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. 


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.
