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Consider the polynomial with (variable) integer coefficients : $ a_0x^n+a_1x^{n-1}+a_2x^{n-2}+...+a_{n-1}x+a_n$, where $x$ varies over the field of complex numbers. The set of all zeros of the polynomial gives the set of all algebraic numbers.

Now, the number of roots of a polynomial of $n$ th degree has at most $n$ roots. Again, [from the hint of Rudin], we can write only in a finite number of ways $n +|a_0| +|a_1|+...|a_n|= N \in \mathbb{N} $, i.e, the combination of $a_i$ 's summing up to $N$ is a finite set; the permutation of those $a_i$ 's also form a countable set. Now, construct the set

$S_n^N =\{ x,\ a_i,\ n, \ N : a_0x^n+a_1x^{n-1}+a_2x^{n-2}+...+a_{n-1}x+a_n = 0$ and $n +|a_0| +|a_1|+...+|a_n|= N\}$, which is evidently finite from the aforementioned arguments.

Now, $\bigcup \limits_{n=1}^{\infty} S_n^N $ is at most countable, being a collection of countable sets. As we further do the union $\bigcup \limits_{N=1}^{\infty}[ \bigcup \limits_{n=1}^{\infty} S_n^N ] $ we get again a countable set, which equals the set of all algebraic numbers.

Is this legit? [ although not fully original ]

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Straight forward proof. Let k = #N.
P(n) is set of polynomials of degree n with interger coefficients.
Whereupon #P(0) = k. #P(1) = k × k = k, etc.
By induction, #P(n) = k.
So the number of roots, including multiple roots,
of P(n) = n × k = k.
Thus the number of roots of $\cup_n$P(n) = k × k = k.

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  • $\begingroup$ Nice. Is mine alright? $\endgroup$ – Subhasis Biswas Oct 30 '18 at 9:55
  • $\begingroup$ @SubhasisBiswas. I don't know. It was so complicated, it was easier to give another proof than to plod through the details of yoursv. $\endgroup$ – William Elliot Oct 30 '18 at 10:16

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