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I know that there are many questions related to this exercise that have been answered, and I have found a couple of different solutions to this exercise as well. But the solutions usually ignore the hint provided by Rudin, and even though some of them do deal with the hint, I am still not quite sure about what we should actually do with it. I wrote up my answer based on my understanding of those solutions used Rudin's hint. I want to know if this is correct. Thanks in advance for any clarification!

The exercise goes like this:

A complex number $z$ is said to be algebraic if there are integers $a_0$, $\dots$, $a_n$, not all zero, such that \begin{equation} a_0z^n + a_1z^{n-1} + \dots + a_{n-1}z + a_n = 0. \end{equation} Prove that the set of all algebraic numbers is countable. Hint: For every positive integer $N$ there are only finitely many equations with \begin{equation} n + |a_0| + |a_1| + \dots + |a_{n-1}| + |a_n| = N. \end{equation}

My attempt based on my understanding of those solutions used Rudin's hint:

Consider the set $A$ of all algebraic numbers, where for each $z \in A$ there are integers $a_0$, $\dots$, $a_n$, not all zero, such that \begin{equation} a_0z^n + a_1z^{n-1} + \dots + a_{n-1}z + a_n = 0. \tag{1} \end{equation} Now fix an $N \in \mathbb{Z}^+$, and consider the subset $A_N$ of $A$, where for each $z \in A_N$ there are integers $a_0$, $\dots$, $a_n$, not all zero, such that (1) holds, and that \begin{equation} n + |a_0| + |a_1| + \dots + |a_{n-1}| + |a_n| = N. \tag{2} \end{equation} Since, by the hint, there are only finitely many such equations as (2); and since, by the Fundamental Theorem of Algebra, for each $(a_0, a_1, \dots, a_{n-1}, a_n)$ satisfying (2) there are $n$ roots, and thus finitely many roots, for the polynomial \begin{equation} a_0x^n + a_1x^{n-1} + \dots + a_{n-1}x + a_n = 0, \tag{3} \end{equation} we have that the set $A_N$ is a finite union of finite sets and hence is finite. Since $A = \bigcup_{N=2}^{\infty} A_N$, we have $A$ is a countable union of finite sets and thus is countable, by the Corollary to Theorem 2.12.

Is my answer correct and rigorous? I really appreciate it!

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    $\begingroup$ Why list equation (4) (mistakenly written with variable $x$) when it’s a duplicate of (2), which is in turn a duplicate of (1)? Can’t you make this more succinct? What you have is correct, otherwise, although it takes a second of thought to understand why every algebraic number is in some $A_N$. $\endgroup$ Jun 8, 2023 at 4:42

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It is correct! However I have some recommendations concering the exposition.

Let $$M = \{ (a_0,\ldots, a_n) \in \mathbb Z^{n+1} \mid (a_0,\ldots, a_n) \ne (0,\ldots,0) \}. $$ This is a countable set.

For $(a_0,\ldots, a_n) \in M$ consider the equation \begin{equation} a_0z^n + a_1z^{n-1} + \dots + a_{n-1}z + a_n = 0 \tag{1} \end{equation} and define $$A(a_0,\ldots, a_n) = \{ z \in \mathbb C \mid z \text{ satisfies } (1)\} ,$$ $$A = \{ z \in \mathbb C \mid \text{There exists } (a_0,\ldots, a_n) \in M \text{ such that } z \text{ satisfies } (1)\}.$$ Clearly $A = \bigcup_{(a_0,\ldots, a_n) \in M } A(a_0,\ldots, a_n)$. Since $(1)$ has at most $n$ distinct roots, $A(a_0,\ldots, a_n)$ is finite.

Thus $A$ is a countable union of finite sets and thus is countable.

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