The question is to prove that the set of all algebraic numbers is countable. He gives the hint that for every positive integer N there are only finitely many equations with n + $\vert a_0 \vert$ + $\vert a_1 \vert$ + ... + $\vert a_n \vert$ = N. I looked at the solution given at https://minds.wisconsin.edu/bitstream/handle/1793/67009/rudin%20ch%202.pdf?sequence=10&isAllowed=y but I'm having trouble understanding it. They let $A_n$ be the set of numbers satisfying the equation above, and then say that the set of algebraic numbers is the union of $A_n$ from 2 to oo. But that's the part I'm having trouble with. How is the set of algebraic numbers equal to this union of the $A_n$? And why do they start at 2 rather than 1?
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https://math.stackexchange.com/a/1127103/469000 – Unknown Jul 01 '20 at 17:00
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@AmanPandey Yeah I saw that when I was searching prior to posting this question. But they basically just addressed my question by saying it was "fairly obvious." – lightnesscaster Jul 01 '20 at 17:02
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$A_N$ (not $A_n$) is not the set of numbers satisfying the equation $n+\vert a_0\vert+\vert a_1\vert+\cdots+\vert a_n \vert=N$. It is the set of numbers satisfying a polynomial equation $\sum_{i=1}^n a_ix^i=0$ for some coefficients $a_0,\ldots,a_n$ which satisfy the equation $n+\vert a_0\vert+\vert a_1\vert+\cdots+\vert a_n \vert=N$.
And such a polynomial equation has at most $n$ solutions, so each $A_N$ is finite.
TonyK
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Oh wow, thanks! I was totally confused by which "equations" he was talking about. – lightnesscaster Jul 01 '20 at 17:11