that this is a sufficient condition.

(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.

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  • $\begingroup$ 1/2+1/3+1/6 = 1. You need more than that. $\endgroup$ – user24142 Apr 22 '15 at 20:22
  • $\begingroup$ @user24142: 1 should also be expressible as an infinite sum $\endgroup$ – ogogmad Apr 22 '15 at 20:34
  • $\begingroup$ That feels like a possible equivalent condition. $\endgroup$ – user24142 Apr 22 '15 at 20:35
  • $\begingroup$ @ user3491648: Can you give me the recent question you mention in (1), please? $\endgroup$ – Piquito Apr 25 '15 at 0:00

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