Is $S = \{ (x,y): x^2+y^2=\frac {1}{n^2}, n \in \mathbb N \wedge (x \in \mathbb Q \vee y \in \mathbb Q) \}$ countable?

Question

I am working on the following problem:

Is the set $S = \left\{ (x,y): x^2+y^2=\frac {1}{n^2}, n \in \mathbb N \wedge (x \in \mathbb Q \vee y \in \mathbb Q) \right\}$ countable?

My answer is yes.

Reason:

For a fixed $n$, let $S_n = \left\{ (x,y): x^2+y^2=\frac {1}{n^2}, n \in \mathbb N \wedge (x \in \mathbb Q \vee y \in \mathbb Q) \right\}$.

Then $S_n = (\mathbb Q \times \mathbb Q) \cup \bigcup\limits_{r\in \mathbb Q}\left(\pm r, \pm\sqrt {\frac {1}{n^2}-r^2}\right)$ where $r <\frac {1}{n^2} $.

Clearly each set forming $S_n$ is countable, its union ($S_n$) is also countable. As $S=\bigcup\limits_{n\in \mathbb N} S_n$, $S$ is also countable.

Am I correct? Thanks in advance!

Answer 1

Actually you have

$$S_n=L_n\cup R_n$$

where

$$L_n=\bigg\{\bigg(r,\pm\sqrt{\frac{1}{n^2}-r^2}\bigg)\ \bigg|\ r\in \mathbb{Q}\text{ and }r^2\leq\frac{1}{n^2}\bigg\}$$ $$R_n=\bigg\{\bigg(\pm\sqrt{\frac{1}{n^2}-r^2},r\bigg)\ \bigg|\ r\in \mathbb{Q}\text{ and }r^2\leq\frac{1}{n^2}\bigg\}$$

Note the different order in $L_n$ and $R_n$ (that's the only difference between them). Those sets correspond to $x\in\mathbb{Q}\vee y\in\mathbb{Q}$ condition which is or, not and. And with this simple modification your reasoning is correct.

Answer 2

You made a minor formal mistake: when you say 'For a fixed $n$...' you should not insert the $(n\in\mathbb N)$ clause in the set definition as it means 'for any natural value of $n$'. Instead you might add it as a requirement to $n$ itself:

For a fixed $n\color{red}{\in \mathbb N}$, let $S_n = \left\{ (x,y): x^2+y^2=\frac {1}{n^2}, x \in \mathbb Q \vee y \in \mathbb Q \right\}$.

Apart from that your solution seems correct to me: there are countably many $n$ values (hence $x,y$ equations of circles), and each circle has countably many points with $x\in\mathbb Q$ and countably many points with $y\in\mathbb Q$ (these two sets not disjoint), hence $S$ is a union of countably infinite family of countably infinite sets. This implies it is equipollent with $\mathbb N^3$ and so it is countably infinite.

Answer 3

Given that $~S = \{ (x,y): x^2+y^2=\frac {1}{n^2}, n \in \mathbb N \wedge (x \in \mathbb Q \vee y \in \mathbb Q) \}~.$
So there doesn't exist any pair of $(x,y)$ other than $(1/n,0)$ and $(0,1/n)$, for all $n \in \mathbb N$, that satisfies the given condition.
Now we know that $~n \in \mathbb N ~$ is countable and countable union of countable sets is countable. Hence $~S=(1/n,0)\cup(0,1/n)~$ is countable.