Proving a set to be countable A set $S = \left\lbrace \left( x, y \right) \vert x^2 + y^2 = \dfrac{1}{n^2}, \text{ where } n \in \mathbb{N} \text{ and either } x \in \mathbb{Q} \text{ or } y \in \mathbb{Q} \right\rbrace$ is given. I need to prove that this is countable.
I have tried looking for a bijection $f: \mathbb{Q} \rightarrow S$ as $f \left( x \right) = \left( x, y \right)$ where $y$ is a fixed real number such that the property of the set is satisfied.
Clearly, this function is not even a surjection.
How should we define our bijection so that we prove $S$ is countable?
 A: Note that $x^2=\frac1{n^2}-y^2$ and $y^2=\frac1{n^2}-x^2$. So if $y$ is rational, then $x^2$ is rational as well, and likewise for $y^2$ if $x$ is rational.
Therefore you can define a surjection from the collection of all roots of rational numbers (which also includes all the rationals). So now comes the question, can you prove that this set is countable? 
A: If you index on $x$ and $n$, you can do the following. Let 
$$
E_n=\mathbb Q\cap \left[-\tfrac1{n^2},\tfrac1{n^2}\right].
$$
This sets $E_n$ are countable, because they are subsets of $\mathbb Q$.
Then $S=S_1\cup S_2$, where 
$$
S_1=\bigcup_{n\in\mathbb N}\bigcup_{x\in E_n}\left\{\left(x,\sqrt{\tfrac1{n^2}-x^2}\right)\right\}\cup\left\{\left(x,-\sqrt{\tfrac1{n^2}-x^2}\right)\right\}
$$
$$
S_2=\bigcup_{n\in\mathbb N}\bigcup_{y\in E_n}\left\{\left(\sqrt{\tfrac1{n^2}-y^2},y\right)\right\}\cup\left\{\left(-\sqrt{\tfrac1{n^2}-y^2},y\right)\right\}
$$
We can map this to a subset of $(\mathbb N\times\mathbb Q\times\{1,2\})^2$. And this last set is countable, so $S$ is countable. 
The key fact is that subsets of countable sets are countable, and that finite cartesian products of countable sets are countable. 
