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);|; qinmathbb{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 $pleftrightarrow (q_1, p)$ for each $pinmathbb{Q}$, 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.

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