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I've been trying to solve this problem for a few days and I feel like I'm missing something big.

Let $ X \subseteq \mathbb{R}_{>0} $ so that there's a $C > 0$ such that for every finite subset $\{x_1,...,x_n\} \subseteq X$ it's true that $ \sum _{i=1} ^n x_i \leq C$. Prove that $X$ is countable.

I managed to prove that $X$ can't have an interval inside of it (i.e $\nexists \, a,b \in \mathbb{R}_{>0}$ so that $[a,b] \subseteq X$), and that same proof can be extended to the case where $X$ is dense inside an interval quite easily, but I can't take it any further. So I now have to either prove that if $\left\vert{S}\right\vert = c $ then one of those holds (I can't think of a counterexample but I can't think of a proof either) or do something else entirely.

I'm pretty sure there's a much better way to look at this, but I can't find it. Any help would be greatly appreciated.

Mikasa
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Bananas
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    The Cantor set is uncountable, but it neither contains an interval, nor is it dense in an interval. Therefore, I don't think the approach you've described here will work. – G Tony Jacobs Aug 31 '13 at 06:03

1 Answers1

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If $X$ is uncountable, there is a countable subset $A$ of $X$ such that every point of $X\setminus A$ is an accumulation point of $X\setminus A$; there’s a proof here (and I’m pretty sure elsewhere on site as well). Now fix $x\in X\setminus A$ and $\epsilon>\frac{x}2$ and consider $(x-\epsilon,x+\epsilon)$.

Alternatively, and even easier, let $X_n=\{x\in X:x\ge 2^{-n}\}$ for each $n\in\Bbb N$ and note that if $X$ is uncountable, some $X_n$ must be uncountable and hence infinite and get an easy contradiction.

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Brian M. Scott
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