# Showing a set is finite or countable

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.

• Can we perhaps get a better title? Dec 4 '17 at 7:52
• Dec 5 '17 at 9:48

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.
• You have showed one example of $B$ that is countable and satisfies the problem statement. But you have to show that no such uncountable $B$ exists. Dec 9 '19 at 2:50
• @ShreyasPimpalgaonkar No. I provided no example. What I did was to prove that, for whatever set $B$ for which the given property holds, $B$ is finite or countable. Dec 9 '19 at 6:57
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. $$\square$$