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 $\mathbb{R}^n$?
Zariski Open Set
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algebraic-geometry
polynomials
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0you have to precise the topology used in $\mathbb{R}^n$. – 2012-04-09
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0The topology on $\mathbb{R}^n$ is the usual topology. – 2012-04-09
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0if $U$ is open for the Zariski topology in $\mathbb{C}^n$, can we conclude that $U$ is open for the usual topology in $\mathbb{C}^n$ ? – 2012-04-09
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0Yes. 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. – 2012-04-09
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0if $U$ a Zariski open set, is $U$ bounded ? – 2012-04-09
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1@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