# 1.4 Difference of sets  (Page 2/2)

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## Interpretation of complement

Proceeding as before we can read the conditional statement for the complement with the help of two ways arrow as :

$x\in A\prime ⇔x\in U\phantom{\rule{1em}{0ex}}and\phantom{\rule{1em}{0ex}}x\notin A$

In terms of minus or difference operation,

$A\prime =U-A$

It is clear from the representation on Venn’s diagram that the universal set comprises of two distinct sets – set A and complement set A’.

$⇒U=A\cup A\prime$

## Compliment of universal set

The complement of universal set is empty set. It is so because difference of union set with itself is the empty set (see Venn's diagram).

$U\prime =\left\{x:\phantom{\rule{1em}{0ex}}x\in U\phantom{\rule{1em}{0ex}}and\phantom{\rule{1em}{0ex}}x\notin U\right\}=\phi$

## Complement of empty set

The complement of the empty set is universal set. It is so because difference of union set with the empty set is universal set (see Venn's diagram).

$\phi \prime =\left\{x:\phantom{\rule{1em}{0ex}}x\in U\phantom{\rule{1em}{0ex}}and\phantom{\rule{1em}{0ex}}x\notin \phi \right\}=U$

## Complement of complement set is set itself

The complement of complement set is set itself. The complement set is defined as :

$A\prime =U-A$

Now, complement of complement set is :

$⇒\left(A\prime \right)\prime =\left(U-A\right)\prime$

Let us consider the example, where :

$U=\left\{1,2,3,4,5,6,7,8\right\}$

$A=\left\{1,2,3,4,5,6\right\}$

Then,

$⇒A\prime =\left\{1,2,3,4,5,6,7,8\right\}-\left\{1,2,3,4,5,6\right\}=\left\{7,8\right\}$

Again taking complement, we have :

$\left(A\prime \right)\prime =\left\{1,2,3,4,5,6,7,8\right\}-\left\{7,8\right\}=\left\{1,2,3,4,5,6\right\}=A$

## Union with complement set

The union of a set with its complement is universal set :

$A\cup A\prime =\left\{x:\phantom{\rule{1em}{0ex}}x\in U\phantom{\rule{1em}{0ex}}and\phantom{\rule{1em}{0ex}}x\in A\right\}\cup \left\{x:\phantom{\rule{1em}{0ex}}x\in U\phantom{\rule{1em}{0ex}}and\phantom{\rule{1em}{0ex}}x\in A\right\}=U$

From Venn’s diagram also, we see that universal set consists of set A and component A’.

$U=A\cup A\prime$

The two sets on the right side of the equation are disjoint sets. Hence,

$A\cup A\prime =U$

## Intersection with complement set

There is nothing common between set A and its component A’. Thus, intersection of a set with its complement yields the empty set,

$A\cap A\prime =\phi$

## De-morgan’s laws

In the real world situation, we want to negate a condition of incidence. For example, consider a class in the school. Some students play either basketball or football or both, but there are students, who play neither basketball nor football. We have to identify later category of students as a set.

Let the set of students playing basketball be “B” and that playing football be “F”. Then, students who do not play basketball is complement set B’ and students who do not play football is complement set F’. We have shown these complement sets separately for visualization. Actually, these complement sets are drawn to the same universal set, "U".

Two complement sets are but overlapping sets. There are students in the set B’ who play football and there are students in the set F’, who play basketball. In order to remove those students playing other game, we intersect two complements. The members of the intersection of two complements, therefore, represent students who play neither basketball nor football. This intersection is shown as third bottom Venn’s diagram in the figure.

Looking at the intersection of two complement sets, however, we observe that this is equal to the complement of union “ $B\cup F$ ”. This conclusion can be derived from basic interpretation as well. We know that union “ $B\cup F$ ” represents students, who play either or both games. The component of the union, therefore, represents, who neither play basketball nor football.

This fact, as a matter of fact, is the first De-morgan’s law. Symbolically,

$B\prime \cap F\prime =\left(B\cup F\right)\prime$

The second De-morgan’s law is :

$B\prime \cup F\prime =\left(B\cap F\right)\prime$

In the parlance of illustration given earlier, let us interpret right hand side of the second De-morgan's law. The intersection “ $B\cap F$ ” represents students playing both games. Its complement, therefore, represents students who do not play both games, but may play one of them.

## Analytical proof

Here, we shall prove first De-morgan’s law in this section. The second law can be proved in similar fashion. Let us consider an arbitrary element “x” belonging to set ( $A\cup B$ )’.

$x\in \left(A\cup B\right)\prime$

$⇒x\notin \left(A\cup B\right)$

Then, by definition of union,

$⇒x\notin \left\{x:\phantom{\rule{1em}{0ex}}x\in A\phantom{\rule{1em}{0ex}}or\phantom{\rule{1em}{0ex}}x\in B\right\}$

Here, “not or” is interpreted same as “and”,

$⇒x\notin A\phantom{\rule{1em}{0ex}}and\phantom{\rule{1em}{0ex}}x\notin B$

$⇒x\in A\prime \phantom{\rule{1em}{0ex}}and\phantom{\rule{1em}{0ex}}x\in B\prime$

$⇒x\in A\prime \cap B\prime$

But, we had started with ( $A\cup B$ )’ and used its definition to show that “x” belongs to another set. It means that the other set consists of the elements of the first set – at the least. Thus,

$\left(A\cup B\right)\prime \subset A\prime \cap B\prime$

Similarly, we can start with $A\prime \cap B\prime$ and reach the conclusion that :

$A\prime \cap B\prime \subset \left(A\cup B\right)\prime$

If sets are subsets of each other, then they are equal. Hence,

$A\prime \cap B\prime =\left(A\cup B\right)\prime$

## Example

Problem 1: In the reference of students in a class, the set “B” represents students, who play basketball. The set “F” represents students, who play football. The set “B” and “F” are left and right circles respectively on the Venn's diagram shown below. Identify regions marked 1 to 8 on the Venn’s diagram. Also interpret regions identified by combination U – (6+7).

Solution : The meaning of regions market 1 – 8 are as given hereunder :

1 : B-F : It represents the difference of “B” and “F”. It consists of students, who play basketball, but not football.

2 : F-B : It represents the difference of “F” and “B”. It consists of students, who play football, but not basketball.

3 : $B\cap F$ : It represents the intersection of two sets. It consists of students, who play both basketball and football.

4 : B: It represents the set “B”. It is union of two disjoint sets “B-F” and “ $B\cup F$ ”. It consists of students, who play basketball.

5 : F: It represents the set “F”. It is union of two disjoint sets “F-B” and “ $B\cap F$ ”. It consists of students, who play football.

6 : B∪F: It represents the union set of set “B” and “F”. Equivalently, it is union of three disjoint sets “B-F”, “ $B\cap F$ ” and “F-B”. It consists of students, who play either of two games or both.

7 : ( $B\cup F$ )’: It represents the component of union set “ $B\cup F$ ”. It consists of students, who play neither basketball nor football.

8 : $\left(B-F\right)\cup \left(F-B\right)$ : It represents union of two disjoint difference sets “B-F” and “F-B”. It consists of students, who play only one game.

The region, identified by U – (6+7), is complement of “ $B\cap F$ ”. It represents students, who do not play both games, but may play one of them.

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