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Claim: Let $M$ be an uncountable subset of the real positive numbers, $M ⊂ \mathbb{R^+}$. Show that for every $r ∈ \mathbb{R^+}$ there is a finite number of different elements $a_1, a_2, . . . , a_n ∈ M$ such that $$\sum_{k=1}^n{a_k} \geq r.$$ Here is what I came up with: Let $M_N = \{a_k ∈ M:a_k\geq \frac{1}{N}\}$ and let $\sum_{k=1}^n{a_k} = r $. Then $$r=\sum_{k=1}^n{a_k}\geq\sum_{k=1\\a_k ∈ M_N}^n{a_k}\geq \frac{q}{N}, $$ where $q$ - number of elements in $M_N$. Hence $M_N$ has at most $rN$ elements.

Are my thoughts correct?

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No, your proof is not correct. Although you're on the right track there are some serious errors.

  • You said "let $\sum_{k = 1}^n a_k = r$." This is assuming the conclusion, and is actually a lot stronger than what you wanted to prove.

  • The sum $\sum_{k = 1, a_k \in M_n}^n a_k$ cannot be bounded from below because $M_n$ is potentially empty for fixed $n$; it's only if we take $n$ sufficiently large (depending on $M$) that it's guaranteed to have elements.

  • You also conclude that $\{k | a_k > 0\}$ is finite (and it's a set of integers - it's not equal to $\bigcup M_N$), but it isn't.


Here is an approach to get you started: There exists an $N$ such that your set $M_N$ is infinite (do you see why?). Choose any $q$ distinct elements $a_{k_1}, ..., a_{k_q}$ of this set, where $q > Nr$. Then

$$\sum_{j = 1}^q a_{k_j} \ge q \cdot \frac 1 N > r$$

as desired.

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