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A subset of a variety is locally closed if it is the intersection of a closed subset with an open subset; it is constructible if it is a finite union of locally closed subsets.

Suppose that the base field is algebraically closed.

Exhibit a subset of $\mathbb A^2$ which is constructible, but not locally closed.

Would you please show me the way of finding such a subset and proving that it satisfies the condition?

Thanks a lot.

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Have you seen this post math.stackexchange.com/q/45667/20998? –  azarel Feb 23 '12 at 7:41
Are you working over an algebraically closed field? –  Bruno Joyal Feb 23 '12 at 7:52
@azarel: Thanks very much for the comment. But the question you referred to considers about the common topology, while I am thinking in Zariski topology. For example, the example given by Matt E in that question is the union of the open subset $\mathbb{A}^2 - \{ (x,y) | x =0 \} $ with the closed subset $ \{ (0,0) \}$. –  ShinyaSakai Feb 23 '12 at 13:16
@Bruno: Thanks very much for reminding me of this. I forgot to write in the question that the base field was algebraically closed. –  ShinyaSakai Feb 23 '12 at 13:19
Dear ShinyaSakai, My answer that @azarel links works either in the Zariski topology (which is what I had in mind when I wrote it) or the usual topology if the ground field is the complex numbers. In particular, it answers your question. Regards, –  Matt E Feb 23 '12 at 13:30
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