Table of Contents

**INTRODUCTION TO SET THEORY**

The idea of set has been intuitively used in mathematics since the time of ancient Greeks now set theory and its associated branches such as Group theory, Ring Theory etc., have far reaching applications.

The systematic development of set theory is attributed to the German mathematician George Cantor (1845 – 1918).

Some elementary definitions of set theory have been studied by students in the high school standard.

**DEFINITION OF SET**

The notion of a set is common; intuitively a set is a well-defined collection of objects. The objects comprising the set are called its member or elements. The sets are usually denoted by the capital letter A, B, X, Y etc., and its elements by small letters a, b, x, y,….

The statement ‘x’ is an element of set A is denoted by x\in A and is read as “x belongs to A”. If ‘x’ is not an element of the set A then it is denoted by x\notin A and read as “x does not belong to A).

Whenever possible a set is written by enclosing its elements by brace brackets { }. For example, A = {a, e, i, o, u}. The other way of specifying the set is stating the characteristic property satisfied by its elements. The above example of the set can also be written as the set of vowels in the English alphabet and is written as, A = {x: x is a vowel in the English alphabet}.

**Finite Set**

If the number of elements in a set is finite then it is said to be a finite set. For example, B = {1, 3, 5, 7, 9} is a finite set.

**Infinite Set**

If the number of elements in a set is infinite then it is said to be an infinite set. For example N = {x: x is a natural number} = {1, 2, 3, 4, …} is an infinite set.

**Singleton Set**

If a set contains only one element it is called a singleton set. For example C = {2} is a singleton set.

**Null Set or Empty Set**

If a set contains no elements it is called a null set of empty set, denoted by ∅. For example D = {x: x^{2} = –9 and x is real} is an empty set.

**Cardinality of a Set**

If A is a finite set, then the cardinality of A is the total number of elements that comprise the set and is denoted by n\left( A \right).

The cardinal number or cardinality of each of the set ∅, {a}, {a, b}, {a, b, c}… is 0, 1, 2, 3, … respectively.

**Universal Set**

In any discussion if all the sets are subsets of a fixed set, then this set is called the universal set and is denoted by ξ.

For example, in the study of theory of numbers the set Z of integers is considered as the universal set.

**Subset**

If every element of a set A is also an element of a set B then A is said to be a subset of B and it is denoted by A\subseteq B\,or\,B\supseteq A.

**Note:**

1. Empty set is a subset of every set.

2. Every set is subset of itself.

For example, let N, Z, Q, R respectively denoted the set of natural numbers; the set of integers; the set of rational numbers; the set of real numbers. Then N\subset Z\subset Q\subset R.

**Proper Subset**

A set A is said to be a proper subset of B if there exists an element of B which is not an element of A. That A is a proper subset of B if A\subset B\,and\,A\ne B.

For example, if A = {1, 3, 5}, B = {1, 3, 5, 7} then A is a proper subset of B.

**Equal Set**

Two sets A and B are said to be equal if and only if A\subset B\,and\,B\subset A.

**Power Set**

The set of all subsets of a set A is called the power set of A and it is denoted by P\left( A \right).

For example, if A = {1, 2, 3} then P\left( A \right)=\left\{ \left\{ \phi \right\},\left\{ 1 \right\},\left\{ 2 \right\},\left\{ 3 \right\},\left\{ 1,2 \right\},\left\{ 2,3 \right\},\left\{ 1,3 \right\},\left\{ 1,2,3 \right\} \right\}

**Note:** There are 2^{3} = 8 elements in P\left( A \right). If a set A has n elements then its power set P\left( A \right) has 2^{n} elements.

**ALGEBRAIC OPERATIONS ON SET**

Let us assume ξ = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, A = {1, 3, 4, 7} and B = {2, 3, 5, 7, 9}

**Union of Sets**

The union of two sets A and B denoted by A⋃B is the set of elements which belong to A or B or both. That is A\cup B=\left\{ x:x\in A\,or\,x\in B \right\}

Then in our example A\cup B=\left\{ 1,2,3,4,5,7,9 \right\}

**Intersection of Sets**

The intersection of two sets A and B denoted by A⋂B is the set of elements which belong to both A and B. That is A\cap B=\left\{ x:x\in A\,and\,x\in B \right\}

