Proving that the algebraic numbers are denumerable

[Mr Davis 97]

Homework Statement

Prove that algebraic numbers are denumerable

Homework Equations

The Attempt at a Solution

This is a very standard exercise, but I haven't looked at its proof and want to see if I can prove it myself.

With each element $(a_n, a_{n-1},...,a_1,a_0 ) \in \mathbb{N}^n$ we can obtain at most $n$ roots of the polynomial $a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0$. We know that $\mathbb{N}^n$ is denumerable (since it is a countable product of denumerable sets). Let $R_n = \text{the set of all roots of polynomials of degree n}$. We can think of $R_n$ as being based on $\mathbb{N}^n$, where each n-tuple is "replaced" by a finite collection of at most $n$ roots. Since in each case we are replacing a single element by a finite number of other elements (roots), the set remains denumerable. So $R_n$ is denumerable. Let $R = R_1 \cup \cdots \cup R_n$. $R$ is denumerable because it is a finite union of denumerable sets.

 

[Mr Davis 97]

Homework Statement

Prove that algebraic numbers are denumerable

Homework Equations

The Attempt at a Solution

This is a very standard exercise, but I haven't looked at its proof and want to see if I can prove it myself.

With each element $(a_n, a_{n-1},...,a_1,a_0 ) \in \mathbb{Z}^n$ we can obtain at most $n$ roots of the polynomial $a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0$. We know that $\mathbb{Z}^n$ is denumerable (since it is a countable product of denumerable sets). Let $R_n = \text{the set of all roots of polynomials of degree n}$. We can think of $R_n$ as being based on $\mathbb{Z}^n$, where each n-tuple is "replaced" by a finite collection of at most $n$ roots. Since in each case we are replacing a single element by a finite number of other elements (roots), the set remains denumerable. So $R_n$ is denumerable. Let $R = R_1 \cup \cdots \cup R_n$. $R$ is denumerable because it is a finite union of denumerable sets.

Edited with corrections

 

[Ray Vickson]

Homework Statement

Prove that algebraic numbers are denumerable

Homework Equations

The Attempt at a Solution

This is a very standard exercise, but I haven't looked at its proof and want to see if I can prove it myself.

With each element $(a_n, a_{n-1},...,a_1,a_0 ) \in \mathbb{Z}^n$ we can obtain at most $n$ roots of the polynomial $a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0$. We know that $\mathbb{Z}^n$ is denumerable (since it is a countable product of denumerable sets). Let $R_n = \text{the set of all roots of polynomials of degree n}$. We can think of $R_n$ as being based on $\mathbb{Z}^n$, where each n-tuple is "replaced" by a finite collection of at most $n$ roots. Since in each case we are replacing a single element by a finite number of other elements (roots), the set remains denumerable. So $R_n$ is denumerable. Let $R = R_1 \cup \cdots \cup R_n$. $R$ is denumerable because it is a finite union of denumerable sets.

Edited with corrections

There is no limit on the degree of a polynomial that defines an algebraic number, so the actual set of algbraic number is
$$ R = \bigcup_{n=1}^{\infty} R_n.$$
Is that set still denumerable?

 

[Mr Davis 97]

There is no limit on the degree of a polynomial that defines an algebraic number, so the actual set of algbraic number is
$$ R = \bigcup_{n=1}^{\infty} R_n.$$
Is that set still denumerable?

You're right. But a denumerable union of denumerable sets is still a denumerable set, right?