Subfield: Algebra


Question: Given any set $S$ of positive integers, show that at least one of the following two assertions holds:

(1) There exist distinct finite subsets $F$ and $G$ of $S$ such that $\sum_{x \in F} 1 / x=\sum_{x \in G} 1 / x$;

(2) There exists a positive rational number $r<1$ such that $\sum_{x \in F} 1 / x \neq r$ for all finite subsets $F$ of $S$.

Solution:
Argue indirectly. Agree, as usual, that the empty sum is 0 to consider rationals in $[0,1)$; adjoining 0 causes no harm, since $\sum_{x \in F} 1 / x=0$ for no nonempty finite subset $F$ of $S$. For every rational $r$ in $[0,1)$, let $F_{r}$ be the unique finite subset of $S$ such that $\sum_{x \in F_{r}} 1 / x=r$. The argument hinges on the lemma below.

Lemma. If $x$ is a member of $S$ and $q$ and $r$ are rationals in $[0,1)$ such that $q-r=1 / x$, then $x$ is a member of $F_{q}$ if and only if it is not one of $F_{r}$.

Proof. If $x$ is a member of $F_{q}$, then

$$
\sum_{y \in F_{q} \backslash\{x\}} \frac{1}{y}=\sum_{y \in F_{q}} \frac{1}{y}-\frac{1}{x}=q-\frac{1}{x}=r=\sum_{y \in F_{r}} \frac{1}{y}
$$

so $F_{r}=F_{q} \backslash\{x\}$, and $x$ is not a member of $F_{r}$. Conversely, if $x$ is not a member of $F_{r}$, then

$$
\sum_{y \in F_{r} \cup\{x\}} \frac{1}{y}=\sum_{y \in F_{r}} \frac{1}{y}+\frac{1}{x}=r+\frac{1}{x}=q=\sum_{y \in F_{q}} \frac{1}{y}
$$

so $F_{q}=F_{r} \cup\{x\}$, and $x$ is a member of $F_{q}$.

Consider now an element $x$ of $S$ and a positive rational $r<1$. Let $n=|r x|$ and consider the sets $F_{r-k / x}, k=0, \ldots, n$. Since $0 \leqslant r-n / x<1 / x$, the set $F_{r-n / x}$ does not contain $x$, and a repeated application of the lemma shows that the $F_{r-(n-2 k) / x}$ do not contain $x$, whereas the $F_{r-(n-2 k-1) / x}$ do. Consequently, $x$ is a member of $F_{r}$ if and only if $n$ is odd.

Finally, consider $F_{2 / 3}$. By the preceding, $\lfloor 2 x / 3\rfloor$ is odd for each $x$ in $F_{2 / 3}$, so $2 x / 3$ is not integral. Since $F_{2 / 3}$ is finite, there exists a positive rational $\varepsilon$ such that $\lfloor(2 / 3-\varepsilon) x\rfloor=\lfloor 2 x / 3\rfloor$ for all $x$ in $F_{2 / 3}$. This implies that $F_{2 / 3}$ is a subset of $F_{2 / 3-\varepsilon}$ which is impossible.
A finite $S$ clearly satisfies (2), so let $S$ be infinite. If $S$ fails both conditions, so does $S \backslash\{1\}$. We may and will therefore assume that $S$ consists of integers greater than 1 . Label the elements of $S$ increasingly $x_{1}<x_{2}<\cdots$, where $x_{1} \geqslant 2$.

We first show that $S$ satisfies (2) if $x_{n+1} \geqslant 2 x_{n}$ for all $n$. In this case, $x_{n} \geqslant 2^{n-1} x_{1}$ for all $n$, so

$$
s=\sum_{n \geqslant 1} \frac{1}{x_{n}} \leqslant \sum_{n \geqslant 1} \frac{1}{2^{n-1} x_{1}}=\frac{2}{x_{1}}
$$

If $x_{1} \geqslant 3$, or $x_{1}=2$ and $x_{n+1}>2 x_{n}$ for some $n$, then $\sum_{x \in F} 1 / x<s<1$ for every finite subset $F$ of $S$, so $S$ satisfies (2); and if $x_{1}=2$ and $x_{n+1}=2 x_{n}$ for all $n$, that is, $x_{n}=2^{n}$ for all $n$, then every finite subset $F$ of $S$ consists of powers of 2 , so $\sum_{x \in F} 1 / x \neq 1 / 3$ and again $S$ satisfies (2).

Finally, we deal with the case where $x_{n+1}<2 x_{n}$ for some $n$. Consider the positive rational $r=1 / x_{n}-1 / x_{n+1}<1 / x_{n+1}$. If $r=\sum_{x \in F} 1 / x$ for no finite subset $F$ of $S$, then $S$ satisfies (2).

We now assume that $r=\sum_{x \in F_{0}} 1 / x$ for some finite subset $F_{0}$ of $S$, and show that $S$ satisfies (1). Since $\sum_{x \in F_{0}} 1 / x=r<1 / x_{n+1}$, it follows that $x_{n+1}$ is not a member of $F_{0}$, so

$$
\sum_{x \in F_{0} \cup\left\{x_{n+1}\right\}} \frac{1}{x}=\sum_{x \in F_{0}} \frac{1}{x}+\frac{1}{x_{n+1}}=r+\frac{1}{x_{n+1}}=\frac{1}{x_{n}}
$$

Consequently, $F=F_{0} \cup\left\{x_{n+1}\right\}$ and $G=\left\{x_{n}\right\}$ are distinct finite subsets of $S$ such that $\sum_{x \in F} 1 / x=\sum_{x \in G} 1 / x$, and $S$ satisfies (1).