# When is a constructible set locally closed?

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.

• When you say "𝑆 contains an open dense subset" then by "open" you must mean "open with respect to the closure of S". Dec 27, 2020 at 6:21
• 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: $\{r=n\} = \{r>=n\} \setminus \{r>n\}$. Dec 27, 2020 at 6:45
• @WlodAA Yes, this is what I meant by “S contains an open dense subset of its closure”
– Ben
Dec 27, 2020 at 13:41
• @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.
– Ben
Dec 27, 2020 at 13:55
• Have you tried to check this for $\mathbb{C}^2 \otimes \mathbb{C}^2 \otimes \mathbb{C}^2$? It might already be a counterexample. Apr 18, 2021 at 8:45