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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 http://math.stackexchange.com/q/45667/20998?2012-02-23
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    Are you working over an algebraically closed field?2012-02-23
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    @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) \}$.2012-02-23
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    @Bruno: Thanks very much for reminding me of this. I forgot to write in the question that the base field was algebraically closed.2012-02-23
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    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,2012-02-23
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    Dear Matt E: Thanks very much. I thought that in a wrong way... Regards,2012-02-27

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