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Let $X$ be a topological space and let $\mathscr{F}$ be the smallest collection of subsets of $X$ which contains all open subsets of $X$ and is closed with respect to the formation of finite intersections and complements.

I solved Ex7.20(i), which says that a subset $E$ of $X$ belongs to $\mathscr{F}$ iff $E$ is a finite union of sets of the form $U \cap C$, where $U$ is open and $C$ is closed.

  1. What is the name of $\mathscr{F}$? I found that if $X$ is a notherian topological space, it is called Constructible sets. Is this name used for any $X$, or is there no special name for the case of non-noetherian space $X$?

  2. Is it true that any finite union $\cup_{i=1}^n (U_i \cap C_i)$(where $U_i$ is open and $C_i$ is closed) can be presented as $U \cap C$? One of the solution of the problem states this, but I doubt it is true. I tried to find a counterexample in $\mathbb{R}^2$ by sketching some sets, but I couldn't find a one.

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2 Answers 2

up vote 10 down vote accepted

Regarding (1), my memory is that the terminology and definitions for constructible sets in the non-Noetherian context is a little delicate. This is confirmed by a quick google search, which turned up, for example, this discussion on Akhil Mathew's blog.

Regarding (2), a set of the form $U\cap C$ is called locally closed (because each of its points has a n.h. in the ambient space whose interesection with the set is closed in that n.h.).

Not every constructible set is locally closed. E.g. consider $\mathbb R^2$ minus a line union a point on the missing line. In coordinates (taking the line to be the $x$-axis and the point to be the origin) this is the set of $(x,y) \in \mathbb R^2$ such that either $x \neq 0$ or else $x = y = 0$.

This is constructible, but is not locally closed: its intersection with any n.h. of the origin is not closed in that n.h. If you found a statement to the contrary in some solution set, then I wouldn't trust that solution set.

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The following is an additional example that shows that $$(U_1 \cap C_1) \cup(U_2 \cap C_2)$$ where the $U_i$ are open, and the $C_i$ are closed, need not be expressible in the form $U \cap C$.

Let $S$ be the set of all real numbers $x$ which are not of the form $1/n$, where $n$ ranges over the positive integers.

Note that $S$ is the union of an open set and the one point set $\{0\}$, so it certainly has the shape $(U_1 \cap C_1) \cup(U_2 \cap C_2)$, with $C_1=U_2=\mathbb{R}$.

We show that $S$ is not of the form $U\cap C$.

Suppose to the contrary that it is. Since $0 \in S$, we have $W \subset U$ for some neighborhood $W$ of $0$. Then for any large enough $n$, $1/n \in W$.

Suppose that $1/k$ and $1/(k+1)$ are in $W$. Since the interval $$\left(\frac{1}{k+1}, \frac{1}{k}\right)$$ is a subset of $S$, it follows that this interval is a subset of $C$. But since $C$ is closed, it follows that $1/k \in C$, contradicting the fact that $1/k\notin S$.

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