Question: The following is the proof of Hurwit'z theorm in [wiki](http://en.wikipedia.org/wiki/Hurwitz%27s_theorem_%28complex_analysis%29)

> 
> Let f be an analytic function on an open subset of the complex plane
>  with a zero of order m at z0, and suppose that {fn} is a sequence of
>  functions converging uniformly on compact subsets to f. Fix some ρ > 0
>  such that f(z) ≠ 0 in 0 < |z−z0| < ρ. Choose δ such that |f(z)| > δ
>  for z on the circle |z−z0| = ρ. Since fk(z) converges uniformly on the
>  disc we have chosen, we can find N such that |fk(z)| ≥ δ/2 for every k
>  ≥ N, ensuring that the quotient fk′(z)/fk(z) is well defined for all z
>  on the circle |z−z0| = ρ. By Morera's theorem we have a uniform
>  convergence:
> 
> 
> $$\frac{f\_{k}'(z)}{f\_{k}(z)} \to \frac{f'(z)}{f(z)}.$$ ..........
> 
> 
> 

But I don't know why it is a uniform convergence. (I think that Morera's theorem is not relevant.) Why does it hold?

Answer: 
I do not have any experience in this, but it seems like you can do this with a diagonal argument. For $P\_{1}$, define $(a,b) := a + bx$ then list the coordinates 

\begin{align\*}
(0,0), &(0,1), (0,2), (0,3)\dots\\
(1,0), &(1,1), (1,2), (1,3) \dots\\
\vdots
\end{align\*} 

This is countable by defining your bijective map down the diagonals. You can then simply reapply this to argument to $(a,b)\times(\mathbb{N}\cup\{0\})$ repeatedly to show it is true for all $P\_{n}$
With the mapping you suggest the polynomials $101 + x$ and $1 + 101x$ would both be mapped to 101010101010 so it won't quite work.

You can find simpler maps based on the same idea though: since digits 0-9 are already being used in the decimal expansion, why not map the polynomials to numbers in base 12 and let "," be the digit with value 10 and "-" the digit with value 11? Now any sequence of decimal digits with embedded commas and hyphens are integers in base 12; for example the two polynomials above would be represented by the base 12 numbers $101,1$ and $1,101$ respectively.