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I came up with the following conjecture, but I'm not sure how to prove it or whether it is true at all.

Conjecture: let some subset of the real numbers $S \subset \mathbb{R}$ have an uncountable amount of elements, all of which are strictly positive, then it is possible to construct some countable subset of this set that diverges when summed.

My intuition says that this is probably true since when you create such an uncountable subset of the real numbers, you somehow have to smuggle in a range of real numbers or some set that is already divergent itself (like the irrationals). However, this is nowhere near a proof and I'm not sure whether I have even learned about the tools to rigorously prove this conjecture (or disprove). Therefore I would greatly appreciate it if someone could point me in the right direction.

Thanks in advance for your time and trouble.

Jori
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1 Answers1

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First note that there must exist $n$ such that there are uncountably many elements of $S$ which are greater than $\frac{1}{n}$. Otherwise we could write $$S = (S \cap \{0\}) \cup \bigcup_{n=1}^{\infty} S \cap \left(\frac{1}{n}, \infty\right)$$ which is countable. (You should work out why.)

Now just pick any sequence from $S \cap \left( \frac{1}{n}, \infty \right)$. Why must its sum diverge?

Clive Newstead
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  • Ah, yes sure. Infinite unions of countable sets are also countable. So if such a $\frac{1}{n}$ would not exist then $S$ would be countable which is contradictory to the assumption that $S$ is not countable. Now we just pick an infinite sequence of numbers in the range $(\frac{1}{n}, \infty)$. All elements of this sequence are strictly larger than $\frac{1}{n}$ and therefore as an application of the order limit theorem it must diverge. Thanks! – Jori May 11 '15 at 15:46
  • @Jori: You got it :) – Clive Newstead May 11 '15 at 15:49