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The expression is this:

$$\bigcup_{n\in\mathbb{N}}\ \bigcup_{a_0\in\mathbb{Z}}\cdots\bigcup_{a_n\in\mathbb{Z}}\big\{z\in\mathbb{C}:a_0z^n+a_1z^{n-1}+\cdots+a_{n-1}z+a_0=0\big\}.$$

I hope it's clear what this is meant to denote (the set of algebraic numbers), but I'm uneasy about it since the number of unions depends on an element in the first union.

I suspect that the set could be rewritten more concretely as a union indexed by $\mathbb{Z}$-valued sequences that eventually terminate, but I'm not sure how that would look.

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This is fine. There is no abuse of notation. Sure, there are other ways of writing the set, but the expression you give is meaningful. You may want to specify that $a_0\ne0$, to avoid by accident allowing all complex numbers in the union. –  Andres Caicedo Jun 16 '12 at 0:25

1 Answer 1

up vote 7 down vote accepted

$$\bigcup_{n\in\Bbb N}\bigcup_{\langle a_0,\dots,a_n\rangle\in\Bbb Z^{n+1}}\left\{z\in\Bbb C:\sum_{k=0}^na_kz^{n-k}=0\right\}$$

Added: I’d overlooked the fact that what was wanted was actually the set of algebraic numbers, which isn’t quite what’s described here or in the question. As Erick Wong points out in the comments, $\langle 0,\dots,0\rangle$ should be removed from the index set $\Bbb Z^{n+1}$, to avoid including $\{z\in\Bbb C:0=0\}=\Bbb C$.

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Thanks. I feel a little silly for not having realized that. –  youler Jun 16 '12 at 0:28
    
However, notice that this is not the set of algebraic numbers. We should remove $\langle 0,\ldots,0 \rangle$ from $\mathbb Z^{n+1}$. –  Erick Wong Jun 16 '12 at 6:40
    
You can just add that $\prod a_k\neq 0$ into the requirements about the polynomial. –  Asaf Karagila Jun 16 '12 at 6:55
    
@Asaf: Yes, that’s certainly one efficient way to do it. –  Brian M. Scott Jun 16 '12 at 6:56
    
@AsafKaragila Thinking about it, I guess $a_0 \ne 0$ is already good enough, and still allows us keep $x^2 - 2$ as a representative of $\sqrt{2}$. –  Erick Wong Jun 16 '12 at 7:30

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