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.