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The sum of an uncountable number of positive numbers

Suppose $f(x)>0$ for all real $x$, and $S$ is a set of uncountable many real numbers, how to prove that $\sum_{x\in S}f(x)=\infty$?

Alternately suppose $\sum_{x\in S}f(x)=k$, how to prove $|S|=N_0$ ?

Community
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jjj
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    hint: look at the cardinality of the set ${x \in S| f(x) > 1/n}$ and let $n\rightarrow \infty$ –  Dec 25 '11 at 15:25

2 Answers2

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Let $A_i=\{ x\in S : f(x)>1/i\}$ what is the cardinality of at least one of the sets $A_i$?

David Mitra
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Suppose the series $\sum_{x\in S_n}f(x)$ converges and consider the sets $S_n=\{x\in S\ |\ f(x)>1/n\}$. Then each of these sets must be finite, because $$ \sum_{x\in S_n}f(x)\geq\sum_{x\in S_n}\frac{1}{n}=\frac{|S_n|}{n}, $$ and $S=\cup S_n$. Therefore if the series $\sum_{x\in S_n}f(x)$ converges, $S$ is countable (since it is a countable union of finite sets).

yoyo
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  • Why would the cardinality of $S_n$ be finite? Is it because $\sum 1/n$ is a lower bound on $\sum_{\alpha \in S} x_{\alpha}$ and we know infinite harmonic series diverges? – Sun Mar 30 '21 at 04:24
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    @Sun No, the harmonic series is irrelevant; see my edits – yoyo Mar 30 '21 at 14:23