Proving the family of constructible sets is a boolean algebra.

Question

A constructible set is a finite union of locally closed sets. A locally closed set is the intersection of an open set and a closed set. I need to show that this is closed under finite unions and complementation to show it is a boolean algebra. The former is trivial to show but I am stuck on the latter. The complement of a union of locally closed sets is the intersection of sets which are a union of a closed set and an open set. I need to show that this is a union of locally closed sets, but I cannot see how this is true.

Answer 1

Indeed, we can write the complement of a finite union of locally closed sets as a finite intersection of sets which are a union of open and closed sets. Consider a single such union: $U \bigcup F$ where $U$ is open and $F$ closed. Suppose we are working with topological space $(X,\tau ) $. Then $U=U\bigcap X$ is locally closed since $U$ is open and $X$ is closed. Equally, $F=F \bigcap X$ is locally closed since $F$ is closed and $X$ open. Hence $U\bigcup F$ is a finite union of locally closed sets: a constructible set.

We now claim that the intersection of two constructible sets is constructible. Suppose $A$ and $B$ are constructible, where $A=\bigcup ^{n} _{i=1} C_i$, and $B = \bigcup ^{m} _{j=1} D_j$, for locally closed $C_i$ and $D_j$. Then $A\bigcap B = \bigcup _{1\le i \le n, 1\le j \le m}(C_i \bigcap D_j)$, the elements of $X$ which are in some $C_i$ and some $D_j$.

Next, $C_i$ and $D_j$ are locally closed so let $C_i=U_i\bigcap F_i $ and $D_j=V_j\bigcap G_j $ where $U_i$ and $V_j$ open, $F_i$ and $G_j$ closed. Then $C_i\bigcap D_j = (U_i\bigcap F_i)\bigcap (V_j \bigcap G_j)=(U_i\bigcap V_j)\bigcap (F_i \bigcap G_j)$, where $U_i\bigcap V_j$ is open and $F_i \bigcap G_j$ is closed, so $C_i\bigcap D_j$ is locally closed.

Therefore $A\bigcap B$ is a finite union of locally closed sets: it's constructible.

Hence, by induction, a finite intersection of contructible sets is constructible.

Hope this helps!