Does the class of all finite unions of closed-open intervals on $\mathbb{R}$ form a ring sets? Does the class of all finite  unions of closed-open intervals on $\mathbb{R}$ form a ring on sets?
By a closed-open interval , I mean  an interval of the form $[x,y)$
A ring of sets is a non-empty class of sets that is closed under symmetric difference of any pair of sets of the class and under intersection of any pair of sets of the class. 
For me, This is untrue. Since $[0,1)$ and $[2,3)$ are in the class but their intersection is not in the class ( since the empty set is not a closed-open interval, is it? )
Am I right? I think that if we added the empty set to the class then the new class forms a boolean algebra of sets.
 A: $\emptyset=\cup\{[a_i,b_i)|i\in\emptyset\}$ shows that $\emptyset$ is a finite union of closed-open intervals on $\mathbb R$, since the empty set serving as index-set is a finite set.

edit (to make things less fuzzy)
In general if $\mathcal C$ is some set of sets (above the set that contains all half-open intervals in $\mathbb R$) then the class that consists of all (finite) unions of sets in $\mathcal C$ contains the empty set as element. These unions are exactly the sets that can be written as $\cup\mathcal V$ where $\mathcal V$ is a (finite) subset of $\mathcal C$. 
Here $a\in\cup\mathcal V\iff a\in V\text{ for some }V\in\mathcal V$. 
One of these subsets $\mathcal V$ is the empty set and clearly $\cup\emptyset=\emptyset$.
A: This follows from the argument in the book (Topology and Modern Analysis) on page 12 talking about finite intersections and unions of classes of sets. The class of all finite unions of closed-open intervals is closed under unions. Since the empty class is necessarily a finite subclass its union the empty set belongs to the class.
