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The expression is this:

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

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.2012-06-16

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

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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    @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 $\sqrt{2}$.2012-06-16