Let $\Lambda$ be a uncountable set and let $\{a_{\alpha}\}_{\alpha\in\Lambda}$ be such that $a_{\alpha}>0$ for all $\alpha \in\Lambda$ proof that, $$\sum_{\alpha\in\Lambda}a_{\alpha}$$ diverges.
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Silva
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7What is true of at least one set $A_n={ x\mid a_x\ge 1/n}$? – David Mitra Dec 13 '15 at 16:16
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2How do you define the sum? What is the process by which the sum is computed? I know how to do this for a countable set. An uncountable set is less certain in my mind. – marty cohen Dec 13 '15 at 16:24
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1You have to read about it, it is defined as a sup. – Silva Dec 13 '15 at 16:26
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1Can you prove it using Riemann integral? – Masacroso Dec 13 '15 at 16:38
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1I do not sure, but I think note! – Silva Dec 13 '15 at 16:40
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You can say that for any $a_{\alpha}>0$ exist some $n\in\Bbb N^*$ such that $a_{\alpha}\cdot n \ge 1$, so the series diverges, maybe? Using hiper naturals :p – Masacroso Dec 13 '15 at 16:46
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See here @EDUARDO, there is a lot more extensive answers and expositions. – Masacroso Dec 13 '15 at 17:38
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One should also define $\sum_{\alpha \in \Lambda}a_{\alpha}$, since this is not a standard thing, perhaps as $\sup \sum_{\alpha \in \Lambda_0}a_{\alpha}$ over all finite $\Lambda_0 \subset \Lambda$. Assume now that $\Lambda' \colon =\{ \lambda\ | \ a_{\lambda}> 0\}$ is uncountable. Now, every positive number is $\ge \frac{1}{n}$ for some $n >0$ natural. Therefore $$\Lambda' = \bigcup_{n\ge 1}\{ \lambda\ | \ a_{\lambda}\ge \frac{1}{n}\} $$ If a countable union of sets is uncountable, then at least one of the terms is uncountable ( since a countable union of countable sets is countable). So there exists $n_0\ge 1$ so that $\{ \lambda\ | \ a_{\lambda}\ge \frac{1}{n_0}\}$ is uncountable, and so infinite. For every $N_0$ natural there exists a finite subset $\Lambda_0\subset \{ \lambda\ | \ a_{\lambda}\ge \frac{1}{n_0}\}$, $| \Lambda_0| \ge N_0$. We get $$\sum_{\alpha \in \Lambda_0 } a_{\alpha} \ge \frac{N_0}{n_0}$$
Since we can take $N_0$ as large as we want, we conclude $$\sum_{\alpha \in \Lambda} a_{\alpha} = \infty$$
orangeskid
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Show the contrapositive, a convergent sequence of non-negative terms has at most countably many positive terms. Let $\sum_{i \in \Lambda} a_i$ be convergent. Then as David's hint, show that $A_n = \{i : a_i > \frac{1}{n}\}$ must be finite. Then use the fact that $A = \bigcup_{n} A_n$ must contain all strictly positive terms, but is a countable union of finite sets.
E.Lim
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Because it is the contrapositive statement and is logically equivalent to what you want to prove. $p \implies q$ is equivalent to $\neg q \implies \neg p$. It shows that if your uncountable sum of positive terms were convergent, then there must be countably many terms, a contradiction.So it must be divergent. – E.Lim Dec 13 '15 at 16:44
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I dont see it @E.Lim, sorry. You are taking the example of harmonic series as some kind of theorem. From an example I dont see how you can justify your assumptions. You must prove first that $a_i>1/n$ is a kind of theorem. – Masacroso Dec 13 '15 at 16:53
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1Sorry, it wasn't my intention of spelling out every step. If there were infinitely many $a_i$ such that $a_i > \frac{1}{n}$, then the series in question must diverge by comparison. Hence there must be finitely many of them. – E.Lim Dec 13 '15 at 16:57