TITLE: Atiyah-Macdonald Ex 7.20 Constructible sets
QUESTION [8 upvotes]: 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.

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 $X$?
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.

REPLY [2 votes]: 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 $1/k\notin S$.