Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
Have you seen this post – 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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.