# Countability of set of positive reals with bounded sum for all finite subsets

Consider a set $B$ of positive real numbers such that the sum of elements in any finite subset of $B$ is always less than or equal to $2$. Show that $B$ is countable.

I'm trying to find a bijection between $\mathbb{N}$ and $B$ but it's not clear how I would do this. I have a feeling I should be using the fact that if you have finite subsets $S$ and $T$, the sum of elements in $S \cup T$ is also $\leq 2$... but I don't see that going anywhere. I'd appreciate a hint, with a full solution in spoiler markdown if you can manage it.

• Hint: Forget about the $2$ stuff. Let $B_0$ be the set of elements of $B$ greater than $1$, and for $k\ge 1$ let $B_k$ be the set of elements of $B$ in the interval $[1/k,1/(k+1))$. Each $B_k$ is finite. – André Nicolas Apr 2 '16 at 0:14
• Simple and brilliant solution. I hope I will someday develop the intuition to come up with answers like this. – rorty Apr 2 '16 at 0:29

Hint: Let $B_0$ be the set of elements of $B$ that are greater than $1$. For every positive integer $n$, let $B_n$ be the set of elements of $B$ that are in the interval $\left[\frac{1}{n},\frac{1}{n+1}\right)$.

The set $B$ has been decomposed into a countable union of finite sets.

• Why must each $B_n$ be finite? – Nagase Apr 3 '16 at 18:26
• Let $n\ge 1$. If there were $2(n+1)$ or more numbers in $B_n$, then the sum of these would be greater than $2$, since each number in $B_n$ is greater than $\frac{1}{n+1}$. And a similar remark holds for $B_0$. – André Nicolas Apr 3 '16 at 20:07
• Thanks, that (including your answer) was very helpful! – Nagase Apr 3 '16 at 20:53
• You are welcome. – André Nicolas Apr 3 '16 at 22:08
• Should $B_{n}=\left (\frac{1}{n+1},\frac{1}{n} \right ]$? – Akash Gaur Dec 5 '17 at 9:39

You can prove it by establishing the following fact:

Let $M$ is an indexing set and for all $j \in M$, let $a_j \in [0,\infty[$, define $$\sum_{j \in M} a_j = \sup\left\{\sum_{j \in N} a_j\mid N \subseteq M, |N| < \aleph_0\right\}$$ Then $\sum_{j \in M} a_j < \infty$ only if only a countable number of $a_j$s is non-zero.

Hint: Consider sets of the form $S_n = \{a_j\mid a_j \geq \frac{1}{n}\}$.

Or, you can convert this to integration w.r.t. counting measure on $B$. Then it immediately follows from a property about integration.