Let $P_n$ be the set of all polynomials of degree n with integer coefficients. Prove that $P_n$ is countable.

So I know to prove that $P_n$ is countable, it must be either infinitely countable or finite. Since we know it's not a finite set, it must be infinitely countable. Therefore, I need to prove that the set of all polynomials is equivalent to the set of natural numbers. This would imply that I need to use induction. My book provides a solution, but I don't really understand it. Can someone provide me with some insight on how to think about this problem?

The book says to set $$ P_n= n + a_0x^n + a_1x^{n-1} +a_2x^{n-2}+...+a_n$$ and this is the part I get lost at, let $$h = n+ a_0 + [a_1] + [a_2]+...+[a_n]$$

They then note $h\ge1$ and each $\lvert a_i \rvert \le h$

After that I'm absolutely confused how they use induction to provide a proof. Please help!

  • $\begingroup$ It is trivial to prove that $P_n$ is not finite: it contains the constant function $p(x)=n$ for each $n\in\Bbb Z$, which is infinite. What is confusing you here? If it is countable and not finite, it must be in bijection with $\Bbb N$ by the Schröder–Bernstein theorem. $\endgroup$ – Mario Carneiro Sep 28 '15 at 21:43
  • $\begingroup$ Apparently they want to use that for given height $h4, there are only finitely many polynomials of that height. Then the set of all polynomials is the union of countably many finite sets ... $\endgroup$ – Hagen von Eitzen Sep 28 '15 at 21:44
  • $\begingroup$ @HagenvonEitzen ... which is totally ordered lexicographically and hence countably infinite. (It bothers me when undergraduate textbooks wordlessly slip in countable choice in "induction proofs" like this, particularly when it is not necessary.) $\endgroup$ – Mario Carneiro Sep 28 '15 at 21:59
  • 2
    $\begingroup$ Walter Rudin, eh? Chapter 2 problem 2? The hint is confusing but what it's getting at is that for each N there are only a finite number of possible polynomials that satisfy the equation. Example: For n = 2, there is n=1, a_1 = 1 to get the polynomial: x For n =3 there is n=2, a_i = 1 to get the polynomials x and x^2 there is n=1 a_i = 2 to get the polynomials 2x and 2x^2 there is n=1 a_i, a_j = 1 to get the polynomials x^2 + x, x^2 + 1, and x+1. Do this for N =3, 4, 5, etc. and you get unique polyonomials. This will eventually yield all polynomials with no repetition. $\endgroup$ – fleablood Sep 28 '15 at 22:02

Note that $P_n$ is isomorphic to $\Bbb Z^{n+1}$ via the correspondence

$$(a_0,a_1,\dots,a_n)\in\Bbb Z^{n+1}\iff a_0+a_1x+\dots+a_nx^n\in P_n,$$

so it is sufficient to prove that $\Bbb Z^{n+1}$ is countably infinite. The easy way to do this is to find an injection from $\Bbb Z^{n+1}$ to $\Bbb N$, since $\Bbb Z^{n+1}$ is clearly not finite, and

$$f(a_0,a_1,\dots,a_n)=p_1^{g(a_0)}p_2^{g(a_1)}\dots p_{n+1}^{g(a_n)}$$

will do the trick (where $p_n$ is the $n$th prime, and $g(n)=2n^2+n$ is an injection $\Bbb Z\to\Bbb N$).


The trick is to see that

$h = n+ |a_0| + |a_1| + |a_2|+...+|a_n|$ ($a_n$ != 0, $n >= 1$)

yields a class of polynomial for each h.

h = 2 has the solutions: $ n =1$, $a_1 = 1$ which yields the polynomial $x$

h=3 has solutions:

$n =2$, $a_2 =+/-1 $ which yields the polynomials $x^2$ and $-x^2$

$n =1$, $a_1 =+/-2$ which yields $2x$ and $-2x$,

$n =1$, $a_1 =+/-1$, $a_0 = +/-1$ which yields $x +1 $, $x -1$, $-x +1$, and $-x - 1$.

and so on.

For each h, there are finite polynomials yield and for all h all polynomials will be yielded. Thus we have a countable union of finite sets of polynomials. Thus there are countably many polynomials.

  • $\begingroup$ Oh, I guess n can equal 0 and a_0 = h, and then you get a constant polynomial: h. Trick is to consider "h" for a specific polynomial. Each polynomial has a distinct "h". For each "h" there are only finitely many polynomials. Thus countably many polynomials. $\endgroup$ – fleablood Sep 28 '15 at 22:51

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