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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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you have to precise the topology used in $\mathbb{R}^n$. – Abdelmajid Khadari Apr 9 '12 at 16:19
The topology on $\mathbb{R}^n$ is the usual topology. – Suresh Apr 9 '12 at 16:26
if $U$ is open for the Zariski topology in $\mathbb{C}^n$, can we conclude that $U$ is open for the usual topology in $\mathbb{C}^n$ ? – Abdelmajid Khadari Apr 9 '12 at 16:33
Yes. It is not difficult to see that a Zariski closed is closed in usual topology. Hence a Zariski open set is open in usual topology. – Suresh Apr 9 '12 at 16:35
@Matrix: No. Consider $U_x = \mathbb{R}^2 \backslash Z(x) $, the complement of a line in $\mathbb{R}^2$. Not bounded. – Fredrik Meyer Apr 9 '12 at 17:12

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 writing could be better, but I think the ideas are correct. – Cantlog Oct 8 '13 at 20:04

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