I am asked to prove the following question:

For each finite set of real numbers, $F$, define $\sum F$ to be the sum of all the numbers in $F$.

For each set $P$ of positive real numbers, define $\sum P=\sup\{\sum F: F\text{ is a finite subset of } P\}$.

Suppose $P$ is a set of positive real numbers such that $\sum P < \infty$. Prove that $P$ is a countable set.

My attempt (just a sketch proof):

Suppose $P$ is uncountable, $\exists \epsilon_1 > 0$, s.t. $$ P_1 = [\inf P + \epsilon_1, \sup P]\cap P $$ is uncountable. Pick $a_1 \in [\inf P, \inf P + \epsilon_1] \cap P \in P$, such a $P$ must exists due to greatest lower bound property.

Now $P_1$ is uncountable, $\exists \epsilon_2 > 0$, s.t. $$ P_2 = [\inf P_1 + \epsilon_2, \sup P_1]\cap P_1 $$ is uncountable. Pick $a_2 \in [\inf P_1, \inf P_1 + \epsilon_2]\cap P_1 \in P$. Carry on this process (this should be formalized using induction), we can find $$ a_1 \leq a_2 \leq a_3 \leq ... $$ with $a_n > 0$ and $a_n \in P$ for all $n$.

Now $\sup\{\sum F_n: F_n = \{a_1,...,a_n\}, n \in \mathbb{N}^+\} = \lim_{n \to \infty}\sum_{i = 1}^n a_i > \lim_{n \to \infty}na_1 = \infty$. This means that $\sum P = \infty$ since $\{\sum F_n: F_n = \{a_1,...,a_n\}, n \in \mathbb{N}^+\} \subset \{\sum F: F\text{ is a finite subset of } P\}$.

I think my proof is a bit clumsy but is it correct?


There are a few problems with your proof. Some are simple typos, but fundamentally, I think the first claim is a problem:

Suppose $P$ is uncountable, $\exists \epsilon_1 > 0$, s.t. $$ P_1 = [\inf P + \epsilon_1, \sup P]\cap P $$

This is certainly a true statement! The issue is that you haven't argued why such an $\epsilon_1$ exists. And, if you do manage to argue that such an $\epsilon_1$ exists, then you are basically done with the proof: just pick a sequence of points $x_n \in P_1$, and $$\sum_{k=1}^n x_k \ge n(\epsilon_1 + \inf P) \ge n\epsilon_1 \to \infty$$ as $n \to \infty$. This immediately shows that $\sum P = \infty$.

So, while I don't doubt the statement is true, buried inside it is the crux of the argument, and is being brushed over, making the argument almost circular.

Instead, I suggest defining $P_n = P \cap [1/n, \infty)$. Argue that $P_n$ is finite, and $P = \bigcup_{n=1}^\infty P_n$.

  • $\begingroup$ Ah, I see your point. It is kind of circular since if I proved that statement, then I immediately proved the problem already! Thanks a lot! $\endgroup$ – davidolohowski Mar 13 at 3:39

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.