Consider the following problem:
Let $(a_{\alpha})_{\alpha\in \mathbb R}$ be non-negative real numbers such that $\sup\{\sum\limits_{\alpha\in F} a_{\alpha}:F\subset \mathbb R \text{ is finite}\}<\infty$,then show that $a_{\alpha}=0$ for all but countably many $\alpha\in \mathbb R$.
I have to prove this statement.Now I started by looking at the sets $F_n=\{\alpha\in \mathbb R:a_\alpha\geq \frac{1}{n}\}$,$n\in \mathbb N$.We claim that each of these sets is finite for if for some $n\in \mathbb N$ the set $F_n$ is infinite then there exists $\alpha_1,\alpha_2,...\in \mathbb R$ such that $\alpha_{k}\geq \frac{1}{n}$ for all $k\in \mathbb N$.So that $\sum\limits_{k=1}^\infty \alpha_k\geq \infty$.So,$F_n$ is finite for each $n\in \mathbb N$.But now choose $\beta_n\in F_n$ so that $a_{\beta_n}\geq \frac{1}{n}$ for all $n\in \mathbb N$.So,the subset $\{a_{\beta_1},a_{\beta_1}+a_{\beta_2},...\}$ of the set $\{\sum\limits_{\alpha\in F}a_{\alpha}:F\subset \mathbb R$ is finite$\}$ is unbounded above hence supremum of the set is $+\infty$ which is a contracdiction.Thus our claim is proved.
Then there is another part of this question:
Show that in the above question,the word 'countable' cannot be replaced by 'finite'.
I can take the example where $a_\alpha=0$ for $\alpha\in \mathbb R\setminus \mathbb N$ and $a_n=\frac{1}{2^{n-1}}$ for $n\in \mathbb N$.Where the supremum of the set in the question is $<\infty$ but not only finitely many elements are non-zero.
Is this approach correct?