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Basically this question has many solutions in MSE for example proof 1, proof 2 etc. I have also tried to prove it and wanted to checked. It is as follows:

  1. My first claim is there are only countably many polynomials in $\mathbb{Z}[x]$, i.e the set $\mathbb{Z}[x]$ is countable. Which is probably easy to see as we can see it as $\mathbb{Z}^n$.
  2. Since each polynomials has at most finite roots so if we consider all the roots in a set that will be nothing but countable union of finite sets, which is countable.

Hence, set of algebraic numbers are countable.

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    $\begingroup$ We can see $\mathbb{Z}[X]$ as $\mathbb{Z}^n$ needs to be thought again. What is $n$? $\endgroup$ – Gribouillis Aug 12 '17 at 7:06
  • $\begingroup$ easy to see as we can see it as Z^n Not obvious what you mean by that (what's $n$?). See however Prove that the set of integer coefficients polynomials is countable. Then step 2 is the easy part. $\endgroup$ – dxiv Aug 12 '17 at 7:07
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    $\begingroup$ You can view $\mathbb{Z}[X]$ as $\bigcup_{n\geq 0}\mathbb{Z}^n$, but not as you claim. $\endgroup$ – Mathematician 42 Aug 12 '17 at 7:08
  • $\begingroup$ Okay thanks, now I got it. $n$ to be chosen. It's not obvious. $\endgroup$ – Sachchidanand Prasad Aug 12 '17 at 7:11
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    $\begingroup$ Essentially right - but the set of polynomials is the union of the set of polynomials of each degree. Each of those sets is countable and a countable union of countable sets is countable. $\endgroup$ – Ethan Bolker Aug 12 '17 at 14:03

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