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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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    Dear Matt E: Thanks very much. I thought that in a wrong way... Regards,2012-02-27

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