Principles of Mathematical Analysis, Rudin, 3th ed, Chapter 2, Problem 2

Problem

A complex number $z$ is said to be algebraic if there are integers $a_0,\cdots,a_n$, not all zero, such that

$$ a_0z^n + a_1z^{n-1} + \cdots + a_{n-1}z + a_n =0. $$

Prove that the set of all algebraic numbers is countable. Hint: For every positive integer $N$ there are only finitely many equations with

$$ n+|a_0| + |a_1| + \cdots + |a_n| = N.$$

Answer

In Rudin's book, the definition of "countable" includes "infinity".

Since the set of algebraic numbers contain all integers, it is an infinite set. It remains to show that it is at most countable.

Define a set $A_N$ as follows:

$$ A_N = \{p(z) = a_0z^n + a_1z^{n-1} + \cdots + a_{n-1}z + a_n \; | \; n + |a_0| + |a_1| + \cdots + |a_n| = N, a_i \in \mathbb{Z}, \text{not all zero}, n \in \mathbb{N}\}.$$

Then $A_N$ is a finite set. For each $p \in A_N$, let $B_p$ be a set of zeros of $p$, which is finite. Hence, the set $C_N:=\bigcup_{p \in A_N} B_p$ is also finite. Note that the set of algebraic number is

$$ \bigcup_{N=0}^\infty C_N, $$

which is a countable union of finite sets, hence is at most countable.


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