Is the set $\{(x,y)| x^2+y^2 = \frac{1}{n^2}, n \in \Bbb{N}, x\in \Bbb{Q}$ or $y \in \Bbb{Q}\}$ countable?

Question

I was thinking about the set $\left\{(x,y)\,\middle|\,x^2+y^2 = \frac{1}{n^2}, n \in \Bbb{N}, x\in \Bbb{Q}\text{ or }y \in \Bbb{Q}\right\}$

Certainly, this set is non-empty as I can find a pair $\left(\frac{1}{n^2},0\right)$ or like $\left(\frac{1}{2n^2},\frac{1}{2n^2}\right)$ or in a more genral sense the points of the form $\left(\frac{a}{n^2},\frac{b}{n^2}\right)$ where $a+b = 1, a,b \in \Bbb{Q}$, so hence I think it's a countable set?

But there may be some irrational coordinates involved which can make the set uncountable?. Can we rule out this possibility?

Answer 1

Your set is a countable union of countable sets:$$\bigcup_{n\in\mathbb N}\left\{(x,y)\in\mathbb{R}^2\,\middle|\,x^2+y^2=\frac1{n^2}\wedge x\in\mathbb{Q}\right\}\cup\left\{(x,y)\in\mathbb{R}^2\,\middle|\,x^2+y^2=\frac1{n^2}\wedge y\in\mathbb{Q}\right\}.$$Therefore, it is countable. Each of these sets is countable because, for instance, for each $x\in\mathbb{Q}$, the equality $x^2+y^2=\frac1{n^2}$ can only occut to finitely many $y$'s.

Answer 2

Denote $C_n$ the set of rational solutions of $x^2+y^2=1/n^2$.

Then, there is a bijection $C_1\rightarrow C_n$ given by $$ (r,s)\mapsto \left(\frac rn,\frac sn\right). $$ Moreover, the set $C_1$ is countable because it's essentialy in bijection with the set of lines through $(1,0)$ with rational slope (you need to add the vertical line to get an exact bijection).

Thus, all $C_n$ are countable.

Therefore $\bigcup_{n\in\Bbb N}C_n$ is countable as countable union of countable sets.