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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?

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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 $P = \bigcup_{n=1}^\infty P_n$.

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  • $\begingroup$ 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

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