Prove that the set of all numbers formed by finite sums of elements of $S$, is countable.

Question

$S$ = {$\sqrt[n]{a}$ : $n\in\mathbb{N}$ and $a\in \mathbb{Q}$}

Prove that the set of all numbers formed by finite sums of elements of $S$, is countable. We know that $S$ is countable.

I am not sure how to approach this one and have been stuck for awhile now. I would appreciate any assistance. Thank you!

Answer 1

You could "cheat" and write the sums formally, i.e. do not consider $\sqrt{2}+\sqrt{3}$ to be the same number as $\sqrt{3}+\sqrt{2}$. That way what you are asking is "Is the collection of finite strings from an alphabet of countably infinite size also countably infinite?", which should follow from some basic theorems like, the union of countably many sets of countable size is countable.

Answer 2

For each $n,m\in\mathbb{N}$, we can see the sum $\sqrt[n]{a_1}+\dots+\sqrt[n]{a_m}$ (where $a_1,\dots,a_m\in\mathbb{Q}$) as a formal sum and it can be represented by the $m$-ary tuple $(a_{n1},...,a_{nm})$ (here $a_{ni}:=a_i$ for $0\leq i\leq m$). So, define the set $A_{nm}:=\{(a_{n1},\dots,a_{nm}):a_i\in\mathbb{Q},1\leq i\leq m\}$. Now is easy to see that each $A_{nm}$ is countable, the the set $$ A_n=\bigcup_{m\in\mathbb{N}}A_{nm} $$ Is countable for each $n\in\mathbb{N}$. In the same form, the set $$ A=\bigcup_{n\in\mathbb{N}}A_n $$ Is countable.

With this, define the bijection $F:A\rightarrow S$ as $F((a_{n1},\dots a_{nm}))= \sqrt[n]{a_1}+\dots+\sqrt[n]{a_m}$. So $|S|=|A|=\aleph_0$.