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Let $X$ be a topological space (or more specifically, $\mathbb{C}^N$ endowed with the Zariski topology), and let $S \subseteq X$ be a constructible set, i.e. $S=\cup_{i=1}^n C_i \cap U_i$, where the $C_i$'s are closed in $X$ and the $U_i$'s are open. Then $S$ contains an open dense subset of its closure $\overline{S}$ (this is true even if $\overline{S}$ is not irreducible-- see Lemma 2.1 in this paper). I would like a list of some known conditions that imply $S$ is itself an open dense subset of its closure (also known as locally closed).

As one example, if there is a transitive topological group action on $S$, then $S$ is locally closed.

One might hope that if $S$ is constructible, irreducible, and connected, then it is locally closed. But this is contradicted by the image of the map on $\mathbb{C}^2$ given by $(x,y)\mapsto(x,xy)$, which satisfies all of these properties, but is not locally closed.

One personal motivation for this question is I am trying to prove that the set of tensors in $\mathbb{C}^{N_1}\otimes\dots\otimes \mathbb{C}^{N_d}$ of tensor rank at most $r$ is locally closed (i.e., an open dense subset of its closure, the set of tensors of border rank at most $r$). I have seen this fact mentioned several times, but have not been able to prove it.

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  • $\begingroup$ When you say "𝑆 contains an open dense subset" then by "open" you must mean "open with respect to the closure of S". $\endgroup$
    – Wlod AA
    Dec 27, 2020 at 6:21
  • $\begingroup$ This looks like a fiber of an upper (or lower) semicontinuous function with values in $\mathbf{Z}$ (the rank) and so it should be locally closed: $\{r=n\} = \{r>=n\} \setminus \{r>n\}$. $\endgroup$ Dec 27, 2020 at 6:45
  • $\begingroup$ @WlodAA Yes, this is what I meant by “S contains an open dense subset of its closure” $\endgroup$
    – Ben
    Dec 27, 2020 at 13:41
  • $\begingroup$ @PiotrAchinger Indeed, the set of two-way tensors (matrices) of rank $r$ is locally closed, but what about multi-way tensors? In contrast to the two-way case, the set of multi-way tensors of tensor rank at most $r$ is not necessarily closed (the closure of this set is known as the set of tensors of border rank at most $r$, or equivalently it is the $r$-th secant to the Segre variety). The set of tensors of tensor rank at most $r$ is easily shown to be constructible, though. I want to prove that it is locally closed. $\endgroup$
    – Ben
    Dec 27, 2020 at 13:55
  • $\begingroup$ Have you tried to check this for $\mathbb{C}^2 \otimes \mathbb{C}^2 \otimes \mathbb{C}^2$? It might already be a counterexample. $\endgroup$ Apr 18, 2021 at 8:45

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