Showing $\bigcup_{k}\{(a_{1},...,a_{k}):a_{j}\in [0,1]\cap \mathbb{Q} ,\sum^{k} a_{j}=1\}$ is countable Showing  $M=\bigcup_{k}\{(a_{1},...,a_{k}):a_{j}\in [0,1]\cap \mathbb{Q} ,\sum^{k}  a_{j}=1\}$ is countable. 
I find this hard to believe because say for fixed k and any $b,c\in [0,1]\cap \mathbb{Q}$, we have $(\frac{1}{k-2},...,\frac{1}{k-2},b,c)\in M$. So I can embed in M any element of $ P([0,1]\cap \mathbb{Q})$,  which is uncountable.
thanks

The cartesian product of two countable sets is countable (think of the zig-zag-across-numerators-and-denominators argument used to show that $\mathbb Q$ itself is countable). Therefore $\mathbb Q^2$, $\mathbb Q^3=\mathbb Q^2\times \mathbb Q$, $\mathbb Q^4=\mathbb Q^3\times \mathbb Q$, etc. are countable. Each $(a_1,\ldots,a_k)$ is certainly in thge countable set  $\mathbb Q^k$. Hence $M$ is the countable union of countable sets, which is once again countable (by the same argument that countable times countable is countable: Each element is in at least one way the $n$th element of the $m$th set on the right ...).
