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.

  • 3
    $\begingroup$ 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. $\endgroup$ – André Nicolas Apr 2 '16 at 0:14
  • $\begingroup$ Simple and brilliant solution. I hope I will someday develop the intuition to come up with answers like this. $\endgroup$ – 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.

  • $\begingroup$ Why must each $B_n$ be finite? $\endgroup$ – Nagase Apr 3 '16 at 18:26
  • $\begingroup$ 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 $B_0$. $\endgroup$ – André Nicolas Apr 3 '16 at 20:07
  • $\begingroup$ Thanks, that (including your answer) was very helpful! $\endgroup$ – Nagase Apr 3 '16 at 20:53
  • $\begingroup$ You are welcome. $\endgroup$ – André Nicolas Apr 3 '16 at 22:08
  • 1
    $\begingroup$ Should $B_{n}=\left (\frac{1}{n+1},\frac{1}{n} \right ]$? $\endgroup$ – 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.