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:

Consider for example the union of the intervals $(3,9)$ and $[7,11]$: $$(3,9) \cup [7,11] = \{ x\in\mathbb{R} \ | \ 3 < x < 9, \ \mbox{or} \ 7 < x < 11\} = $$ $$= \{x\in\mathbb{R} \ | \ 3 < x \leq 11\}=(3,11]$$

So, $(3,9) \cup [7,11] = (3,11].$

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

Another example, let's do the union of the intervals $(-1,0)$ and $(0,+\infty)$: $$(-1,0) \cup (0,+\infty) = \{ x\in\mathbb{R} \ | \ -1 < x < 0, \ \mbox{or} \ 0 < x \}$$ And this expression can not be simplified, such that the union of the intervals $(-1,0)$ and $(0,+\infty)$ remains $$(-1,0) \cup (0,+\infty)$$

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.

An example of intersection.

Consider the intervals $[0,+\infty)$ and $(-\infty,1)$.

Then its intersection is: $$[0,+\infty) \cap (-\infty,1) = \{ x\in\mathbb{R} \ | \ 0 \leq x \ \mbox{and} \ x < 1\} =$$ $$=\{x\in\mathbb{R} \ | \ 0 \leq x < 1 \} =$$ $$=[0,1)$$

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: $\mathbb{R}=(-\infty,+\infty)$.