Union, intersection and complementary of intervals

Union of intervals

Given two any real intervals, its union is a set that consists of all the elements that belong to the first interval and all the elements that belong to the second one.

The union of the intervals $$(a,b)$$ and $$(c,d)$$ is denoted as $$(a,b)\cup (c,d)$$ and is calculated this way:

$$$(a,b)\cup (c,d) = \{ x\in\mathbb{R} \ | \ x\in(a,b) \ \mbox{or} \ x\in(c,d)\}=$$$ $$$=\{ x\in\mathbb{R} \ | \ a < x < b \ \mbox{or} \ c < x < d\}$$$

Depending on the order in which numbers $$a, b, c$$ and $$d$$ are, the result will be different. Being $$(a,b)$$ and $$(c,d)$$ two intervals, we have $$a < b$$ and $$c < d$$, but the relative position of the endpoints of an interval may change regarding the endpoints of the other one. Thus, we find the cases, as follows:

• if $$a < b < c < d$$ then the union $$(a,b) \cup (c,d)$$ results in the set formed by two intervals: $$$(a,b) \cup (c,d) = \{ x\in\mathbb{R} \ | \ x\in(a,b) \ \mbox{or} \ x\in(c,d) \} $$$ The result is the same if $$c < d < a < b.$$

• if $$a < c < d < b$$, then the interval $$(c,d)$$ is included in $$(a,b)$$, so, $$$(a,b) \cup (c,d) = \{ x\in\mathbb{R} \ | \ a < x < b, \ \mbox{or} \ c < x < d\} = $$$ $$$= \{ x\in\mathbb{R} \ | \ a < x < b\}=$$$ $$$=(a,b)$$$

Similarly, if $$c < a < b < d$$, we obtain $$(a,b) \cup (c,d)=(c,d)$$. That is, if an interval is included into another one, the union of the two is equal to the greater one.

• if $$c < a < d < b$$, then we have $$$(a,b) \cup (c,d) = \{ x\in\mathbb{R} \ | \ a < x < b, \ \mbox{or} \ c < x < d\}$$$ But as $$c < a$$ and $$d < b$$, we have two intervals that overlap so we have a unique interval: $$$(a,b) \cup (c,d) = \{ x\in\mathbb{R} \ | \ c < x < b\} = (c,b)$$$

In the same way, if $$a < c < b < d$$ we obtain: $$$(a,b) \cup (c,d) = \{ x\in\mathbb{R} \ | \ a < x < d\} = (a,d)$$$

Note now that the union of intervals is not always an interval. Furthermore, in the case of open intervals, either closed or mixed, the result is analogous:

In this case the union of two intervals has given us an interval.

Intersection of intervals

Given any two real intervals, their intersection is the set of all elements that belong to both intervals.

The intersection of intervals $$(a,b)$$ and $$(c,d)$$ is denoted as $$$(a,b)\cap(c,d)$$$ and is calculated as: $$$(a,b)\cap (c,d) = \{ x\in\mathbb{R} \ | \ x\in(a,b) \ \mbox{and} \ x\in(c,d)\}=$$$ $$$=\{ x\in\mathbb{R} \ | \ a < x < b \ \mbox{and} \ c < x < d\}$$$

Depending on the order in which the numbers $$a, b, c$$ and $$d$$ are, the result will be one or another. As in the union, we have that $$a < b$$ and $$c < d$$, but the relative position of the endpoints may change compared to the extremes of the other interval. Thus, we find the cases, as follows:

• if $$a < c < d < b$$, we have that the interval $$(c,d)$$ is included in $$(a,b)$$, then, $$$(a,b) \cap (c,d) = \{ x\in\mathbb{R} \ | \ a < x < b, \ \mbox{and} \ c < x < d\} = $$$ $$$= \{ x\in\mathbb{R} \ | \ a < c < x < d < b\}=$$$ $$$=(c,d)$$$

Similarly, if $$c < a < b < d$$, we obtain $$(a,b) \cap (c,d)=(a,b)$$. Namely, if an interval is included into another, the intersection of the two is equal to the lower one.

• if $$c < a < d < b$$, then we have that $$$(a,b) \cap (c,d) = \{ x\in\mathbb{R} \ | \ a < x < b, \ \mbox{and} \ c < x < d\} =$$$ $$$=\{x\in\mathbb{R} \ | \ c < a < x < d < b\} =$$$ $$$=(a,d)$$$

In the same way, if $$a < c < b < d$$, we obtain: $$$(a,b) \cap (c,d) = \{ x\in\mathbb{R} \ | \ b < x < c\} = (b,c)$$$

• if $$a < b < c < d$$ then the intersection $$(a,b) \cap (c,d)$$ is: $$$(a,b) \cap (c,d) = \{ x\in\mathbb{R} \ | \ x\in(a,b) \ \mbox{and} \ x\in(c,d) \} $$$
  but as $$b < c$$, there's no value $$x$$ that belongs to the two intervals at the same time. In this case we say that the intersection is empty and it is denoted by the symbol $$\emptyset$$: $$$(a,b) \cap (c,d)=\emptyset.$$$ The result is the same if $$c < d < a < b$$.

In the case of two intervals which intersection is empty, we say that these intervals are disjoint.

The concept empty, $$\emptyset$$, is also considered as an interval, because $$\emptyset=(a,a)$$ for any real number $$a$$, so, unlike the union, the intersection of intervals is always an interval, but you can obtain the particular case of the empty interval.

Complementary

The complementary step is an operation that affects a single interval.

Given an interval any, its complementary is the set of numbers that do not belong to the interval.

We denote the complementary of the interval $$J=(a,b)$$ as $$$\overline{J}=\overline{(a,b)}$$$

To calculate it we will see the different cases if it is a bounded or unbounded interval:

• if the interval is bounded, we have: $$$\overline{(a,b)}= \{ x\in\mathbb{R} \ | \ x\notin (a,b)\} =$$$ $$$=\{ x\in\mathbb{R} \ | \ x\leq a, \ \mbox{or} \ b\leq x\}=$$$ $$$=(-\infty,a]\cup [b,+\infty)$$$

• if the interval is unbounded, we have: $$$\overline{(-\infty,b)}= \{ x\in\mathbb{R} \ | \ x\notin (-\infty,b)\} =$$$ $$$=\{ x\in\mathbb{R} \ | \ b\leq x\}=$$$ $$$=[b,+\infty)$$$

Similarly,

$$$\overline{(a,+\infty)}= \{ x\in\mathbb{R} \ | \ x\notin (a,+\infty)\} =$$$ $$$=\{ x\in\mathbb{R} \ | \ x\leq a\}=$$$ $$$=(-\infty,a]$$$

In the particular case of the empty interval, $$\emptyset$$, we have that its complementary are all the elements that do not belong to $$\emptyset$$, but as it doesn't have any element $$\emptyset$$, then the complementary is the total: $$$\overline{\emptyset}=\{ x\in\mathbb{R} \ | \ x\notin \emptyset= \mathbb{R}\}$$$

It should be noted that the total is also an interval, as: $$\mathbb{R}=(-\infty,+\infty)$$.