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Consider the Zariski Topology on $\mathbb{C}^n.$ Then is it true that for every non-empty Zariski open set $U,$ $U \cap \mathbb{R}^n$ is open dense in $\mathbb{R}^n$?

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    @Matrix: No. Consider $U_x = \mathbb{R}^2 \backslash Z(x) $, the complement of a line in $\mathbb{R}^2$. Not bounded.2012-04-09

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Well I believe this should be the way to go ahead. Suppose $V:=U \cap \mathbb{R}^n$ is not open dense in $\mathbb{R}^n,$ then there is an open set $\mathcal{O}\subset\mathbb{R}^n$ such that $V\cap \mathcal{O} =\emptyset.$ In particular, $\mathcal{O} \subset U^c. $

Let $U^c = \bigcap_{m=1} ^ M \{f_m = 0\}.$ Consider an arbitrary $m$ and let $f_m= \sum_{a}c_ax^a = \sum_a \{Re(c_a)x^a + i \times Im(c_a)c^ax^a\}.$

The fact that $f_m = 0$ on $\mathcal{O}$ now essentially shows that $c_a = 0$ $ \forall a.$ This implies $f_m$ is the identically zero polynomial. Since $f$ was arbitrary, this means $U^c = C^n,$ contradicting $U$ is non-empty.

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    The writi$n$g could be better, but I think the ideas are correct.2013-10-08