Then in our example A\cap B=\left\{ 3,7 \right\}

**Properties of Union and Intersection**

**1. Consistency Property**

The three relations *B* ⊂ *A*, *A *⋃ *B* = *A* and *A* ⋂ **B** = *B* are mutually equivalent. The following properties can be easily deduced from consistency property:

i) *A* ⋃ *Ø* = *A*, *A* ⋂ *Ø* = *Ø* ⇒ *Ø* ⊂ *A*

ii) *A* ⋃ *ξ* = *ξ*, *A* ⋂ *ξ* = *A* ⇒ *A* ⊂ *ξ*

iii) *A* ⋃ *A* = *A*, *A* ⋂ *A* = *A* ⇒ *A* ⊂ *A* **Idempotent Property**

iv) *A* ⋃ (*A* ⋂ *B*) = *A*, *A* ⋂ (*A* ⋃ *B*) = *A* **Absorption Property**

**2. Commutative Property**

i)* A* ⋃ *B* = *B* ⋃ *A*

ii) *A* ⋂ *B* = *B* ⋂ *A*

**3. Associative Property**

i) *A* ⋃ (*B* ⋃ *C*) = (*A* ⋃ *B*) ⋃ *C*

ii) *A* ⋂ (*B* ⋂ *C*) = (*A* ⋂ *B*) ⋂ *C*

**4. Distributive Property**

i)* A* ⋃ (*B* ⋂ *C*) = (*A* ⋃ *B*) ⋂ (*A* ⋃ *C*)

ii) *A* ⋂ (*B* ⋃ *C*) = (*A* ⋂ *B*) ⋃ (*A* ⋂ *C*)

**Difference of Sets**

The difference of two sets A and B, denoted by A – B is the set of elements of A which are not the elements of B. That is A-B=\left\{ x:x\in A,\,x\notin B \right\}

Then in our example A-B=\left\{ 1,4 \right\}

**Properties of Difference**

i) A – B ⊂ A

ii) A – B ≠ B – A

iii) A – B, A⋂B, B – A are mutually disjoint sets.

**Complement of a Set**

The complement of a set A with respect to the universal set ξ is defined as ξ – A and is denoted by A’ or A^{c}. That is {{A}^{c}}=\left\{ x:x\in \xi ,\,x\notin A \right\}

Then in our example {{A}^{c}}=\left\{ 2,5,6,8,9,10 \right\}

**Properties of Complement**

\[i){{\left( {{A}^{c}} \right)}^{c}}=A\]

\[ii){{\phi }^{c}}=\xi \]

\[iii){{\xi }^{c}}=\phi \]

**CARTESIAN PRODUCTS OF TWO SETS**

Let A and B be two sets. Then the Cartesian product of A and B are defined as the set of all ordered pairs of (x, y) where x ∈ A and y ∈ B and is denoted by A\times B. Thus A\times B=\left\{ \left( x,y \right):x\in A\,and\,y\in B \right\}

Two ordered pairs (a, b) and (c, d) are equal if and only if a = c and b = d. If A contains ‘m’ elements and B contains ‘n’ elements then A\times Bcontains mn ordered pairs.

If A = {a, b} and B = {1, 2, 3} then A\times B=\left\{ \left( a,1 \right),\left( a,2 \right),\left( a,3 \right),\left( b,1 \right),\left( b,2 \right),\left( b,3 \right) \right\}

Example 01 |

**Q. If A = {2, 3, 4}, B = {4, 5, 6} and C = {6, 7} evaluate the following:**

\[(a)\left( A\cap B \right)\times \left( B-C \right)\]

\[(b)A\times \left( C-B \right)\]

\[(c)\left( A-B \right)\times B\]

\[(d)\left( A-B \right)\times \left( B-C \right)\]

\[(e)\left( A\times B \right)-\left( B\times C \right)\]

**Solution:**

\[(a)A\cap B=\left\{ 2,3,4 \right\}\cap \left\{ 4,5,6 \right\}=\left\{ 4 \right\}\]

\[B-C=\left\{ 4,5,6 \right\}-\left\{ 6,7 \right\}=\left\{ 4,5 \right\}\]

\[\therefore \left( A\cap B \right)\times \left( B-C \right)=\left\{ 4 \right\}\times \left\{ 4,5 \right\}=\left\{ \left( 4,4 \right),\left( 4,5 \right) \right\}\]

\[(b)C-B=\left\{ 6,7 \right\}-\left\{ 4,5,6 \right\}=\left\{ 7 \right\}\]

\[\therefore A\times \left( C-B \right)=\left\{ 2,3,4 \right\}\times \left\{ 7 \right\}=\left\{ \left( 2,7 \right),\left( 3,7 \right),\left( 4,7 \right) \right\}\]

\[(c)A-B=\left\{ 2,3,4 \right\}-\left\{ 4,5,6 \right\}=\left\{ 2,3 \right\}\]

\[\therefore \left( A-B \right)\times B=\left\{ 2,3 \right\}\times \left\{ 4,5,6 \right\}=\left\{ \left( 2,4 \right),\left( 2,5 \right),\left( 2,6 \right),\left( 3,4 \right),\left( 3,5 \right),\left( 3,6 \right) \right\}\]

