Problem: Show that the set of polynomials with rational coefficients is countable.

Idea: We know that the set of rational numbers is denumerable. This implies that the set of rational numbers is countable. We also know that the degree that each polynomial can be is a natural number (I think, and I'm not sure how to word this.) Therefore, I think we can reason through this somehow, by showing that the plane $Q X N$ Is denumerable. I'm just not sure how to do this. Any ideas.

Note: this is for my introduction to proofs class study quide. So, I would prefer not to go to over the top.


3 Answers 3


Let us begin by stating that the set of rationals is countable, as Z2 is countable and every rational number can be expressed as an ordered pair of integers.

Then we can enumerate the polynomials as follows:

1.Start with n=1
2.For every m on [1,n] list the next order-m polynomial
3.Increase n by 1 and return to step 2.

For a proper enumeration of each order of polynomials (which must exist as, Q being countable, Qn is countable for all finite n and every order-n rational-coefficient polynomial can be expressed as an ordered n-tuple), every polynomial of rational coefficients will eventually appear exactly once.


Write $\Bbb N_0=\Bbb N\cup\{0\}$

Theorem: Let $X\subset\Bbb C$ be a countable set. Then the set $$X[z]=\left\{\sum_{j=0}^{n}a_jz^j:n\in\Bbb N_0,a_j\in X\right\}$$ is countable.

Proof: Since $X$ is countable, so is $\mu(z)=\{az:a\in X\}$. Thus the sets $$S_n(z)=\prod_{k=0}^{n}\mu(z^k)=\left\{(a_0,a_1z,a_2z^2,..,a_nz^n): (a_0,a_1,...,a_n)\in X^{n+1}\right\}$$ are countable for all $n\in\Bbb N_0$, because they are each a finite product of countable sets. Then for any $p=(p_1,...,p_n)\in\Bbb C^n$, write $\phi_n(p)=\sum_{k=1}^{n}p_k$. We then have that every set of degree-$n$ polynomials with coefficients in $X$, given by $$X_n[z]=\{\phi_n(p):p\in S_n(z)\},$$ is countable. And since $$X[z]=\bigcup_{n\in\Bbb N_0}X_n[z]$$ is a countable union of countable sets, it must also be countable. $\square$

Since $\Bbb Q$ is a countable subset of $\Bbb C$, we get that $\Bbb Q[x]$ is countable via the theorem.


Let $A_n = $ set of n-degree polynomials with rational coefficients, and let $A$ be set of all polynomials with rational coefficients. So, $1+x+\cdots + x^n \in A_n$. Firstly, we need a way to "represent" any n-degree polynomial. To simplify I will say that by n-degree I mean with n coefficients, otherwise it gets cumbersome to write. So we say this: $$ 2+x+x^2+x^3+\cdots +4x^{n-1} \equiv (2,1,1,1,\cdots ,4) \in A_n $$ Next through mathematical induction we proof that $A_n$ is a countable set for any $n$.

Let $n=1$ then $A_1$ would just become a set of constant polynomials. It just contains all the rational numbers. And we know that $|\mathbb{Q}|=|\mathbb{N}|$. So $A_1$ is also countable because it just have pairs with one element which is a rational. So $A_1$ can be written like so: $A_1 = \{ (a_1), (a_2), \cdots \}$. Since all the elements of it are countable.

Assuming $A_n$ to be countable for any arbritary $n$ of our choosing.

Testing if $A_{n+1}$ is countable or not. Now we know $A_n$ and $A_1$ are countable so, we can map them to natural numbers and thus written in form of a sequence. So I choose to write them like a sequence and then construct $A_{n+1}$ in such a way which shows that it is also countable by naturals.

$$ A_n = \{ (a_{11}, \cdots, a_{1n}), (a_{21}, \cdots, a_{2n}), (a_{31}, \cdots, a_{3n}), \cdots \} $$

Since both $A_n$ and $Q$ are countable there is no problem in writing them with the following representations

$$ \begin{matrix} A_n\downarrow\ \mathbb Q \rightarrow& r_1& r_2& \cdots&\\ (a_{11}, \cdots, a_{1n})& (r_1, a_{11}, \cdots, a_{1n})& (r_2, a_{11}, \cdots, a_{1n})& \\ (a_{21}, \cdots, a_{2n})& (r_1, a_{21}, \cdots, a_{2n})& (r_2, a_{21}, \cdots, a_{2n})& \\ \vdots& & & \ddots& \end{matrix} $$

In this way we can make a mapping from naturals to the set $A_{n+1}$ similiar to how we make from naturals to rationals.

So, we have proved that a set n-degree polynomials with rational coefficients is countable.

Now, through a proof we know that, the union of any number (even countably infinite) of countable sets is also countable. We know that our set of all polynomials is a countable union of $A_n$'s since again if we take $A_n$ to be one element of set $A$ then A can be mapped to natural numbers quite easily since $n$ of $A_n$ is from naturals itself. So, $A$ is union of countably infinite $A_n$'s. $$ A = \bigcup^{\infty}_{n=1} A_n $$ And so the set $A$ is also countable. Hence, the set of all polynomials with rational coefficients is countable.


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