Show that the set is unbounded above 
Suppose $A$ is an uncountable subset of $[0,\infty)$. Show that the set $$\displaystyle \left \{\sum_{i=1}^n a_i\ \Big|\ a_1,\ldots,a_n \in A;\ n \in \mathbb{Z}^+\right \}$$ is unbounded above.

Since $A$ is uncountable, it can't be put into a one-to-one correspondence with $\mathbb{Z^+}$. Therefore, for any $M>0$, there exists $a \in A$ such that $a > M$ and so the partial sums seem to increase without bound. This seems to be a not very rigorous argument, but can the idea be used to solve the question?
 A: There needn't be $a \in A$ such that $a>M$. It's quite possible that every element of $A$ is bounded above by $1$ or $\frac{1}{2}$ or something even smaller.
Consider $A_n:=\left\{a \in A \mid a \geq \frac{1}{n}\right\}$ for $n \in \mathbb{N}$. If each $A_n$ is finite, then either $A=\bigcup_{n=1}^\infty A_n$ or $A=\bigcup_{n=1}^\infty A_n \cup \{0\}$, both of which are countable sets. So there must exist $n$ such that $A_n$ is infinite. Now given $M>0$, can you find a partial sum of elements in $A$ (or specifically $A_n$) that is greater than $M$?
A: I can prove it by contradiction.
Suppose $B=\displaystyle \left \{\sum_{i=1}^n a_i\Big|a_1,\cdots,a_n \in A,\ n \in \mathbb{Z}^+\right \}$ is bounded, here $a_1,\cdots,a_n$ need to be distinct. In other words, there exists a positive $M$ such that $\forall n\ge1,a_1,\cdots,a_n,\displaystyle\sum_{i=1}^n a_i\le M$. So that all series from $B$ have upper bound $M$, let $M$ be its supremum, then $\forall \{a_i\}\subset A,\displaystyle\sum_{i=1}^{\infty} a_i \le M$.
Since $A$ is uncountable, the series $\displaystyle\sum_{i=1}^{\infty} a_i$ in $A$ is also uncountable, let $C=\displaystyle \left \{\sum_{i=1}^{\infty} a_i\Big|a_i\in A\right \}$, then $C$ is a uncountable set in $[0,M]$. For convenience, let $M=1$.
Now, choose series $\{c_{jk}=\displaystyle\sum_{i=1}^{\infty} a_{ijk} \Big| c_{jk} \in C\}^{\infty}_{k=1} \subset [\frac{1}{j+1},\frac{1}{j}]$, $j\in \mathbb{Z}^+$, where $\# \{c_{jk}\}^{\infty}_{k=1}$ may be finite or zero.
Then, we can construct a new serie $\displaystyle\sum_{j=1}^{\infty} \sum_{k=1}^{\infty} c_{jk} = \sum_{j=1}^{\infty} \sum_{k=1}^{\infty} \sum_{i=1}^{\infty} a_{ijk}$, which is an element of $C$, but has value of $\infty$.
Contradiction.
