In the same way that we have defined the union and the intersection of two intervals, we can define this two operations on a family or collection of intervals.
If we have a collection of $$n$$ intervals, and being $$n$$ any natural number, we define the union and the intersection of the family in a recursive way.
Namely, given a finite family of intervals, $$\{J_1,J_2,\ldots,J_n\}$$ we define their union as: $$$\bigcup_{k\leq n}J_k = \Big( \bigcup_{k\leq n-1} J_k \Big)\cup J_n$$$
And similarly, we define the intersection as: $$$\bigcap_{k\leq n}J_k = \Big( \bigcap_{k\leq n-1} J_k \Big)\cap J_n$$$
Using an example, we will see more clearly:
Suppose we have a family of intervals: $$$J_1=\Big(-\dfrac{1}{2},\dfrac{1}{2}\Big),J_2=\Big(-\dfrac{1}{3},\dfrac{1}{3}\Big), J_3=\Big(\dfrac{2}{5},\dfrac{8}{3}\Big) \ \mbox{and}$$$ $$$\ J_4=\Big[-\dfrac{2}{7},\dfrac{1}{8}\Big]$$$ And we want to calculate the union and intersection.
Let's start with the union: According to the definition, to calculate the union of four intervals it is necessary to know what is the union of three. And to calculate this union, we need the union of the first two. This process is what is known as recursion.
We begin with the union as the first two:
$$\Big(-\dfrac{1}{2},\dfrac{1}{2}\Big)\cup\Big(-\dfrac{1}{3},\dfrac{1}{3}\Big)=\Big(-\dfrac{1}{2},\dfrac{1}{2}\Big)$$ since $$\Big(-\dfrac{1}{3},\dfrac{1}{3}\Big) \subseteq \Big(-\dfrac{1}{2},\dfrac{1}{2}\Big).$$
Below is the union of three intervals:
$$$J_1 \cup J_2\cup J_3=(J_1 \cup J_2)\cup J_3=\Big(-\dfrac{1}{2},\dfrac{1}{2}\Big)\cup \Big(\dfrac{2}{5},\dfrac{8}{3}\Big) =$$$ $$$= \Big(-\dfrac{1}{2},\dfrac{8}{3}\Big)$$$
Finally, we only need to do the union with the fourth interval:
$$$\bigcup_{n\leq 4} J_n = J_1\cup J_2\cup J_3\cup J_4=\Big(-\dfrac{1}{2},\dfrac{8}{3}\Big)\cup\Big[-\dfrac{2}{7},\dfrac{1}{8}\Big]=$$$ $$$=\Big(-\dfrac{1}{2},\dfrac{8}{3}\Big)$$$
To do the intersection we proceed in the same way. First we intersect the first two. Since the second is included in the first one, the intersection is the second interval:
$$J_1 \cap J_2=\Big(-\dfrac{1}{3},\dfrac{1}{3}\Big)$$
Below is the intersection with the third interval:
$$(J_1 \cap J_2)\cap J_3 = \Big(-\dfrac{1}{3},\dfrac{1}{3}\Big) \cap \Big(\dfrac{2}{5},\dfrac{8}{3}\Big)=\emptyset$$
Since $$\dfrac{1}{3} < \dfrac{2}{5}$$. Finally, as the intersection with the vacuum is always empty, we have:
$$\bigcap_{n\leq 4} J_n = J_1\cap J_2\cap J_3\cap J_4=\emptyset$$
We can also make unions and intersections of not finite families.
This time the problem that arises is that we can not define it recursively, but we adapt the definition looking for where the succession of finite unions or intersections tends.
To do it, we look at where the extremes of the intervals tend and we choose the supremum or the infimum in the sequence in each case:
For the union,
- To find the lower extreme we must find the infimum in the sequence of the lower extremes.
- To find the upper extreme we must find the supremum in the sequence of the upper extremes.
And for the intersection,
-
To find the lower extreme we must find the supremum in the sequence of the lower extremes.
- To find the upper extreme we must find the infimum in the sequence of the upper extremes.
To see it clearly, let's see an example:
Consider the family of intervals as: $$$\Big(-\dfrac{1}{n},\dfrac{1}{n}\Big)$$$ and we want to study the union and the intersection of all, taking different natural values of $$n$$. We denote it as:
$$$\bigcup_{n\geq 1}\Big(-\dfrac{1}{n},\dfrac{1}{n}\Big) \ \mbox{and} \ \bigcap_{n\geq 1}\Big(-\dfrac{1}{n},\dfrac{1}{n}\Big)$$$
Then, if we consider the sequence of lower extremes: $$\Big\{-\dfrac{1}{n}\Big\}_n$$ we see that it is a strictly increasing sequence with limit $$0$$.
Also, for the sequence of upper extremes: $$\Big\{\dfrac{1}{n}\Big\}_n$$ it is a strictly decreasing sequence with limit $$0$$.
Then, our sequence of intervals $$$\Big\{\Big(-\dfrac{1}{n},\dfrac{1}{n}\Big)\Big\}_n$$$ is constructing smaller intervals in every step (the lower extreme increase while the upper one decrease) and it tends to an interval with extremes $$0$$: $$$(0,0)=\emptyset$$$
Having studied the two sequences, we construct the union of the family of intervals finding the infimum and the supremum of $$\Big\{-\dfrac{1}{n}\Big\}_n$$ and $$\Big\{\dfrac{1}{n}\Big\}_n$$ respectively, which in two cases is the first term $$$a_1=-\dfrac{1}{1}=-1 \ \mbox{and} \ b_1=\dfrac{1}{1}=1$$$ because of the increasing behaviour of the first sequence and the decreasing of the second one. Summarizing, we have: $$$\bigcup_{n\geq 1}\Big(-\dfrac{1}{n},\dfrac{1}{n}\Big)=(-1,1)$$$
Similarly, to calculate the intersection, we find the supremum and the infimum of the two sequences, which in this case is equal to find its limits because of the increasing behaviour of the first sequence and the decreasing of the second one. So, we obtain: $$$\bigcap_{n\geq 1}\Big(-\dfrac{1}{n},\dfrac{1}{n}\Big)=(0,0)=\emptyset$$$