不存在$ℛ$，每个正有理数唯一表为$ℛ$的有限子集的元素和

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ICMC 4.1 (20 Dec 2020) Problem 4.
不存在正有理数的子集$ℛ$使得每个正有理数唯一表为$ℛ$有限子集的元素之和

Solution 1

$ℛ$是非空的。缩放 $ℛ$ 中的所有值，我们可以假设 $1 \in ℛ$。为了得出矛盾，现在假设 $x, y \in ℛ$ 和 $x \in(y, 2 y)$。然后注意 $x-y$ 是唯一子集 $A \subset ℛ$ 的总和，但 $y \notin A$ 为 $y>x-y$。因此 $\{x\}$ 和 $\{y\} \cup A$ 是两个不同的集合，其元素之和为 $x$，矛盾。因此 $ℛ$ 中不可能存在两个商∈(1,2) 的元素，我们将在下一段中使用这个引理。

现在假设 $r_1, r_2, r_3, \ldots$ 是区间 $(0,1)$ 中 $ℛ$ 的元素。根据上面的引理，对于 $a>0$，在任何区间 $(a, b)$ 中只存在有限数量的元素，不失一般性假设它们是有序的 $r_1>r_2>r_3>\cdots$。注意 $1 / 2^n \geq r_n$，否则我们与上面的引理矛盾。但是，如果对于任何 $m$，$r_m=1 / 2^m-\varepsilon$ 对于 $\varepsilon \in\left(0,1 / 2^m\right)$，那么数字 $1-\varepsilon / 2$ 不能表示为 $ℛ$ 的有限子集的元素之和，因为
$$
\sup _{A \subset ℛ \cap(0,1)} \leq \sum_{i=1}^{\infty} r_i \leq \sum_{i=1}^{\infty} \frac{1}{2^i}-\varepsilon=1-\varepsilon 。
$$
因此，对于所有 $m$，我们必须有 $r_m=1 / 2^m$。但现在 $1 / 3=0.\overline{01}_2$ 不能表示为 2 的幂的有限和，因此不能表示为 $ℛ$ 子集的有限和，因此这样的集合 $ℛ$ 不可能存在。

Solution 2

Similarly to the first solution, we show that the answer is no, by assuming for the sake of contradiction that there does exist such a set $ℛ$. Similarly, the set must be non-empty, so we assume WLOG that $1 \in ℛ$ and we prove that if $y \in ℛ$, then $(y, 2 y) \cap ℛ=\emptyset$.

We will show that if $y \in ℛ$, then $2 y \in ℛ$. Assume for a contradiction that $2 y \notin ℛ$. Then $2 y$ can be written as a sum of the elements of a finite subset of $ℛ$, say $A=\left\{x_1, x_2, \ldots, x_n\right\}$. If $y \in A$, then $A \backslash\{y\}$ and $\{y\}$ will be two finite subsets of $ℛ$ that sum to $y$, a contradiction. Hence $y \notin A$. Since $2 y \notin ℛ$, we must have that $n \geq 2$, so there exists at least one element of $A$ which is strictly smaller than $y$. So assume for a contradiction that $x_1<y$. Then $2 y-x_1=\sum_{k=2}^n x_k$ is one representation of $2 y-x_1$. Since $x_1<y$, we have that $y-x_1>0$, so suppose that $B$ is a subset of $ℛ$ that sums to $y-x_1$. Then $y \notin B$ (as $y-x_1<y$ ), so $B \cup\{y\}$ and $A \backslash\left\{x_1\right\}$ are two representations of $2 y-x_1$ (which are different since one contains $y$, while the other doesn't). This shows a contradiction, so $2 y$ must be in $ℛ$. A simple inductive argument shows that if $y \in ℛ$, then for any $n \in \mathbb{Z}_{\geq 1}, 2^n y \in ℛ$.

Now let $c \in ℛ$ such that $c<1$. Then there exists $n \in \mathbb{Z}_{\geq 1}$ such that $c \in\left[1 / 2^n, 1 / 2^{n-1}\right)$. The above paragraph implies that $2^n c \in ℛ$, but $2^n c \in[1,2)$, so by the first paragraph $2^n c=1$. Hence the only possible elements of $ℛ$ smaller than 1 are negative powers of 2 . Similarly as in the first solution, $1 / 3$ is not expressible as a finite sum of powers of two, which shows a contradiction, so no such set $ℛ$ can exist.