Let $B$ be a set of positive real numbers with the property that adding together any finite subset of elements of $B$ always gives a sum of 2 or less. Show $B$ must be finite or countable.
I do not know where to start with this proof. Any help is appreciated.
For each $n\in\mathbb N$, let$$B_n=\left\{b\in B\,\middle|\,b\geqslant\frac2n\right\}\subset B.$$Of course, $B_n$ can have no more than $n-1$ distinct elements; otherwise, the sum of $n$ distinct elements of $B_n$ would be grater than $2$.
But$$B=\bigcup_{n\in\mathbb N}B_n.$$Since $\mathbb N$ is countable and each $B_n$ is finite, $B$ is countable.
To proceed by contradiction, assume that $B$ is not countable. Choose some $b \in B$, since B is not countable, there must exist a sum of elements , call it c such that $c\gt nb$ for some $n\in \mathbb{N}$. (Simply consider the next n elements that are greater than b, their sum is greater than nb.) Choosing $n \geq \frac{2}{b}$. This implies $\exists~c\in B$ with $c\gt2$. This is absurd since summing this with anything else in B is clearly not less than 2, as the supposition states it should be. $\square$