\[(d)A-B=\left\{ 2,3,4 \right\}-\left\{ 4,5,6 \right\}=\left\{ 2,3 \right\}\]

\[B-C=\left\{ 4,5,6 \right\}-\left\{ 6,7 \right\}=\left\{ 4,5 \right\}\]

\[\therefore \left( A-B \right)\times \left( B-C \right)=\left\{ 2,3 \right\}\times \left\{ 4,5 \right\}=\left\{ \left( 2,4 \right),\left( 2,5 \right),\left( 3,4 \right),\left( 3,5 \right) \right\}\]

\[(e)A\times B=\left\{ 2,3,4 \right\}\times \left\{ 4,5,6 \right\}=\left\{ \left( 2,4 \right),\left( 2,5 \right),\left( 2,6 \right),\left( 3,4 \right),\left( 3,5 \right),\left( 3,6 \right),\left( 4,4 \right),\left( 4,5 \right),\left( 4,6 \right) \right\}\]

\[B\times C=\left\{ 4,5,6 \right\}\times \left\{ 6,7 \right\}=\left\{ \left( 4,6 \right),\left( 4,7 \right),\left( 5,6 \right),\left( 5,7 \right),\left( 6,6 \right),\left( 6,7 \right) \right\}\]

\[\therefore \left( A\times B \right)-\left( B\times C \right)=\left\{ \left( 2,4 \right),\left( 2,5 \right),\left( 2,6 \right),\left( 3,4 \right),\left( 3,5 \right),\left( 3,6 \right),\left( 4,4 \right),\left( 4,5 \right) \right\}\]

Example 02 |

**Q. If ξ = {a, b, c, d, e}, A = {a, c, d}, B = {b, c, e} C = {b, c, e} Evaluate the following:**

\[(a){{A}^{c}}\times \left( B-C \right)\]

\[(b){{\left( A\cup B \right)}^{c}}\times \left( B\cap C \right)\]

\[(c)\left( A-B \right)\times \left( B-C \right)\]

\[(d){{\left( B\cup C \right)}^{c}}\times A\]

\[(e)\left( B-A \right)\times {{C}^{c}}\]

**Solution:**

\[(a){{A}^{c}}=\xi -A=\left\{ a,b,c,d,e \right\}-\left\{ a,c,d \right\}=\left\{ b,e \right\}\]

\[B-C=\left\{ d,e \right\}-\left\{ b,c,e \right\}=\left\{ d \right\}\]

\[\therefore {{A}^{c}}\times \left( B-C \right)=\left\{ b,e \right\}\times \left\{ d \right\}=\left\{ \left( b,d \right),\left( e,d \right) \right\}\]

\[(b){{\left( A\cup B \right)}^{c}}=\xi -\left( A\cup B \right)=\left\{ a,b,c,d,e \right\}-\left\{ a,c,d,e \right\}=\left\{ b \right\}\]

\[B\cap C=\left\{ d,e \right\}\cap \left\{ b,c,e \right\}=\left\{ e \right\}\]

\[\therefore {{\left( A\cup B \right)}^{c}}\times \left( B\cap C \right)=\left\{ b \right\}\times \left\{ e \right\}=\left\{ \left( b,e \right) \right\}\]

\[(c)A-B=\left\{ a,c,d \right\}-\left\{ d,e \right\}=\left\{ a,c \right\}\]

\[B-C=\left\{ d,e \right\}-\left\{ b,c,e \right\}=\left\{ d \right\}\]

\[\therefore \left( A-B \right)\times \left( B-C \right)=\left\{ a,c \right\}\times \left\{ d \right\}=\left\{ \left( a,d \right),\left( c,d \right) \right\}\]

\[(d){{\left( B\cup C \right)}^{c}}=\xi -\left( B\cup C \right)=\left\{ a,b,c,d,e \right\}-\left\{ b,c,d,e \right\}=\left\{ a \right\}\]

\[\therefore {{\left( B\cup C \right)}^{c}}\times A=\left\{ a \right\}\times \left\{ a,c,d \right\}=\left\{ \left( a,a \right),\left( a,c \right),\left( a,d \right) \right\}\]

\[(e)B-A=\left\{ d,e \right\}-\left\{ a,c,d \right\}=\left\{ e \right\}\]

\[{{C}^{c}}=\xi -C=\left\{ a,b,c,d,e \right\}-\left\{ b,c,e \right\}=\left\{ a,d \right\}\]

\[\therefore \left( B-A \right)\times {{C}^{c}}=\left\{ e \right\}\times \left\{ a,d \right\}=\left\{ \left( e,a \right),\left( e,d \right) \right\}\]

