If $\sum_{i\in F} a_i$ is finite then $a_i =0$ except for countably many $i\in F$

Question

This question was asked in my real analysis quiz and I couldnot solve it .

Question : Let $a_{i} , i\in \mathbb{R}$ be non-negative real numbers such that $$\sup\left\{\sum_{i\in F}a_i\bigg| F\subseteq \mathbb{R} \text{ a finite subset}\right\}$$ is finite. Show that $a_{i}=0$ except for countably many $i\in \mathbb{R}$. Also give reasons if 'countably' can be replaced by 'finite'?

I am badly struck on this. As supremum of sum is finite so I think there would be only finite $a_{i}$ which would be non-zero instead of countable.

I feel it hard to give rigorous arguments in such kind of questions. So , it is my humble request to give a rigorous answer and tell in general on how such problems should be approached.

Thanks!!

You can't replace countable with finite. A simple counterexample: $a_{i}=1/2^i$ when i is a non-negative integer and 0 otherwise. — ypercubeᵀᴹ, Sep 06 '20 at 08:02
Does this answer your question? Proving $E$ is countable if the $\sup$ is finite — TheSilverDoe, Sep 06 '20 at 08:52

score 4 · Accepted Answer · answered Sep 05 '20 at 08:02

4

Let $M$ be the value of that supremum over sums.

Let $S_n = \{i : a_i \ge 1/n\}$. Then $M \ge \sum_{i \in S_n} a_i \ge |S_n|/n$, so $S_n$ must be finite.

Thus $\{i : a_i > 0\} = \bigcup_{n \ge 1} S_n$ must be countable.

answered Sep 05 '20 at 08:02

angryavian

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If $\sum_{i\in F} a_i$ is finite then $a_i =0$ except for countably many $i\in F$

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