Let $A$ be an infinite set of positive real numbers. Define $S := \sup_{B \in \mathcal{B}}\sum_{x\in B}x$, where $\mathcal{B}$ is the collection of all finite subsets of $A$. I want to show that $A$ is countable if $S < \infty$.
Since $S < \infty$, for all $\epsilon > 0$, there exists a finite subset $B$ of $A$ such that $S - \epsilon < \sum_{x \in B} x < S$. My ultimate idea is to show that $A_n := A \cap (1/n, \infty)$ is finite and therefore $A = \cup_n A_n$ is countable. But it is not clear to me that the existence of such an finite subset $B$ implies that $A_n$ is finite. Is there something I am missing? Thanks in advance!
$\sum_{x \in A_n} x \leq S$. But $x \in A_n$ implies $x >\frac 1 n$ so $\sum_{x \in A_n} x >\frac 1n Card(A_n)$. So we get $Card (A_n) \leq nS$. In poarticular, $Card (A_n) <\infty$ for each $n$. Since $A =\bigcup A_n$ we are done.