Example 03 |

**Q. If A=\left\{ x:{{x}^{2}}-5x+6=0 \right\},B=\left\{ 0,3,4 \right\},C=\left\{ x:x\in N\,and\,\,x<4 \right\} Evaluate the following:**

\[(a)A\times \left( B\cap C \right)\]

\[(b)\left( A\cup B \right)\times \left( B-C \right)\]

\[(c)\left( A-B \right)\times \left( C-B \right)\]

**Solution:**

\[A=\left\{ x:{{x}^{2}}-5x+6=0 \right\}=\left\{ x:\left( x-2 \right)\left( x-3 \right)=0 \right\}=\left\{ 2,3 \right\}\]

\[B=\left\{ 0,3,4 \right\}\]

\[C=\left\{ x:x\in N\,and\,\,x<4 \right\}=\left\{ 1,2,3 \right\}\]

\[(a)B\cap C=\left\{ 0,3,4 \right\}\cap \left\{ 1,2,3 \right\}=\left\{ 3 \right\}\]

\[\therefore A\times \left( B\cap C \right)=\left\{ 2,3 \right\}\times \left\{ 3 \right\}=\left\{ \left( 2,3 \right),\left( 3,3 \right) \right\}\]

\[(b)A\cup B=\left\{ 2,3 \right\}\cup \left\{ 0,3,4 \right\}=\left\{ 0,2,3,4 \right\}\]

\[B-C=\left\{ 0,3,4 \right\}-\left\{ 1,2,3 \right\}=\left\{ 0,4 \right\}\]

\[\therefore \left( A\cup B \right)\times \left( B-C \right)=\left\{ 0,2,3,4 \right\}\times \left\{ 0,4 \right\}\]

\[=\left\{ \left( 0,0 \right),\left( 0,4 \right),\left( 2,0 \right),\left( 2,4 \right),\left( 3,0 \right),\left( 3,4 \right),\left( 4,0 \right),\left( 4,4 \right) \right\}\]

\[(c)A-B=\left\{ 2,3 \right\}-\left\{ 0,3,4 \right\}=\left\{ 2 \right\}\]

\[C-B=\left\{ 1,2,3 \right\}-\left\{ 0,3,4 \right\}=\left\{ 1,2 \right\}\]

\[\therefore \left( A-B \right)\times \left( C-B \right)=\left\{ 2 \right\}\times \left\{ 1,2 \right\}=\left\{ \left( 2,1 \right),\left( 2,2 \right) \right\}\]

Example 04 |

**Q. If A=\left\{ x:x\in Z,0\le x\le 10 \right\},\,B=\left\{ x:x\in Z,5<x\le 15 \right\}\,\And \,C=\left\{ x:x\in Z,x>5 \right\}\, then find A – B and (B⋂C) – A, where Z is the set of all integers.**

**Solution:**

\[A=\left\{ x:x\in Z,0\le x\le 10 \right\}=\left\{ 0,1,2,3,4,5,6,7,8,9,10 \right\}\]

\[B=\left\{ x:x\in Z,5<x\le 15 \right\}=\left\{ 6,7,8,9,10,11,12,13,14,15 \right\}\]

\[C=\left\{ x:x\in Z,x>5 \right\}\,=\left\{ 6,7,8,9,10,……. \right\}\]

\[\therefore A-B=\left\{ 0,1,2,3,4,5 \right\}\]

\[B\cap C=\left\{ 6,7,8,9,10,11,12,13,14,15 \right\}\]

\[\therefore \left( B\cap C \right)-A=\left\{ 11,12,13,14,15 \right\}\]

Example 05 |

**Q. In a survey concerning the smoking habits of consumers, it was found that 55% smoke cigarette A, 50% smoke cigarette B, 42% smoke cigarette C, 28% smoke cigarette A & B, 20% smoke cigarette A & C, 12% smoke cigarette B & C and 10% smoke all the three cigarettes. What percentage does not smoke?**

**Solution:**

According to the question n(A) = 55, n(B) = 50, n(c) = 42, n(A⋂B) = 28, n(A⋂C) = 20, n(B⋂C) = 12 and n(A⋂B⋂C) = 10.

Now the percentage of smokers who smoke any of the cigarettes are n(A⋃B⋃C).

n(A⋃B⋃C) = n(A) + n(B) + n(C) – n(A⋂B) – n(A⋂C) – n(B⋂C) + n(A⋂B⋂C) = 55 + 50 + 42 – 28 – 20 – 12 + 10 = 97%

Therefore percentage does not smoke = (100 – 97)% = 3%

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