Constructible sets contain a dense open subset of their closure

[Hello Kitty]

I recently came across the following remark in a book: "Notice that a constructible set contains a dense open subset of its closure." Now this doesn't seem at all obvious to me.

Let us recall the definitions first. A locally closed set is the intersection of a closed and an open set. A constructible set is a finite union of locally closed sets.

Let H denote our constructible set in a topological space X. Let H^c, H^{\circ}, H' denote closure, interior and complement respectively.

Now H^{\circ} \subseteq H \subseteq H^c is open so this seems like an obvious choice. We just need to show that it is dense in the latter. Well any set is dense inside its closure so it would be sufficient to show that (H^{\circ})^c = H^c.

Note that LHS \subseteq RHS is obvious.

The following facts follow from the definitions:

1) The closure of a union is the union of the closures.
2) The closure of an intersection is a subset of the intersection of the closures.
3) The interior of an intersection is the intersection of the interiors.
4) The interior of a union is a subset of the union of the interiors.

So let H = (O_1 \cap C_1) \cup \dots \cup (O_n \cap C_n) for C_i, O_i closed and open respectively.

H^c = (O_1 \cap C_1)^c \cup \dots \cup (O_n \cap C_n)^c \subseteq (O_1^c \cap C_1) \cup \dots \cup (O_n^c \cap C_n)\ \ \ (*)

On the other hand

H^{\circ} \supseteq (O_1 \cap C_1)^{\circ} \cup \dots \cup (O_n \cap C_n)^{\circ} = (O_1 \cap C_1^{\circ}) \cup \dots \cup (O_n \cap C_n^{\circ})

hence

(H^{\circ})^c \supseteq (O_1 \cap C_1^{\circ})^c \cup \dots \cup (O_n \cap C_n^{\circ})^c.

If this were to equal (*), we'd be done. A sufficient condition would seem to be that

(U_1 \cap U_2)^c = U_1^c \cap U_2^c for any U_1, U_2 open.

I've tried this but I can't quite nail it.

Am I on the right track or am I making life difficult for myself? Thanks in advance!

 

[morphism]

It's false. It's not even true for locally closed sets; counterexample: \{0\} = \{0\} \cap \mathbb{R} with the usual topology.

 

[Hello Kitty]

You mean to say that the sufficient condition I proposed is false, not that the original claim is false. (If our constructible set is closed then that set is dense and open in itself.)

So any suggestion on how to prove the original claim?

 

[adriank]

morphism is saying that the original claim is false, and here's why: In \mathbb{R} with the usual topology, \{0\} = \{0\} \cap \mathbb{R} is constructible. Its closure is {0}, and the only open set it contains is the empty set, which is clearly not dense in {0}.

Does the book qualify that statement more? (For example, is it only considering certain kinds of constructible sets?)Additionally, your sufficient condition is false. Consider the open sets (0, 1) and (1, 2) in \mathbb{R}; the closure of their intersection is empty, but the intersection of their closures is {1}.

 

[mathwonk]

adriank, you are mistaken. the set {0} is open in itself hence is an open subset of its closure. your mistake is in not understanding the use here of the relative topology. this fact is true and is contained in an exercise (3.18?) on page 94 of hartshorne's algebraic geometry.

 

[adriank]

I understand the difference. morphism was going by the assumption that the open subset of H^c the statement refers to is open in the topology of X, rather than the subspace topology of H^c, and I was clarifying it.

One more thing: fact 4 which Hello Kitty claims follows from the definitions is false: the interior of the union of [0, 1] and [1, 2] is (0, 2), while the union of their interiors is the union of (0, 1) and (1, 2); the former is not a subset of the latter. I believe it is true that the union of the interiors is a subset of the interior of the union.

 

[mathwonk]

the last assertion in your last sentence seems obviously true.

so your contention is that if P is a false statement uttered by X, then your statement that "X's point was the following: P is true", is a true statement? Maybe, but it is confusing.

Or maybe you are contending that even if your statement is also false that you do understand that it is, without having said so?