given a set $A \subseteq \mathbb{R}$ which follows:
1) $\mathbb{N} \subseteq A$
2) $\forall x \in A \ \forall n \in \mathbb{N} \exists y\in A \ : y^n = +x \ or \ y^n = -x $
3)if a set $A'$ follows the conditions 1 and 2 than $A \subseteq A' $
show that $|A| = \aleph _0$.
thanks ahead.
Let $${\Bbb N}^{\frac{1}{k}} = \{ l^\frac{1}{k} \;|\; l \in \Bbb N\}$$ be the set of all k-th roots from the naturals which is obviously countable for all $k\in \Bbb N$ and $$A' = \bigcup_{k\in \Bbb N} {\Bbb N}^{\frac{1}{k}}$$ be the union of all those sets which is countable as a countable union of countable sets.
Then $A'$ fulfills 1.) and 2.) by definition because 1.) is clear by construction and for $x \in A'$ it has to hold $x\in {\Bbb N}^{\frac{1}{k}}$ for some $k$ hence $$x=l^\frac{1}{k}$$ for some $l,k \in \Bbb N$. Taking $$y = l^\frac{1}{kn} \in {\Bbb N}^{\frac{1}{kn}} \subseteq A'$$ gives us $$y^n = l^\frac{n}{kn} = l^\frac{1}{k} = x$$ as wanted.
Hence $A'$ fulfills 1.) & 2.) and is countable so we get as a result $$\Bbb N \subseteq A \subseteq A'$$ and $A$ is sandwiched by countable sets so it's countable itself.