sum of finite numbers of elements from the uncountable set of positive reals

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?

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.