Is $T$, the set of all polynomials with integer coefficients, countable? 
Is $T$, the set of all polynomials with integer coefficients, countable?

Here is my attempt at a proof:
Let $S$ = {$ (a_0, a_1, ..., a_k)$ | $a_i\in \Bbb Z$, $k \in \Bbb N$}
$T$ is the set in question. That is, $T$ = {$ (a_0, a_1x, ..., a_kx^{k})$ | $a_i \in \Bbb Z$, $k \in \Bbb N$}.
Then by the enumeration principle $S$ is countable, i.e., |$S$| = |$\Bbb N$|, because it can be labelled by
$L =$ {$0, 1, 2, 3, 4, 5, 6, 7, 8, 9, , , (, )$} [Note: I've included "," as an element].
Now define an injection $f: T \to S$ by:
$f(a_0 + a_1x + ... +a_kx^{k}$) = ($a_0, a_1, ..., a_k$).
Then $|T| \le |S|$.
We can define a similar injection $g: S \to T$, establishing |$S| \le |T$|. So |$T$| = |$S$| = |$\Bbb N$| which is what we wanted to prove. 
Is this good? I'm pretty new to set theory so I want to make sure these proofs are valid. 
 A: Any polynomial $p(x)$ over $\Bbb Z$ can be written as $p(x)=a_0+a_1x+\ldots +a_nx^n,n\in \Bbb N$.
Since $a_i\in \Bbb Z$ ,and $\Bbb Z$ is countable so for each $a_i$ we have countable choices .
EDIT:By "choices" I mean number of options available for us to select the $a_i's$ from.Since the polynomial is over $\Bbb Z$ so we have the option of selecting the $a_i's$ from the set of integers.
And since countable union of countable sets is countable so the set of all polynomials over $\Bbb Z$ is countable also
A: I would say you are on the right track. I don't think you have to make such a long argument that $|S| = |T|$ ... you can just say that each $S$-member is uniquely determined by a choice of $k$ and of $a_i$ with $i \leq k$, and likewise for each $T$-member, so there is a clear bijection between $S$ and $T$, so $|S| = |T|$.
But as someone else just noted in an answer, you need to construct a bijection between $\mathbb{N}$ and $S$ (or $T$). I would do something like this: starting at stage 0, enumerate the polynomials with $k \leq 0$ and $|a_i| \leq 0$ for each $i \leq k$; then, at stage 1, enumerate those polynomials with $k \leq 1$ and $|a_i| \leq 1$ for each $i \leq k$ (that have not been already listed), and so on for each stage $> 1$. (That is, at stage $n$, "tack on" to your enumeration all polynomials that were not previously listed and that have $k \leq n$ and the absolute value of each $a_i$ less than or equal to $n$). At each stage you add only finitely many polynomials to your enumeration, and every polynomial gets listed at some stage, so there are only countably many.
A: Your result only holds if you show that $|S| = |\Bbb N|$ which you have not done because you could not use the enumeration principle. $S$ has infinitely many elements of the form $(a, 0, \cdots, 0) $ and then infinitely many terms of the form $(0, b, 0, \cdots) $ and then infinitely many of the form $(a, b, 0, \cdots) $ and so on and so forth and therefore you cannot enumerate them with that ease. That is the problem with the demonstration. You would still have to build an explicit injection from $S $ on to $\Bbb N $ to prove $|S| = |\Bbb N|$.
A: Send (a_0,a_1x^1,...a_nx^n) to (p_0^(a_0),p_1^(a_1),....) where p_i is the i+1th prime.
