Divergence of a positive series over a uncountable set. 
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. 

 A: 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 
$$\sum_{\alpha \in \Lambda} a_{\alpha} = \infty$$
A: 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.
