Question: Let $X$ be an affine variety over $\Bbb C$, and let $Y\subseteq X$ be a constructible set (i.e. $Y$ is a finite union of locally closed sets). Is it true that the Zariski closure of $Y$ is the same as the closure of $Y$ in the standard Euclidean topology inherited from the inclusion $X\subseteq\Bbb C^n$?

In this question I asked whether these two closures were the same when $Y$ is the orbit of an algebraic group action. Then, this answer says that the answer is yes because orbits are constructible sets. However, I don't know a proof of the fact that these orbit closures are the same for constructible sets.

If $\bar{Y}^E$ denotes the euclidean closure and $\bar{Y}^Z$ the Zariski one, then it is clear that $$\bar{Y}^E\subseteq \bar{Y}^Z$$ since the Euclidean topology is finer than the Zariski topology. But for the converse, I am clueless.


1 Answer 1


Yes it is true. You probably know that a non-empty Zariski-open set $U\subseteq X$ is Zariski-dense, and it is a standard fact that $U$ is also Euclidean-dense. (See this mathoverflow answer for a slick proof of that fact.) Assuming this, the proof of your claim is simple:

Wlog, $Y$ is locally closed because the closure of a finite union is the union of the closures. Then, by definition, $Y$ is Zariski-open in $\bar{Y}^Z$ and hence $Y$ is Euclidean-dense in $\bar{Y}^Z$, i.e. $\bar{Y}^E\cap\bar{Y}^Z=\bar{Y}^Z$. Since $\bar{Y}^E\subseteq\bar{Y}^Z$, this concludes the proof.

  • $\begingroup$ Does the second part of your argument not need that $ \bar{Y}^Z$ is irreducible ? Because i do not see why that would be true $\endgroup$
    – Backfisch
    Oct 12, 2022 at 16:11
  • $\begingroup$ Never mind, i just realized that in such a case we could decompose $\bar{Y}^Z$ into its irreducible components and thus $Y$ into smaller locally closed sets whose closures are then said components. $\endgroup$
    – Backfisch
    Oct 12, 2022 at 16:19

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