# The set of all rational points in the plane is a countable set

From Kolmogorov's Introductory Real Analysis. I am doing some self-study and would like some feedback on whether my proof is correct.

I am using that the set of rational numbers is countable as given, and I am invoking the following Theorem which is proved in the book.

Theorem 2. The union of a finite or countable number of countable sets $$A_1, A_2, \ldots$$ is itself countable.

Claim. The set of all rational points in the plane (points with rational coordinates) is countable.

Proof.
Since $$\mathbb{Q}$$ is a countable set, we can write $$\mathbb{Q} = \{q_1, q_2, q_3, \ldots\}$$. If we fix $$q_1$$ we can define the following set: $$Q_1 = \{(q_1, q)\;|\; q\in\mathbb{Q}\},$$ which are all the rational points in the plane with $$q_1$$ in the $$x$$ position. We can create a one-to-one correspondence with $$\mathbb{Q}$$ by simply setting $$p\leftrightarrow (q_1, p)$$ for each $$p\in\mathbb{Q}$$, which shows that $$Q_1$$ is countable. We can now define the following union which includes all rational points in the plane: $$\mathcal{Q} = \bigcup_{n=1}^\infty Q_n,$$ which is a countable union of countable sets. By Theorem 2, $$\mathcal{Q}$$ is a countable set. ■

• This seems fine except that you defined $Q_1$ but didn't actually define $Q_i$ for any other $i$. This should be explicitly stated that $Q_i = \{(q_i,q)\mid q\in \Bbb Q\}$ Commented Apr 21, 2021 at 14:10
• As a stylistic choice, it may help as well to not overuse the letter $q$ here. You have five different types of $q$ appearing $(\Bbb Q, \mathcal{Q}, Q_i, q_i, q)$ which may make it difficult for some readers to follow along if they weren't well familiar with what you are doing. For such a short proof as this it might not matter, but in a longer more involved proof this could cause several headaches. Commented Apr 21, 2021 at 14:16
• @CoveredInChocolate Reopened. Good to go! Commented Apr 21, 2021 at 15:28
• You don't need to talk of writing out $\mathbb Q$ as $\{q_1, q_2, .... \}$. You could simply note that: $\mathbb Q$ is countable. And the for each $q\in\mathbb Q$ define set $Q_q = \{(q,p): p\in \mathbb Q\}$. Show $\mathbb Q\leftrightarrow Q_q$ via $p\leftrightarrow (p,q)$ is a bijection. Then $\mathscr Q = \bigcup_{q\in \mathbb Q} Q_q$ is a countable union. Commented Apr 21, 2021 at 15:45
• Your proof is correct. Key point is noting $p \leftrightarrow (p,q)$ is bijection between $\{(p,q)|p\in \mathbb Q\}\leftrightarrow \mathbb Q$ (depending on how picky your professor wants you to demonstrate knowledge of the definitions you may be asked to prove that but that just be a matter of stating the definitions) And that therefore $\mathbb Q \times \mathbb Q = \bigcup_{q\in Q}\{(p,q)|p\in \mathbb Q\}$ is a countable union of countable sets.... It's a good proof. Commented Apr 21, 2021 at 16:32